DIFFERENTIAL EQUATIONS First Order Differential Equations. Paul Dawkins

Transcription

1 DIFFERENTIAL EQUATIONS First Order Paul Dawkins

2 Table of Contents Preface... First Order... 3 Introduction... 3 Linear... 4 Separable... 7 Eact... 8 Bernoulli Substitutions Intervals of Validit Modeling with First Order Equilibrium Solutions... 7 Euler s Method Paul Dawkins i

3 Preface Here are m online notes for m differential equations course that I teach here at Lamar Universit. Despite the fact that these are m class notes, the should be accessible to anone wanting to learn how to solve differential equations or needing a refresher on differential equations. I ve tried to make these notes as self contained as possible and so all the information needed to read through them is either from a Calculus or Algebra class or contained in other sections of the notes. A couple of warnings to m students who ma be here to get a cop of what happened on a da that ou missed.. Because I wanted to make this a fairl complete set of notes for anone wanting to learn differential equations I have included some material that I do not usuall have time to cover in class and because this changes from semester to semester it is not noted here. You will need to find one of our fellow class mates to see if there is something in these notes that wasn t covered in class.. In general I tr to work problems in class that are different from m notes. However, with Differential Equation man of the problems are difficult to make up on the spur of the moment and so in this class m class work will follow these notes fairl close as far as worked problems go. With that being said I will, on occasion, work problems off the top of m head when I can to provide more eamples than just those in m notes. Also, I often don t have time in class to work all of the problems in the notes and so ou will find that some sections contain problems that weren t worked in class due to time restrictions. 3. Sometimes questions in class will lead down paths that are not covered here. I tr to anticipate as man of the questions as possible in writing these up, but the realit is that I can t anticipate all the questions. Sometimes a ver good question gets asked in class that leads to insights that I ve not included here. You should alwas talk to someone who was in class on the da ou missed and compare these notes to their notes and see what the differences are. 4. This is somewhat related to the previous three items, but is important enough to merit its own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Using these notes as a substitute for class is liable to get ou in trouble. As alread noted not everthing in these notes is covered in class and often material or insights not in these notes is covered in class. 007 Paul Dawkins ii

4 First Order Introduction In this chapter we will look at solving first order differential equations. The most general first order differential equation can be written as, d f ( t, ) dt = () As we will see in this chapter there is no general formula for the solution to (). What we will do instead is look at several special cases and see how to solve those. We will also look at some of the theor behind first order differential equations as well as some applications of first order differential equations. Below is a list of the topics discussed in this chapter. Linear Equations Identifing and solving linear first order differential equations. Separable Equations Identifing and solving separable first order differential equations. We ll also start looking at finding the interval of validit from the solution to a differential equation. Eact Equations Identifing and solving eact differential equations. We ll do a few more interval of validit problems here as well. Bernoulli In this section we ll see how to solve the Bernoulli Differential Equation. This section will also introduce the idea of using a substitution to help us solve differential equations. Substitutions We ll pick up where the last section left off and take a look at a couple of other substitutions that can be used to solve some differential equations that we couldn t otherwise solve. Intervals of Validit Here we will give an in-depth look at intervals of validit as well as an answer to the eistence and uniqueness question for first order differential equations. Modeling with First Order Using first order differential equations to model phsical situations. The section will show some ver real applications of first order differential equations. Equilibrium Solutions We will look at the behavior of equilibrium solutions and autonomous differential equations. Euler s Method In this section we ll take a brief look at a method for approimating solutions to differential equations. 007 Paul Dawkins iii

5 Linear The first special case of first order differential equations that we will look is the linear first order differential equation. In this case, unlike most of the first order cases that we will look at, we can actuall derive a formula for the general solution. The general solution is derived below. However, I would suggest that ou do not memorize the formula itself. Instead of memorizing the formula ou should memorize and understand the process that I'm going to use to derive the formula. Most problems are actuall easier to work b using the process instead of using the formula. So, let's see how to solve a linear first order differential equation. Remember as we go through = t. It's sometimes this process that the goal is to arrive at a solution that is in the form eas to lose sight of the goal as we go through this process for the first time. In order to solve a linear first order differential equation we MUST start with the differential equation in the form shown below. If the differential equation is not in this form then the process we re going to use will not work. d p() t g() t dt + = () Where both p(t) and g(t) are continuous functions. Recall that a quick and dirt definition of a continuous function is that a function will be continuous provided ou can draw the graph from left to right without ever picking up our pencil/pen. In other words, a function is continuous if there are no holes or breaks in it. Now, we are going to assume that there is some magical function somewhere out there in the m t, called an integrating factor. Do not, at this point, worr about what this function world, is or where it came from. We will figure out what m ( t ) is once we have the formula for the general solution in hand. So, now that we have assumed the eistence of m ( t ) multipl everthing in () b m ( t ) will give.. This d m() t + m() t p() t = m() t g() t () dt Now, this is where the magic of m ( t ) comes into pla. We are going to assume that whatever m ( t ) is, it will satisf the following. m( t) pt m ( t) Again do not worr about how we can find a ( t ) = (3) m that will satisf (3). As we will see, provided p(t) is continuous we can find it. So substituting (3) into () we now arrive at. d m() t + m () t = m() t g() t (4) dt 007 Paul Dawkins 4

6 At this point we need to recognize that the left side of (4) is nothing more than the following product rule. d m() t + m () t = ( m() t () t ) dt So we can replace the left side of (4) with this product rule. Upon doing this (4) becomes m t t = m t g t (5) ( () ()) () () Now, recall that we are after (t). We can now do something about that. All we need to do is integrate both sides then use a little algebra and we'll have the solution. So, integrate both sides of (5) to get. m t t dt = m t g t dt ( () ()) () () m m Ú Ú t t + c=ú t g t dt (6) Note the constant of integration, c, from the left side integration is included here. It is vitall important that this be included. If it is left out ou will get the wrong answer ever time. The final step is then some algebra to solve for the solution, (t). m t t = m t g t dt-c () t = Ú Ú () () m() t m t g t dt- c Now, from a notational standpoint we know that the constant of integration, c, is an unknown constant and so to make our life easier we will absorb the minus sign in front of it into the constant and use a plus instead. This will NOT affect the final answer for the solution. So with this change we have. () t m() t Ú m t g t dt+ c = (7) Again, changing the sign on the constant will not affect our answer. If ou choose to keep the minus sign ou will get the same value of c as I do ecept it will have the opposite sign. Upon plugging in c we will get eactl the same answer. There is a lot of plaing fast and loose with constants of integration in this section, so ou will need to get used to it. When we do this we will alwas to tr to make it ver clear what is going on and tr to justif wh we did what we did. So, now that we ve got a general solution to () we need to go back and determine just what this m t is. This is actuall an easier process than ou might think. We ll start magical function with (3). ( t) pt = ( t) m m 007 Paul Dawkins 5

7 Divide both sides b m ( t ), m m ( t) () t = () p t Now, hopefull ou will recognize the left side of this from our Calculus I class as nothing more than the following derivative. ln m t = p t ( ()) () As with the process above all we need to do is integrate both sides to get. ln m t + k = p t dt Ú Ú () () ln m t = p t dt+ k You will notice that the constant of integration from the left side, k, had been moved to the right side and had the minus sign absorbed into it again as we did earlier. Also note that we re using k here because we ve alread used c and in a little bit we ll have both of them in the same equation. So, to avoid confusion we used different letters to represent the fact that the will, in all probabilit, have different values. Eponentiate both sides to get m ( t ) out of the natural logarithm. m () t Ú = e () ptdt+ k Now, it s time to pla fast and loose with constants again. It is inconvenient to have the k in the eponent so we re going to get it out of the eponent in the following wa. ptdt () + k m t = e Ú () k ptdt () a+ b a b = ee Ú Recall =! Now, let s make use of the fact that k is an unknown constant. If k is an unknown constant then k so is e so we might as well just rename it k and make our life easier. This will give us the following. ptdt () m t = ke Ú (8) () So, we now have a formula for the general solution, (7), and a formula for the integrating factor, (8). We do have a problem however. We ve got two unknown constants and the more unknown constants we have the more trouble we ll have later on. Therefore, it would be nice if we could find a wa to eliminate one of them (we ll not be able to eliminate both.). This is actuall quite eas to do. First, substitute (8) into (7) and rearrange the constants. 007 Paul Dawkins 6

8 () t = ptdt () keú g t dt+ c keú () () ptdt ptdt () k Ú Ú e g() t dt+ c = ptdt () k eú = Ú Ú ptdt () c eú g() t dt + k Ú ptdt () e So, (7) can be written in such a wa that the onl place the two unknown constants show up is a ratio of the two. Then since both c and k are unknown constants so is the ratio of the two constants. Therefore we ll just call the ratio c and then drop k out of (8) since it will just get absorbed into c eventuall. The solution to a linear first order differential equation is then where, () t m() t Ú m t g t dt+ c = (9) m () t Ú ptdt () =e (0) Now, the realit is that (9) is not as useful as it ma seem. It is often easier to just run through the process that got us to (9) rather than using the formula. We will not use this formula in an of m eamples. We will need to use (0) regularl, as that formula is easier to use than the process to derive it. Solution Process The solution process for a first order linear differential equation is as follows.. Put the differential equation in the correct initial form, (). m t, using (0).. Find the integrating factor, 3. Multipl everthing in the differential equation b m ( t) and verif that the left side becomes the product rule m ( t) ( t) and write it as such. ' 4. Integrate both sides, make sure ou properl deal with the constant of integration. 5. Solve for the solution (t). Let s work a couple of eamples. Let s start b solving the differential equation that we derived back in the Direction Field section. 007 Paul Dawkins 7

9 Eample Find the solution to the following differential equation. dv v dt = - Solution First we need to get the differential equation in the correct form. dv 0.96v 9.8 dt + = From this we can see that p(t)=0.96 and so m ( t) is then. m () t = eú = e 0.96dt 0.96t Note that officiall there should be a constant of integration in the eponent from the integration. However, we can drop that for eactl the same reason that we dropped the k from (8). Now multipl all the terms in the differential equation b the integrating factor and do some simplification. 0.96t dv 0.96t 0.96t e e v = 9.8e dt 0.96t 0.96t e v = 9.8e Integrate both sides and don't forget the constants of integration that will arise from both integrals. Û ( 0.96 t ) t e v dt = dt ı Ú e e v+ k = 50e + c 0.96t 0.96t Oka. It s time to pla with constants again. We can subtract k from both sides to get. 0.96t 0.96t e v= 50e + c-k Both c and k are unknown constants and so the difference is also an unknown constant. We will therefore write the difference as c. So, we now have 0.96t 0.96t e v= 50e + c From this point on we will onl put one constant of integration down when we integrate both sides knowing that if we had written down one for each integral, as we should, the two would just end up getting absorbed into each other. 0.96t The final step in the solution process is then to divide both sides b e or to multipl both -0.96t sides b e. Either will work, but I usuall prefer the multiplication route. Doing this gives the general solution to the differential equation. t vt = 50 + ce Paul Dawkins 8

10 From the solution to this eample we can now see wh the constant of integration is so important in this process. Without it, in this case, we would get a single, constant solution, v(t)=50. With the constant of integration we get infinitel man solutions, one for each value of c. Back in the direction field section where we first derived the differential equation used in the last eample we used the direction field to help us sketch some solutions. Let's see if we got them correct. To sketch some solutions all we need to do is to pick different values of c to get a solution. Several of these are shown in the graph below. So, it looks like we did prett good sketching the graphs back in the direction field section. Now, recall from the Definitions section that the Initial Condition(s) will allow us to zero in on a particular solution. Solutions to first order differential equations (not just linear as we will see) will have a single unknown constant in them and so we will need eactl one initial condition to find the value of that constant and hence find the solution that we were after. The initial condition for first order differential equations will be of the form t = 0 0 Recall as well that a differential equation along with a sufficient number of initial conditions is called an Initial Value Problem (IVP). Eample Solve the following IVP. dv v v( 0) 48 dt = - = Solution To find the solution to an IVP we must first find the general solution to the differential equation and then use the initial condition to identif the eact solution that we are after. So, since this is the same differential equation as we looked at in Eample, we alread have its general solution. 0.96t v= 50 + ce - Now, to find the solution we are after we need to identif the value of c that will give us the solution we are after. To do this we simpl plug in the initial condition which will give us an equation we can solve for c. So let's do this 007 Paul Dawkins 9

11 48= v 0 = 50+ c fi c=- So, the actual solution to the IVP is. v = 50-e -0.96t A graph of this solution can be seen in the figure above. Let s do a couple of eamples that are a little more involved. Eample 3 Solve the following IVP. 3 Êp ˆ p cos( ) + sin( ) = cos ( ) sin( ) - Á = 3, 0 < Ë 4 Solution : Rewrite the differential equation to get the coefficient of the derivative a one. sin ( ) + = cos ( ) sin ( ) - cos cos Now find the integrating factor. + tan = cos sin -sec m () t tan d lnsec lnsec = e Ú = e = e = sec Can ou do the integral? If not rewrite tangent back into sines and cosines and then use a simple substitution. Note that we could drop the absolute value bars on the secant because of the limits on. In fact, this is the reason for the limits on. Also note that we made use of the following fact. ln f ( ) e = f () This is an important fact that ou should alwas remember for these problems. We will want to simplif the integrating factor as much as possible in all cases and this fact will help with that simplification. Now back to the eample. Multipl the integrating factor through the differential equation and verif the left side is a product rule. Note as well that we multipl the integrating factor through the rewritten differential equation and NOT the original differential equation. Make sure that ou do this. If ou multipl the integrating factor through the original differential equation ou will get the wrong solution! sec + sec tan = sec cos sin -sec Integrate both sides. ( sec( ) ) = cos( ) sin( ) - sec ( ) 007 Paul Dawkins 0

12 Note the use of the trig formula Ú ( sec ) = cos sin-sec Ú Ú d d = - sec sin sec d sec( ) ( ) =- cos( ) - tan ( ) + c sin q = sinqcosq that made the integral easier. Net, solve for the solution. cos cos cos tan c cos =- cos ( ) cos ( ) - sin ( ) + c cos ( ) =- - + Finall, appl the initial condition to find the value of c. Êp ˆ Êp ˆ Êp ˆ Êp ˆ Êp ˆ 3 = Á =- cosá cosá - siná + ccosá Ë 4 Ë 4 Ë Ë 4 Ë 4 3 =- + c c = 7 The solution is then. cos cos sin 7cos =- - + Below is a plot of the solution. 007 Paul Dawkins

13 Eample 4 Find the solution to the following IVP. t + = t - t+ = Solution First, divide through b the t to get the differential equation into the correct form. + = t- + t t Now let s get the integrating factor, m () t. m () () t ÛÙ dt ı t = e = e Now, we need to simplif m () t. However, we can t use () et as that requires a coefficient of one in front of the logarithm. So, recall that r ln = rln and rewrite the integrating factor in a form that will allow us to simplif it. () ln t ln t ln t m t = e = e = t = t We were able to drop the absolute value bars here because we were squaring the t, but often the can t be dropped so be careful with them and don t drop them unless ou know that ou can. Often the absolute value bars must remain. Now, multipl the rewritten differential equation (remember we can t use the original differential equation here ) b the integrating factor. 3 t = t - t + t Integrate both sides and solve for the solution. 3 t = t - t + tdt () Ú = c = t t t t 4 3 t t t c Finall, appl the initial condition to get the value of c. = () = c fi c= 4 3 The solution is then, () t = t - t t Here is a plot of the solution. 007 Paul Dawkins

14 Eample 5 Find the solution to the following IVP. 3 t - = t sin( t) - t + 4t ( p) = p Solution First, divide through b t to get the differential equation in the correct form. - = t 4 sin( t) - t + 4t 3 t Now that we have done this we can find the integrating factor, m ( t). m () t ÛÙ - dt ı t = e = e -ln t Do not forget that the - is part of p(t). Forgetting this minus sign can take a problem that is ver eas to do and turn it into a ver difficult, if not impossible problem so be careful! Now, we just need to simplif this as we did in the previous eample. () - -ln t ln t - - m t = e = e = t = t Again, we can drop the absolute value bars since we are squaring the term. Now multipl the differential equation b the integrating factor (again, make sure it s the rewritten one and not the original differential equation). Integrate both sides and solve for the solution. - t = t sin t - + 4t 007 Paul Dawkins 3

15 Ú - t t = t sin t dt tdt Ú () 4 () t =- t cos( t) + t sin( t) + t cos( t) - t + t + ct 4 - t t =- t cos t + tsin t + cos t - t+ t + c Appl the initial condition to find the value of c p = ( p) =- p + p - p + p + cp = p - p + p + cp p - p = cp 4 c = p - 4 The solution is then () Ê ˆ t =- t cos t + t sin t + t cos t - t + t + Áp - t 4 Ë 4 Below is a plot of the solution. Let s work one final eample that looks more at interpreting a solution rather than finding a solution. Eample 6 Find the solution to the following IVP and determine all possible behaviors of the solution as t Æ. If this behavior depends on the value of 0 give this dependence. - = 4sin 3t 0 = 0 Solution First, divide through b a to get the differential equation in the correct form. - = sin( 3t) Now find m ( t). 007 Paul Dawkins 4

16 m () t Û t - dt - ı = e = e Multipl m() t through the differential equation and rewrite the left side as a product rule. Integrate both sides and solve for the solution. e t t - - t t Ê - ˆ - Áe = e sin 3 Ë = Ûe sin 3t dt+ c ı ( t) t t t e =- e cos( 3t) - e sin( 3t) + c t 4 4 () t =- cos( 3t) - sin( 3t) + ce Appl the initial condition to find the value of c and note that it will contain 0 as we don t have a value for that = ( 0) =- + c fi c= So the solution is 4 4 Ê 4 ˆ t t t Á Ë 37 e () =- cos( 3 )- sin( 3 ) + + t Now that we have the solution, let s look at the long term behavior (i.e. t Æ ) of the solution. The first two terms of the solution will remain finite for all values of t. It is the last term that will determine the behavior of the solution. The eponential will alwas go to infinit as t Æ, however depending on the sign of the coefficient c (es we ve alread found it, but for ease of this discussion we ll continue to call it c). The following table gives the long term behavior of the solution for all values of c. Range of c Behavior of solution ast Æ c < 0 () t Æ- c = 0 () t remains finite c > 0 () t Æ This behavior can also be seen in the following graph of several of the solutions. 007 Paul Dawkins 5

17 Now, because we know how c relates to 0 we can relate the behavior of the solution to 0. The following table give the behavior of the solution in terms of 0 instead of c. Range of 0 Behavior of solution ast Æ <- =- 0 >- 0 t Æ- t remains finite t Æ 4 Note that for 0 =- 37 the solution will remain finite. That will not alwas happen. Investigating the long term behavior of solutions is sometimes more important than the solution itself. Suppose that the solution above gave the temperature in a bar of metal. In this case we would want the solution(s) that remains finite in the long term. With this investigation we would now have the value of the initial condition that will give us that solution and more importantl values of the initial condition that we would need to avoid so that we didn t melt the bar. 007 Paul Dawkins 6

18 Separable We are now going to start looking at nonlinear first order differential equations. The first tpe of nonlinear first order differential equations that we will look at is separable differential equations. A separable differential equation is an differential equation that we can write in the following form. d N( ) M ( ) d = () Note that in order for a differential equation to be separable all the 's in the differential equation must be multiplied b the derivative and all the 's in the differential equation must be on the other side of the equal sign. Solving separable differential equation is fairl eas. We first rewrite the differential equation as the following N d = M d Then ou integrate both sides. = ÚN d Ú M d () So, after doing the integrations in () ou will have an implicit solution that ou can hopefull solve for the eplicit solution, (). Note that it won't alwas be possible to solve for an eplicit solution. Recall from the Definitions section that an implicit solution is a solution that is not in the form = while an eplicit solution has been written in that form. We will also have to worr about the interval of validit for man of these solutions. Recall that the interval of validit was the range of the independent variable, in this case, on which the solution is valid. In other words, we need to avoid division b zero, comple numbers, logarithms of negative numbers or zero, etc. Most of the solutions that we will get from separable differential equations will not be valid for all values of. Let s start things off with a fairl simple eample so we can see the process without getting lost in details of the other issues that often arise with these problems. Eample Solve the following differential equation and determine the interval of validit for the solution. d 6 () d = = 5 Solution It is clear, hopefull, that this differential equation is separable. So, let s separate the differential equation and integrate both sides. As with the linear first order officiall we will pick up a constant of integration on both sides from the integrals on each side of the equal sign. The two can be moved to the same side and absorbed into each other. We will use the convention that puts the single constant on the side with the s. 007 Paul Dawkins 7

19 - d - Ú d = Ú = 6d 3 6d - = + c So, we now have an implicit solution. This solution is eas enough to get an eplicit solution, however before getting that it is usuall easier to find the value of the constant at this point. So appl the initial condition and find the value of c. - = 3 () + c c=- 8 5 Plug this into the general solution and then solve to get an eplicit solution. - = 3-8 = 8-3 Now, as far as solutions go we ve got the solution. We do need to start worring about intervals of validit however. Recall that there are two conditions that define an interval of validit. First, it must be a continuous interval with no breaks or holes in it. Second it must contain the value of the independent variable in the initial condition, = in this case. 8 So, for our case we ve got to avoid two values of. Namel, ± 3 ª± since these will give us division b zero. This gives us three possible intervals of validit < <- - < < < < However, onl one of these will contain the value of from the initial condition and so we can see that < < 3 3 must be the interval of validit for this solution. Here is a graph of the solution. 007 Paul Dawkins 8

20 Note that this does not sa that either of the other two intervals listed above can t be the interval of validit for an solution. With the proper initial condition either of these could have been the interval of validit. We ll leave it to ou to verif the details of the following claims. If we use an initial condition of (- 4) =- 0 we will get eactl the same solution however in this case the interval of validit would be the first one. - < <- Likewise, if we use ( 6) =- 80 as the initial condition we again get eactl the same solution and in this case the third interval becomes the interval of validit. 8 < < 3 So, simpl changing the initial condition a little can give an of the possible intervals. Eample Solve the following IVP and find the interval of validit for the solution = () = 3-4 Solution This differential equation is clearl separable, so let's put it in the proper form and then integrate both sides Paul Dawkins 9

21 ( - 4) = ( ) Ú( - 4) = Ú ( ) d d d d = c We now have our implicit solution, so as with the first eample let s appl the initial condition at this point to determine the value of c. 3 () () () 3-43 = c c=- The implicit solution is then 3-4 = We now need to find the eplicit solution. This is actuall easier than it might look and ou alread know how to do it. First we need to rewrite the solution a little = 0 To solve this all we need to recognize is that this is quadratic in and so we can use the quadratic formula to solve it. However, unlike quadratics ou are used to, at least some of the constants will not actuall be constant, but will in fact involve s. So, upon using the quadratic formula on this we get. 3 () ( ) 4± = 3 ( ) 4± = Net, notice that we can factor a 4 out from under the square root (it will come out as a ) and then simplif a little. 3 ( ) 4± = = ± We are almost there. Notice that we ve actuall got two solutions here (the ± ) and we onl want a single solution. In fact, onl one of the signs can be correct. So, to figure out which one is correct we can reappl the initial condition to this. Onl one of the signs will give the correct value so we can use this to figure out which one of the signs is correct. Plugging = into the solution gives. 3= = ± = ± = 3, In this case it looks like the + is the correct sign for our solution. Note that it is completel possible that the could be the solution so don t alwas epect it to be one or the other. The eplicit solution for our differential equation is. 007 Paul Dawkins 0

22 3 = To finish the eample out we need to determine the interval of validit for the solution. If we were to put a large negative value of in the solution we would end up with comple values in our solution and we want to avoid comple numbers in our solutions here. So, we will need to determine which values of will give real solutions. To do this we will need to solve the following inequalit In other words, we need to make sure that the quantit under the radical stas positive. Using a computer algebra sstem like Maple or Mathematica we see that the left side is zero at = as well as two comple values, but we can ignore comple values for interval of validit computations. Finall a graph of the quantit under the radical is shown below. So, in order to get real solutions we will need to require because this is the range of s for which the quantit is positive. Notice as well that this interval also contains the value of that is in the initial condition as it should. Therefore, the interval of validit of the solution is Here is graph of the solution. 007 Paul Dawkins

23 Eample 3 Solve the following IVP and find the interval of validit of the solution. 3 = ( 0) =- + Solution First separate and then integrate both sides ( ) d = + d Ú -3 - ( ) d = ÛÙ ı + d c - = + + Appl the initial condition to get the value of c. 3 - = + c c=- The implicit solution is then, Now let s solve for (). 3 - = + - = 3- + = ( ) =± Paul Dawkins

24 Reappling the initial condition shows us that the is the correct sign. The eplicit solution is then, ( ) = Let s get the interval of validit. That s easier than it might look for this problem. First, since + 0 the inner root will not be a problem. Therefore all we need to worr about is division b zero and negatives under the outer root. We can take care of both be requiring > 3> + ( ) 9> 4+ 9 > > 4 Note that we were able to square both sides of the inequalit because both sides of the inequalit are guaranteed to be positive in this case. Finall solving for we see that the onl possible range of s that will not give division b zero or square roots of negative numbers will be, < < and nicel enough this also contains the initial condition =0. This interval is therefore our interval of validit. Here is a graph of the solution. 007 Paul Dawkins 3

25 Eample 4 Solve the following IVP and find the interval of validit of the solution. - = e = 0 Solution This differential equation is eas enough to separate, so let's do that and then integrate both sides. e d = -4 d Ú e ( 4) Ú d = - d e = c Appling the initial condition gives = 5-0+ c c=- 4 This then gives an implicit solution of. e = We can easil find the eplicit solution to this differential equation b simpl taking the natural log of both sides. = ln -4-4 Finding the interval of validit is the last step that we need to take. Recall that we can't plug negative values or zero into a logarithm, so we need to solve the following inequalit -4-4> 0 The quadratic will be zero at the two points = ±. A graph of the quadratic (shown below) shows that there are in fact two intervals in which we will get positive values of the polnomial and hence can be possible intervals of validit. So, possible intervals of validit are - < < - + < < From the graph of the quadratic we can see that the second one contains = 5, the value of the 007 Paul Dawkins 4

26 independent variable from the initial condition. Therefore the interval of validit for this solution is. + < < Here is a graph of the solution. Eample 5 Solve the following IVP and find the interval of validit for the solution. dr r = r() = dq q Solution This is actuall a fairl simple differential equation to solve. I m doing this one mostl because of the interval of validit. So, get things separated out and then integrate. dr = dq r q Û dr = Û Ù dq Ù ı r ıq - = ln q + c r Now, appl the initial condition to find c. - = ln() + c c=- So, the implicit solution is then, - = ln q - r Solving for r gets us our eplicit solution. r = - ln q 007 Paul Dawkins 5

27 Now, there are two problems for our solution here. First we need to avoid q = 0 because of the natural log. Notice that because of the absolute value on the q we don t need to worr about q being negative. We will also need to avoid division b zero. In other words, we need to avoid the following points. - ln q = 0 lnq = eponentiate both sides q = e q =± e So, these three points break the number line up into four portions, each of which could be an interval of validit. - < q <- - e < q < 0 0 < q < e < q < The interval that will be the actual interval of validit is the one that contains q =. Therefore, the interval of validit is 0 < q < e. Here is a graph of the solution. e e 007 Paul Dawkins 6

28 Eample 6 Solve the following IVP. d -t = e sec( )( + t ) ( 0) = 0 dt Solution This problem will require a little work to get it separated and in a form that we can integrate, so let's do that first. -t d ee = + t dt cos e t ( ) d = e ( + t ) dt cos - - Now, with a little integration b parts on both sides we can get an implicit solution. - -t e cos d = e + t dt Ú - e Ú -t ( sin cos) e ( 3) - =- t + t+ + c Appling the initial condition gives. (- ) =- ( 3) + c c= 5 Therefore, the implicit solution is. - e -t 5 ( sin( ) - cos( ) ) =- e ( t + t+ 3) + It is not possible to find an eplicit solution for this problem and so we will have to leave the solution in its implicit form. Finding intervals of validit from implicit solutions can often be ver difficult so we will also not bother with that for this problem. As this last eample showed it is not alwas possible to find eplicit solutions so be on the lookout for those cases. 007 Paul Dawkins 7

29 Eact The net tpe of first order differential equations that we ll be looking at is eact differential equations. Before we get into the full details behind solving eact differential equations it s probabl best to work an eample that will help to show us just what an eact differential equation is. It will also show some of the behind the scenes details that we usuall don t bother with in the solution process. The vast majorit of the following eample will not be done in an of the remaining eamples and the work that we will put into the remaining eamples will not be shown in this eample. The whole point behind this eample is to show ou just what an eact differential equation is, how we use this fact to arrive at a solution and wh the process works as it does. The majorit of the actual solution details will be shown in a later eample. Eample Solve the following differential equation ( + + ) d = 0 d Solution Let s start off b supposing that somewhere out there in the world is a function Y(,) that we can find. For this eample the function that we need is Y, = Do not worr at this point about where this function came from and how we found it. Finding the function, Y(,), that is needed for an particular differential equation is where the vast majorit of the work for these problems lies. As stated earlier however, the point of this eample is to show ou wh the solution process works rather than showing ou the actual solution process. We will see how to find this function in the net eample, so at this point do not worr about how to find it, simpl accept that it can be found and that we ve done that for this particular differential equation. Now, take some partial derivatives of the function. Y = -9 Y = + + Now, compare these partial derivatives to the differential equation and ou ll notice that with these we can now write the differential equation as. d Y +Y = 0 () d Now, recall from our multi-variable calculus class (probabl Calculus III) that () is nothing more than the following derivative (ou ll need the multi-variable chain rule for this ). d ( (, ( ) )) d Y So, the differential equation can now be written as 007 Paul Dawkins 8

30 d ( (, ( ) )) 0 d Y = Now, if the ordinar (not partial ) derivative of something is zero, that something must have Y, = c. Or, been a constant to start with. In other words, we ve got to have = c 3 This then is an implicit solution for our differential equation! If we had an initial condition we could solve for c. We could also find an eplicit solution if we wanted to, but we ll hold off on that until the net eample. Oka, so what did we learn from the last eample? Let s look at things a little more generall. Suppose that we have the following differential equation. d M (, ) + N(, ) = 0 () d Note that it s important that it be in this form! There must be an = 0 on one side and the sign separating the two terms must be a +. Now, if there is a function somewhere out there in the world, Y(,), so that, Y = M, and Y = N, then we call the differential equation eact. In these cases we can write the differential equation as d Y +Y = 0 (3) d Then using the chain rule from Calculus III we can further reduce the differential equation to the following derivative, d ( (, ( ) )) 0 d Y = The (implicit) solution to an eact differential equation is then Y, = c (4) Well, it s the solution provided we can find Y(,) anwa. Therefore, once we have the function we can alwas just jump straight to (4) to get an implicit solution to our differential equation. Finding the function Y(,) is clearl the central task in determining if a differential equation is eact and in finding its solution. As we will see, finding Y(,) can be a somewhat length process in which there is the chance of mistakes. Therefore, it would be nice if there was some simple test that we could use before even starting to see if a differential equation is eact or not. This will be especiall useful if it turns out that the differential equation is not eact, since in this case Y(,) will not eist. It would be a waste of time to tr and find a noneistent function! 007 Paul Dawkins 9

31 So, let's see if we can find a test for eact differential equations. Let's start with () and assume that the differential equation is in fact eact. Since it s eact we know that somewhere out there is a function Y(,) that satisfies Y = M Y = N Now, provided Y(,) is continuous and its first order derivatives are also continuous we know that Y =Y However, we also have the following. Y = Y = M = M ( ) ( N) Y = Y = = N Therefore, if a differential equation is eact and Y(,) meets all of its continuit conditions we must have. M = N (5) Likewise if (5) is not true there is no wa for the differential equation to be eact. Therefore, we will use (5) as a test for eact differential equations. If (5) is true we will assume that the differential equation is eact and that Y(,) meets all of its continuit conditions and proceed with finding it. Note that for all the eamples here the continuit conditions will be met and so this won t be an issue. Oka, let s go back and rework the first eample. This time we will use the eample to show how to find Y(,). We ll also add in an initial condition to the problem. Eample Solve the following IVP and find the interval of validit for the solution ( + + ) d = 0, ( 0) =- 3 d Solution First identif M and N and check that the differential equation is eact. M = - 9 M = N = + + N = So, the differential equation is eact according to the test. However, we alread knew that as we have given ou Y(,). It s not a bad thing to verif it however and to run through the test at least once however. Now, how do we actuall find Y(,)? Well recall that Y = M Y = N 007 Paul Dawkins 30

32 We can use either of these to get a start on finding Y(,) b integrating as follows. Y= Md OR Y= Nd Ú Ú However, we will need to be careful as this won t give us the eact function that we need. Often it doesn t matter which one ou choose to work with while in other problems one will be significantl easier than the other. In this case it doesn t matter which one we use as either will be just as eas. So, I ll use the first one. (, ) 9 d 3 3 h( ) Ú Y = - = - + Note that in this case the constant of integration is not reall a constant at all, but instead it will be a function of the remaining variable(s), in this case. Recall that in integration we are asking what function we differentiated to get the function we are integrating. Since we are working with two variables here and talking about partial differentiation with respect to, this means that an term that contained onl constants or s would have differentiated awa to zero, therefore we need to acknowledge that fact b adding on a function of instead of the standard c. Oka, we ve got most of Y(,) we just need to determine h() and we ll be done. This is actuall eas to do. We used Y = M to find most of Y(,) so we ll use Y = N to find h(). Differentiate our Y(,) with respect to and set this equal to N (since the must be equal after all). Don t forget to differentiate h()! Doing this gives, Y = + h ( ) = + + = N From this we can see that h ( ) = + Note that at this stage h() must be onl a function of and so if there are an s in the equation at this stage we have made a mistake somewhere and it s time to go look for it. We can now find h() b integrating. h = + d = + + k Ú You ll note that we included the constant of integration, k, here. It will turn out however that this will end up getting absorbed into another constant so we can drop it in general. So, we can now write down Y(,). 3 3 Y, = k = k With the eception of the k this is identical to the function that we used in the first eample. We can now go straight to the implicit solution using (4) k = c 007 Paul Dawkins 3

33 We ll now take care of the k. Since both k and c are unknown constants all we need to do is subtract one from both sides and combine and we still have an unknown constant = c-k = c Therefore, we ll not include the k in anmore problems. This is where we left off in the first eample. Let s now appl the initial condition to find c = c fi c= 6 The implicit solution is then = 0 Now, as we saw in the separable differential equation section, this is quadratic in and so we can solve for () b using the quadratic formula. ( ) ( ) 4 ()( 3 3 6) - + ± = = () ± Now, reappl the initial condition to figure out which of the two signs in the ± that we need. - ± 5 - ± 5-3= ( 0) = = =- 3, So, it looks like the - is the one that we need. The eplicit solution is then. = Now, for the interval of validit. It looks like we might well have problems with square roots of negative numbers. So, we need to solve = 0 Upon solving this equation is zero at = and = Note that ou ll need to use some form of computational aid in solving this equation. Here is a graph of the polnomial under the radical. 007 Paul Dawkins 3

34 So, it looks like there are two intervals where the polnomial will be positive. - < < However, recall that intervals of validit need to be continuous intervals and contain the value of that is used in the initial condition. Therefore the interval of validit must be < Here is a quick graph of the solution. That was a long eample, but mostl because of the initial eplanation of how to find Y(,). The remaining eamples will not be as long. 007 Paul Dawkins 33

35 Eample 3 Find the solution and interval of validit for the following IVP. + 4= 3- - = 8 Solution Here, we first need to put the differential equation into proper form before proceeding. Recall that it needs to be = 0 and the sign separating the two terms must be a plus! = 0 So we have the following = 0 M = + M = 4 4 N = - N = 6 4 and so the differential equation is eact. We can either integrate M with respect to or integrate N with respect to. In this case either would be just as eas so we ll integrate N this time so we can sa that we ve got an eample of both down here. Y, = - 6d = - 6+ h Ú This time, as opposed to the previous eample, our constant of integration must be a function of since we integrated with respect to. Now differentiate with respect to and compare this to M. Y = + h ( ) = + 4= M So, it looks like h = 4 fi h = 4 Again, we ll drop the constant of integration that technicall should be present in h() since it will just get absorbed into the constant we pick up in the net step. Also note that, h() should onl involve s at this point. If there are an s left at this point a mistake has been made so go back and look for it. Writing everthing down gives us the following for Y(,). Y, = So, the implicit solution to the differential equation is = c Appling the initial condition gives, = c c = The solution is then = 0 Using the quadratic formula gives us 007 Paul Dawkins 34

36 = ( ) 6± ± = 3 6± = 3 3± = 3 Reappling the initial condition shows that this time we need the + (we ll leave those details to ou to check). Therefore, the eplicit solution is = 3 Now let s find the interval of validit. We ll need to avoid = 0 so we don t get division b zero. We ll also have to watch out for square roots of negative numbers so solve the following equation = 0 The onl real solution here is = Below is a graph of the polnomial. So, it looks like the polnomial will be positive, and hence oka under the square root on - < < Now, this interval can t be the interval of validit because it contains = 0 and we need to avoid that point. Therefore, this interval actuall breaks up into two different possible intervals of validit. - < < 0 0< < The first one contains = -, the value from the initial condition. Therefore, the interval of 007 Paul Dawkins 35

37 validit for this problem is - < < 0. Here is a graph of the solution. Eample 4 Find the solution and interval of validit to the following IVP. t -t-( - ln( t + )) = 0 ( 5) = 0 t + Solution So, first deal with that minus sign separating the two terms. t - t+ ( ln( t + ) - ) = 0 t + Now, find M and N and check that it s eact. t t M = - t M = t + t + t N = ln( t + ) - Nt = t + So, it s eact. We ll integrate the first one in this case. t Y t, = Û Ù - tdt = ln t + - t + h ı t + Differentiate with respect to and compare to N. Y = ln t + + h = ln t + - = N So, it looks like we ve got. This gives us h =- fi h =- Y = t, ln t t 007 Paul Dawkins 36

38 The implicit solution is then, Appling the initial condition gives, ln t + -t - = c - 5 = c The implicit solution is now, ( ) t t ln =- 5 This solution is much easier to solve than the previous ones. No quadratic formula is needed this time, all we need to do is solve for. Here s what we get for an eplicit solution. t -5 () t = ln t + - Alright, let s get the interval of validit. The term in the logarithm is alwas positive so we don t need to worr about negative numbers in that. We do need to worr about division b zero however. We will need to avoid the following point(s). ( t ) ( t ) ln + - = 0 ln + = We now have three possible intervals of validit. e t + = e t e =± - - < <- e < t < - e t e - < t < The last one contains t = 5 and so is the interval of validit for this problem is Here s a graph of the solution. e - < t <. 007 Paul Dawkins 37

39 Eample 5 Find the solution and interval of validit for the following IVP e - + e + 3 e = 0 0 = Solution Let s identif M and N and check that it s eact. M = 3 e - M = 9 e + 9 e N = e + 3 e N = 9 e + 9 e So, it s eact. With the proper simplification integrating the second one isn t too bad. However, the first is alread set up for eas integration so let s do that one Y (, ) = Ú 3 e - d= e - + h( ) Differentiate with respect to and compare to N Y = e + 3 e + h ( ) = e + 3 e = N So, it looks like we ve got h = 0 fi h = 0 Recall that actuall h() = k, but we drop the k because it will get absorbed in the net step. That gives us h() = 0. Therefore, we get. Y, = e - 3 The implicit solution is then Appl the initial condition. The implicit solution is then 3 e - = c e 3 = c - = This is as far as we can go. There is no wa to solve this for and get an eplicit solution. 007 Paul Dawkins 38

40 Bernoulli In this section we are going to take a look at differential equations in the form, + p = q n where p() and q() are continuous functions on the interval we re working on and n is a real number. Differential equations in this form are called Bernoulli Equations. First notice that if n = 0 or n = then the equation is linear and we alread know how to solve it in these cases. Therefore, in this section we re going to be looking at solutions for values of n other than these two. In order to solve these we ll first divide the differential equation b n -n - + p = q n to get, We are now going to use the substitution v= - to convert this into a differential equation in terms of v. As we ll see this will lead to a differential equation that we can solve. We are going to have to be careful with this however when it comes to dealing with the n derivative,. We need to determine just what is in terms of our substitution. This is easier to do than it might at first look to be. All that we need to do is differentiate both sides of our substitution with respect to. Remember that both v and are functions of and so we ll need to use the chain rule on the right side. If ou remember our Calculus I ou ll recall this is just implicit differentiation. So, taking the derivative gives us, -n v = - n Now, plugging this as well as our substitution into the differential equation gives, v p( ) v q( ) -n + = This is a linear differential equation that we can solve for v and once we have this in hand we can also get the solution to the original differential equation b plugging v back into our substitution and solving for. Let s take a look at an eample. Eample Solve the following IVP and find the interval of validit for the solution = ( ) =-, > 0 Solution So, the first thing that we need to do is get this into the proper form and that means dividing everthing b. Doing this gives, = 007 Paul Dawkins 39

41 The substitution and derivative that we ll need here is, v= - v =- - With this substitution the differential equation becomes, v + v= So, as noted above this is a linear differential equation that we know how to solve. We ll do the details on this one and then for the rest of the eamples in this section we ll leave the details for ou to fill in. If ou need a refresher on solving linear differential equations then go back to that section for a quick review. Here s the solution to this differential equation Ú- d -4ln -4 v - v=- fi m = e = e = Û -4 ( v - ) d= - d ı Ú =- + fi = v ln c v c ln Note that we dropped the absolute value bars on the in the logarithm because of the assumption that > 0. Now we need to determine the constant of integration. This can be done in one of two was. We can can convert the solution above into a solution in terms of and then use the original initial condition or we can convert the initial condition to an initial condition in terms of v and use that. Because we ll need to convert the solution to s eventuall anwa and it won t add that much work in we ll do it that wa. So, to get the solution in terms of all we need to do is plug the substitution back in. Doing this gives, 4 - = c- ln At this point we can solve for and then appl the initial condition or appl the initial condition and then solve for. We ll generall do this with the later approach so let s appl the initial condition to get, = c - fi c= - ln ln 6 Plugging in for c and solving for gives, -6-6 ( ) = = = ln- - ln + 6ln- 6ln + 6ln 6 Note that we did a little simplification in the solution. This will help with finding the interval of validit. Before finding the interval of validit however, we mentioned above that we could convert the 007 Paul Dawkins 40

42 original initial condition into an initial condition for v. Let s briefl talk about how to do that. To do that all we need to do is plug = into the substitution and then use the original initial condition. Doing this gives, - - v = = - =- So, in this case we got the same value for v that we had for. Don t epect that to happen in general if ou chose to do the problems in this manner. Oka, let s now find the interval of validit for the solution. First we alread know that > 0 and that means we ll avoid the problems of having logarithms of negative numbers and division b zero at = 0. So, all that we need to worr about then is division b zero in the second term and this will happen where, + 6ln = 0 ln = = e fi = e ª.8788 The two possible intervals of validit are then, < < e e < < and since the second one contains the initial condition we know that the interval of validit is - 6 then e < <. Here is a graph of the solution. Let s do a couple more eamples and as noted above we re going to leave it to ou to solve the linear differential equation when we get to that stage. 007 Paul Dawkins 4

43 Eample Solve the following IVP and find the interval of validit for the solution. - - = 5+ e 0 = Solution The first thing we ll need to do here is multipl through b and we ll also do a little rearranging to get things into the form we ll need for the linear differential equation. This gives, =e The substitution here and its derivative is, 3 v= v = 3 Plugging the substitution into the differential equation gives, 3 v - 5v= e fi v - 5v= 3e m( ) = e We rearranged a little and gave the integrating factor for the linear differential equation solution. Upon solving we get, 3 - v = ce - e 5 7 Now go back to s = ce - 7 e Appling the initial condition and solving for c gives, 8 = c- fi c= Plugging in c and solving for gives, Ê39e - 3e ˆ = Á Ë 7 There are no problem values of for this solution and so the interval of validit is all real numbers. Here s a graph of the solution. 007 Paul Dawkins 4

44 Eample 3 Solve the following IVP and find the interval of validit for the solution = 0 =- Solution First get the differential equation in the proper form and then write down the substitution = fi v= - v =- 3 - Plugging the substitution into the differential equation gives, -v - v= fi v + v=- m =e Again, we ve rearranged a little and given the integrating factor needed to solve the linear differential equation. Upon solving the linear differential equation we have, v =- - + ce - Now back substitute to get back into s. 3 =- - + ce - - Now we need to appl the initial condition and solve for c. - = + c fi c=- Plugging in c and solving for gives, = e 3 Net, we need to think about the interval of validit. In this case all we need to worr about it is division b zero issues and using some form of computational aid (such as Maple or Mathematica) we will see that the denominator of our solution is never zero and so this solution will be valid for all real numbers. Here is a graph of the solution. 007 Paul Dawkins 43

45 To this point we ve onl worked eamples in which n was an integer (positive and negative) and so we should work a quick eample where n is not an integer. Eample 4 Solve the following IVP and find the interval of validit for the solution. + - = 0 () = 0 Solution Let s first get the differential equation into proper form. - + = fi + = The substitution is then, - v= v = Now plug the substitution into the differential equation to get, v + v = fi v + v = m = As we ve done with the previous eamples we ve done some rearranging and given the integrating factor needed for solving the linear differential equation. Solving this gives us, 3 v = + c - In terms of this is, = 3 + c - Appling the initial condition and solving for c gives, 0 = + c fi c=- 3 3 Plugging in for c and solving for gives us the solution Ê ˆ - + ( ) = Á - = Ë 9 Note that we multiplied everthing out and converted all the negative eponents to positive eponents to make the interval of validit clear here. Because of the root (in the second term in the numerator) and the in the denominator we can see that we need to require > 0 in order for the solution to eist and it will eist for all positive s and so this is also the interval of validit. Here is the graph of the solution. 007 Paul Dawkins 44

46 007 Paul Dawkins 45

47 Substitutions In the previous section we looked at Bernoulli Equations and saw that in order to solve them we n needed to use the substitution v= -. Upon using this substitution we were able to convert the differential equation into a form that we could deal with (linear in this case). In this section we want to take a look at a couple of other substitutions that can be used to reduce some differential equations down to a solvable form. The first substitution we ll take a look at will require the differential equation to be in the form, Ê ˆ = F Á Ë First order differential equations that can be written in this form are called homogeneous differential equations. Note that we will usuall have to do some rewriting in order to put the differential equation into the proper form. Once we have verified that the differential equation is a homogeneous differential equation and we ve gotten it written in the proper form we will use the following substitution. v( ) = We can then rewrite this as, = v and then remembering that both and v are functions of we can use the product rule (recall that is implicit differentiation from Calculus I) to compute, = v+ v Under this substitution the differential equation is then, v+ v = F v dv d v = F( v) -v fi = F v -v As we can see with a small rewrite of the new differential equation we will have a separable differential equation after the substitution. Let s take a quick look at a couple of eamples of this kind of substitution. Eample Solve the following IVP and find the interval of validit for the solution = 0 =- 7, > 0 Solution Let s first divide both sides b to rewrite the differential equation as follows, Ê 4 4 ˆ =- - =- - Á Ë 007 Paul Dawkins 46

48 Now, this is not in the officiall proper form as we have listed above, but we can see that everwhere the variables are listed the show up as the ratio, / and so this is reall as far as we need to go. So, let s plug the substitution into this form of the differential equation to get, v v+ v =-4- v Net, rewrite the differential equation to get everthing separated out. vv =-4-v 4+ v v =- v v dv=- d 4+ v Integrating both sides gives, 4 ln 4 + v =- ln + c We need to do a little rewriting using basic logarithm properties in order to be able to easil solve this for v. 4 - ln 4+ v = ln + c Now eponentiate both sides and do a little rewriting - - ln ln ( 4 ) 4 + c c c + v = e = ee = Note that because c is an unknown constant then so is we did above. c e and so we ma as well just call this c as Finall, let s solve for v and then plug the substitution back in and we ll pla a little fast and loose with constants again. 4 c c 4+ v = = 4 4 Ê c ˆ v = Á -4 4 Ë 4 Ê c-4 ˆ = Á 4 Ë Êc-4 ˆ c-4 Ë 4 4 = Á = 4 At this point it would probabl be best to go ahead and appl the initial condition. Doing that gives, c-46 49= fi c= Paul Dawkins 47

49 Note that we could have also converted the original initial condition into one in terms of v and then applied it upon solving the separable differential equation. In this case however, it was probabl a little easier to do it in terms of given all the logarithms in the solution to the separable differential equation. Finall, plug in c and solve for to get, 8-8- = fi ( ) =± 4 4 The initial condition tells us that the must be the correct sign and so the actual solution is, =- 8- For the interval of validit we can see that we need to avoid = 0 and because we can t allow negative numbers under the square root we also need to require that, fi So, we have two possible intervals of validit, < 0 0< and the initial condition tells us that it must be 0< The graph of the solution is, 4 Eample Solve the following IVP and find the interval of validit for the solution. = ln- ln = 4, > Paul Dawkins 48

50 Solution On the surface this differential equation looks like it won t be homogeneous. However, with a quick logarithm propert we can rewrite this as, Ê ˆ = ln Á Ë In this form the differential equation is clearl homogeneous. Appling the substitution and separating gives, ʈ v+ v = vln Á Ëv v dv ( ln( v) -) Ê Ê ˆ ˆ v = válná - v Ë Ë d = Integrate both sides and do a little rewrite to get, -ln ln - = ln + c ( ( v ) ) ( v ) ln ln - = c-ln u = ln -. You were able to do the integral on the left right? It used the substitution Now, solve for v and note that we ll need to eponentiate both sides a couple of times and pla fast and loose with constants again. - - ln( ) + c c ln( ) c ln( v) - = e = ee = c ln( v) = + c c = e fi v = e v Plugging the substitution back in and solving for gives, c = e fi ( ) = e c Appling the initial condition and solving for c gives, -c- 4= e fi c =- + ln4 v The solution is then, = e + ln4-007 Paul Dawkins 49

51 We clearl need to avoid = 0 to avoid division b zero and so with the initial condition we can see that the interval of validit is > 0. The graph of the solution is, For the net substitution we ll take a look at we ll need the differential equation in the form, = G a+ b In these cases we ll use the substitution, v= a+ b fi v = a+ b Plugging this into the differential equation gives, b( v - a ) = G ( v ) dv v = a+ bg( v) fi = d a+ bg v So, with this substitution we ll be able to rewrite the original differential equation as a new separable differential equation that we can solve. Let s take a look at a couple of eamples. Eample 3 Solve the following IVP and find the interval of validit for the solution = 0 0 = Solution In this case we ll use the substitution. v= 4- v = 4- Note that we didn t include the + in our substitution. Usuall onl the a + b part gets included in the substitution. There are times where including the etra constant ma change the difficult of the solution process, either easier or harder, however in this case it doesn t reall make much difference so we won t include it in our substitution. 007 Paul Dawkins 50

52 So, plugging this into the differential equation gives, ( v ) ( v ) 4-v - + = 0 ( v ) v = 4- + dv =-d + -4 As we ve shown above we definitel have a separable differential equation. Also note that to help with the solution process we left a minus sign on the right side. We ll need to integrate both sides and in order to do the integral on the left we ll need to use partial fractions. We ll leave it to ou to fill in the missing details and given that we ll be doing quite a bit of partial fraction work in a few chapters ou should reall make sure that ou can do the missing details. Û dv Û dv Ù = = -d ı Ù v + v- 3 ı ( v+ 3)( v-) Ú 4 Û Ù - dv= -d 4ı v- v+ 3 Ú ( ln( ) ln( 3) ) v- - v+ =- + c Ê v- lná ˆ= c - 4 Ëv+ 3 Note that we plaed a little fast and loose with constants above. The net step is fairl mess but needs to be done and that is to solve for v and note that we ll be plaing fast and loose with constants again where we can get awa with it and we ll be skipping a few steps that ou shouldn t have an problem verifing. v- = c-4-4 e = ce v+ 3-4 v- = ce v+ 3 ( e ) -4-4 v - c = + 3ce At this stage we should back awa a bit and note that we can t pla fast and loose with constants anmore. We were able to do that in first step because the c appeared onl once in the equation. At this point however the c appears twice and so we ve got to keep them around. If we absorbed the 3 into the c on the right the new c would be different from the c on the left because the c on the left didn t have the 3 as well. So, let s solve for v and then go ahead and go back into terms of c e v = -4 - ce + 3c e 4- = - ce 007 Paul Dawkins ce = 4- - ce -4-4

53 The last step is to then appl the initial condition and solve for c. + 3c = ( 0) =- fi c=-3 -c The solution is then, -9e = e -4-4 Note that because eponentials eist everwhere and the denominator of the second term is alwas positive (because eponentials are alwas positive and adding a positive one onto that won t change the fact that it s positive) the interval of validit for this solution will be all real numbers. Here is a graph of the solution. Eample 4 Solve the following IVP and find the interval of validit for the solution. 9- = e 0 = 0 Solution Here is the substitution that we ll need for this eample. v= 9- v = 9 - Plugging this into our differential equation gives, v v + = e 9 v v = 9e - dv e dv = d fi = d v -v 9e - 9-e Note that we did a little rewrite on the separated portion to make the integrals go a little easier. -v B multipling the numerator and denominator b e we can turn this into a fairl simpl substitution integration problem. So, upon integrating both sides we get, -v 007 Paul Dawkins 5

54 -v ( e ) ln 9- = + c Solving for v gives, -v c 9 - e = ee = ce e -v = 9 -ce ( ce ) v=-ln 9- Plugging the substitution back in and solving for gives us, = 9 -ln( 9- e ) c Net, appl the initial condition and solve for c. 0= 0 =- ln 9-c fi c= 8 9 The solution is then, = 9 -ln( 9-8e ) Now, for the interval of validit we need to make sure that we onl take logarithms of positive numbers as we ll need to require that, e > 0 fi e < fi < ln = 0.78 Here is a graph of the solution. 8 8 In both this section and the previous section we ve seen that sometimes a substitution will take a differential equation that we can t solve and turn it into one that we can solve. This idea of substitutions is an important idea and should not be forgotten. Not ever differential equation can be made easier with a substitution and there is no wa to show ever possible substitution but remembering that a substitution ma work is a good thing to do. If ou get stuck on a differential equation ou ma tr to see if a substitution if some kind will work for ou. 007 Paul Dawkins 53

55 Intervals of Validit I ve called this section Intervals of Validit because all of the eamples will involve them. However, there is a lot more to this section. We will see a couple of theorems that will tell us when we can solve a differential equation. We will also see some of the differences between linear and nonlinear differential equations. First let's take a look at a theorem about linear first order differential equations. This is a ver important theorem although we re not going to reall use it for its most important aspect. Theorem Consider the following IVP. + p t = g t t = 0 0 If p(t) and g(t) are continuous functions on an open interval a < t < b and the interval contains t o, then there is a unique solution to the IVP on that interval. So, just what does this theorem tell us? First, it tells us that for nice enough linear first order differential equations solutions are guaranteed to eist and more importantl the solution will be unique. We ma not be able to find the solution, but do know that it eists and that there will onl be one of them. This is the ver important aspect of this theorem. Knowing that a differential equation has a unique solution is sometimes more important than actuall having the solution itself! Net, if the interval in the theorem is the largest possible interval on which p(t) and g(t) are continuous then the interval is the interval of validit for the solution. This means, that for linear first order differential equations, we won't need to actuall solve the differential equation in order to find the interval of validit. Notice as well that the interval of validit will depend onl partiall on the initial condition. The interval must contain t o, but the value of o, has no effect on the interval of validit. Let s take a look at an eample. Eample Without solving, determine the interval of validit for the following initial value problem. t = ln 0-4t 4 =- 3 Solution First, in order to use the theorem to find the interval of validit we must write the differential equation in the proper form given in the theorem. So we will need to divide out b the coefficient of the derivative. ln 0-4t + = t -9 t -9 Net, we need to identif where the two functions are not continuous. This will allow us to find all possible intervals of validit for the differential equation. So, p(t) will be discontinuous at t =± 3 since these points will give a division b zero. Likewise, g(t) will also be discontinuous at t =± 3 as well as t = 5 since at this point we will have the natural logarithm of zero. Note that in 007 Paul Dawkins 54

56 this case we won't have to worr about natural log of negative numbers because of the absolute values. Now, with these points in hand we can break up the real number line into four intervals where both p(t) and g(t) will be continuous. These four intervals are, - < t <-3-3< t < 3 3< t < 5 5< t < The endpoints of each of the intervals are points where at least one of the two functions is discontinuous. This will guarantee that both functions are continuous everwhere in each interval. Finall, let's identif the actual interval of validit for the initial value problem. The actual interval of validit is the interval that will contain t o = 4. So, the interval of validit for the initial value problem is. 3< t < 5 In this last eample we need to be careful to not jump to the conclusion that the other three intervals cannot be intervals of validit. B changing the initial condition, in particular the value of t o, we can make an of the four intervals the interval of validit. The first theorem required a linear differential equation. There is a similar theorem for non-linear first order differential equations. This theorem is not as useful for finding intervals of validit as the first theorem was so we won t be doing all that much with it. Here is the theorem. Theorem Consider the following IVP. (, ) = f t t = 0 0 f If f(t,) and are continuous functions in some rectangle t a < < b, g < < d containing the point (t o, o ) then there is a unique solution to the IVP in some interval t o h < t < t o + h that is contained in a < t < b. That s it. Unlike the first theorem, this one cannot reall be used to find an interval of validit. So, we will know that a unique solution eists if the conditions of the theorem are met, but we will actuall need the solution in order to determine its interval of validit. Note as well that for non-linear differential equations it appears that the value of 0 ma affect the interval of validit. Here is an eample of the problems that can arise when the conditions of this theorem aren t met. Eample Determine all possible solutions to the following IVP. = 3 0 = 0 Solution First, notice that this differential equation does NOT satisf the conditions of the theorem. df 3 f ( ) = = d Paul Dawkins 55

57 So, the function is continuous on an interval, but the derivative is not continuous at = 0 and so will not be continuous at an interval containing = 0. In order to use the theorem both must be continuous on an interval that contains o = 0 and this is problem for us since we do have o = 0. Now, let s actuall work the problem. This differential equation is separable and is fairl simple to solve. Ú 3-3 d = dt 3 = t+ c Appling the initial condition gives c = 0 and so the solution is. 3 3 = t 3 = t 3 () t 007 Paul Dawkins 56 Ú 3 Ê ˆ = Á t Ë3 Ê ˆ =± Á t Ë3 So, we ve got two possible solutions here, both of which satisf the differential equation and the initial condition. There is also a third solution to the IVP. (t) = 0 is also a solution to the differential equation and satisfies the initial condition. In this last eample we had a ver simple IVP and it onl violated one of the conditions of the theorem, et it had three different solutions. All the eamples we ve worked in the previous sections satisfied the conditions of this theorem and had a single unique solution to the IVP. This eample is a useful reminder of the fact that, in the field of differential equations, things don t alwas behave nicel. It s eas to forget this as most of the problems that are worked in a differential equations class are nice and behave in a nice, predictable manner. Let s work one final eample that will illustrate one of the differences between linear and nonlinear differential equations. Eample 3 Determine the interval of validit for the initial value problem below and give its dependence on the value of o = 0 = Solution Before proceeding in this problem, we should note that the differential equation is non-linear and meets both conditions of the Theorem and so there will be a unique solution to the IVP for each possible value of o. Also, note that the problem asks for an dependence of the interval of validit on the value of o. This immediatel illustrates a difference between linear and non-linear differential equations. 3 0

58 Intervals of validit for linear differential equations do not depend on the value of o. Intervals of validit for non-linear differential can depend on the value of o as we pointed out after the second theorem. So, let s solve the IVP and get some intervals of validit. First note that if o = 0 then (t) = 0 is the solution and this has an interval of validit of - < t < So for the rest of the problem let's assume that 0 0. Now, the differential equation is separable so let's solve it and get a general solution. Ú - d = Ú - = t+ c dt Appling the initial condition gives c =- 0 The solution is then. - = t - () t = 0 () t = - t 0 Now that we have a solution to the initial value problem we can start finding intervals of validit. From the solution we can see that the onl problem that we ll have is division b zero at t = This leads to two possible intervals of validit. - < t < 0 This leads to the following possible intervals of validit, depending on the value of o. 007 Paul Dawkins t 0 0 < t < That actual interval of validit will be the interval that contains t o = 0. This however, depends on the value of o. If o < 0 then < 0 and so the second interval will contain t o = 0. Likewise if o > 0 then > and in this case the first interval will contain t o = 0.

59 If 0 > 0 - < t < is the interval of validit. If = 0 - < t < is the interval of validit. 0 If 0 < 0 < t < is the interval of validit. On a side note, notice that the solution, in its final form, will also work if o = 0. 0 So what did this eample show us about the difference between linear and non-linear differential equations? First, as pointed out in the solution to the eample, intervals of validit for non-linear differential equations can depend on the value of o, whereas intervals of validit for linear differential equations don t. Second, intervals of validit for linear differential equations can be found from the differential equation with no knowledge of the solution. This is definitel not the case with non-linear differential equations. It would be ver difficult to see how an of these intervals in the last eample could be found from the differential equation. Knowledge of the solution was required in order for us to find the interval of validit Paul Dawkins 58

60 Modeling with First Order We now move into one of the main applications of differential equations both in this class and in general. Modeling is the process of writing a differential equation to describe a phsical situation. Almost all of the differential equations that ou will use in our job (for the engineers out there in the audience) are there because somebod, at some time, modeled a situation to come up with the differential equation that ou are using. This section is not intended to completel teach ou how to go about modeling all phsical situations. A whole course could be devoted to the subject of modeling and still not cover everthing! This section is designed to introduce ou to the process of modeling and show ou what is involved in modeling. We will look at three different situations in this section : Miing Problems, Population Problems, and Falling Bodies. In all of these situations we will be forced to make assumptions that do not accuratel depict realit in most cases, but without them the problems would be ver difficult and beond the scope of this discussion (and the course in most cases to be honest). So let s get started. Miing Problems In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank ma or ma not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at an time t we want to develop a differential equation that, when solved, will give us an epression for Q(t). Note as well that in man situations we can think of air as a liquid for the purposes of these kinds of discussions and so we don t actuall need to have an actual liquid, but could instead use air as the liquid. The main assumption that we ll be using here is that the concentration of the substance in the liquid is uniform throughout the tank. Clearl this will not be the case, but if we allow the concentration to var depending on the location in the tank the problem becomes ver difficult and will involve partial differential equations, which is not the focus of this course. The main equation that we ll be using to model this situation is : Rate of change of Q(t) = Rate at which Q(t) enters the tank 007 Paul Dawkins 59 Rate at which Q(t) eits the tank where, dq Rate of change of Q(t) = = Q () t dt Rate at which Q(t) enters the tank = (flow rate of liquid entering) (concentration of substance in liquid entering) Rate at which Q(t) eits the tank = (flow rate of liquid eiting) (concentration of substance in liquid eiting) Let s take a look at the first problem.

61 Eample A 500 gallon tank initiall contains 600 gallons of water with 5 lbs of salt dissolved in it. Water enters the tank at a rate of 9 gal/hr and the water entering the tank has a salt concentration of ( 5 + cos ( t )) lbs/gal. If a well mied solution leaves the tank at a rate of 6 gal/hr, how much salt is in the tank when it overflows? Solution First off, let s address the well mied solution bit. This is the assumption that was mentioned earlier. We are going to assume that the instant the water enters the tank it somehow instantl disperses evenl throughout the tank to give a uniform concentration of salt in the tank at ever point. Again, this will clearl not be the case in realit, but it will allow us to do the problem. Now, to set up the IVP that we ll need to solve to get Q(t) we ll need the flow rate of the water entering (we ve got that), the concentration of the salt in the water entering (we ve got that), the flow rate of the water leaving (we ve got that) and the concentration of the salt in the water eiting (we don t have this et). So, we first need to determine the concentration of the salt in the water eiting the tank. Since we are assuming a uniform concentration of salt in the tank the concentration at an point in the tank and hence in the water eiting is given b, Amount of salt in the tank at an time, t Concentration = Volume of water in the tank at an time, t The amount at an time t is eas it s just Q(t). The volume is also prett eas. We start with 600 gallons and ever hour 9 gallons enters and 6 gallons leave. So, if we use t in hours, ever hour 3 gallons enters the tank, or at an time t there is t gallons of water in the tank. So, the IVP for this situation is, Ê Qt ˆ Q Ê ˆ () t = ( 9) Á ( + cos() t ) - ( 6) Á Ë5 Ë600+ 3t Q( 0) = 5 9 Q() t Q () t = ( + cos () t ) t Q( 0 ) = 5 This is a linear differential equation and it isn t too difficult to solve (hopefull). We will show most of the details, but leave the description of the solution process out. If ou need a refresher on solving linear first order differential equations go back and take a look at that section. Q( t) 9 Q () t + = ( + cos () t ) 00+ t 5 Û Ù ı ÛÙ dt ı 00+ t ln( 00+ t) 00 m () t = e = e = ( + t) (( 00+ ) ()) = Û Ù 9 ( 00+ )( + cos() ) t Q t dt t t dt ı Paul Dawkins 60

62 9Ê ˆ + t Qt = Á + t + + t t + + t t - t + c 5Ë3 3 ( 00 ) () ( 00 ) ( 00 ) sin() 00 cos() sin () () t sin () t ˆ c t ( t) ( t) 9Ê cos Q() t = ( 00 + t) + sin () t Á Ë So, here s the general solution. Now, appl the initial condition to get the value of the constant, c. 9Ê ˆ c 5= Q( 0) = Á ( 00) + + c= Ë So, the amount of salt in the tank at an time t is. 9Ê cos() t sin () t ˆ Q() t = ( 00+ t) + sin () t Á3 00 t ( 00 t) Ë + + ( 00 + t) Now, the tank will overflow at t = 300 hrs. The amount of salt in the tank at that time is. Q 300 = lbs Here s a graph of the salt in the tank before it overflows. Note that the whole graph should have small oscillations in it as ou can see in the range from 00 to 50. The scale of the oscillations however was small enough that the program used to generate the image had trouble showing all of them. The work was a little mess with that one, but the will often be that wa so don t get ecited about it. This first eample also assumed that nothing would change throughout the life of the process. That, of course will usuall not be the case. Let s take a look at an eample where something changes in the process. 007 Paul Dawkins 6

63 Eample A 000 gallon holding tank that catches runoff from some chemical process initiall has 800 gallons of water with ounces of pollution dissolved in it. Polluted water flows into the tank at a rate of 3 gal/hr and contains 5 ounces/gal of pollution in it. A well mied solution leaves the tank at 3 gal/hr as well. When the amount of pollution in the holding tank reaches 500 ounces the inflow of polluted water is cut off and fresh water will enter the tank at a decreased rate of gallons while the outflow is increased to 4 gal/hr. Determine the amount of pollution in the tank at an time t. Solution Oka, so clearl the pollution in the tank will increase as time passes. If the amount of pollution ever reaches the maimum allowed there will be a change in the situation. This will necessitate a change in the differential equation describing the process as well. In other words, we ll need two IVP s for this problem. One will describe the initial situation when polluted runoff is entering the tank and one for after the maimum allowed pollution is reached and fresh water is entering the tank. Here are the two IVP s for this problem. ÊQ( t) ˆ Q () t = ( 3)( 5) -( 3) Á Q( 0) = 0 t t Ë 800 () t ( t t ) Ê Q ˆ Q () t = ( )( 0) -( 4) Á Q( tm) = 500 tm t te 800 Ë - - m The first one is fairl straight forward and will be valid until the maimum amount of pollution is reached. We ll call that time t m. Also, the volume in the tank remains constant during this time so we don t need to do anthing fanc with that this time in the second term as we did in the previous eample. We ll need a little eplanation for the second one. First notice that we don t start over at t = 0. We start this one at t m, the time at which the new process starts. Net, fresh water is flowing into the tank and so the concentration of pollution in the incoming water is zero. This will drop out the first term, and that s oka so don t worr about that. Now, notice that the volume at an time looks a little funn. During this time frame we are losing two gallons of water ever hour of the process so we need the - in there to account for that. However, we can t just use t as we did in the previous eample. When this new process starts up there needs to be 800 gallons of water in the tank and if we just use t there we won t have the required 800 gallons that we need in the equation. So, to make sure that we have the proper volume we need to put in the difference in times. In this wa once we are one hour into the new process (i.e t - t m = ) we will have 798 gallons in the tank as required. Finall, the second process can t continue forever as eventuall the tank will empt. This is denoted in the time restrictions as t e. We can also note that t e = t m since the tank will empt 400 hours after this new process starts up. Well, it will end provided something doesn t come along and start changing the situation again. Oka, now that we ve got all the eplanations taken care of here s the simplified version of the IVP s that we ll be solving. m 007 Paul Dawkins 6

64 () () 3Q t Q () t = 5- Q( 0) = 0 t tm 800 Q Q t =- Q t = 500 t t t () t ( t t ) m m m e The first IVP is a fairl simple linear differential equation so we ll leave the details of the solution to ou to check. Upon solving ou get. () Q t = e Now, we need to find t m. This isn t too bad all we need to do is determine when the amount of pollution reaches 500. So we need to solve. () Q t - 3t 800-3t 800 = e = 500 fi t = So, the second process will pick up at hours. For completeness sake here is the IVP with this information inserted. Q() t Q () t =- Q( ) = t t This differential equation is both linear and separable and again isn t terribl difficult to solve so I ll leave the details to ou again to check that we should get. Q () t = ( t) 30 So, a solution that encompasses the complete running time of the process is 3t Ï Ô e 0 t Q() t =Ì( t) Ô t Ó 30 Here is a graph of the amount of pollution in the tank at an time t. m 007 Paul Dawkins 63

65 As ou can surel see, these problems can get quite complicated if ou want them to. Take the last eample. A more realistic situation would be that once the pollution dropped below some predetermined point the polluted runoff would, in all likelihood, be allowed to flow back in and then the whole process would repeat itself. So, realisticall, there should be at least one more IVP in the process. Let s move on to another tpe of problem now. Population These are somewhat easier than the miing problems although, in some was, the are ver similar to miing problems. So, if P(t) represents a population in a given region at an time t the basic equation that we ll use is identical to the one that we used for miing. Namel, Rate of change of P(t) = Rate at which P(t) enters the region - Rate at which P(t) eits the region Here the rate of change of P(t) is still the derivative. What s different this time is the rate at which the population enters and eits the region. For population problems all the was for a population to enter the region are included in the entering rate. Birth rate and migration into the region are eamples of terms that would go into the rate at which the population enters the region. Likewise, all the was for a population to leave an area will be included in the eiting rate. Therefore things like death rate, migration out and predation are eamples of terms that would go into the rate at which the population eits the area. Here s an eample. Eample 3 A population of insects in a region will grow at a rate that is proportional to their current population. In the absence of an outside factors the population will triple in two weeks time. On an given da there is a net migration into the area of 5 insects and 6 are eaten b the local bird population and 7 die of natural causes. If there are initiall 00 insects in the area will the population survive? If not, when do the die out? Solution Let s start out b looking at the birth rate. We are told that the insects will be born at a rate that is proportional to the current population. This means that the birth rate can be written as rp where r is a positive constant that will need to be determined. Now, let s take everthing into account and get the IVP for this problem. P = rp P 0 = 00 P = rp- 8 P 0 = 00 Note that in the first line we used parenthesis to note which terms went into which part of the differential equation. Also note that we don t make use of the fact that the population will triple in two weeks time in the absence of outside factors here. In the absence of outside factors means that the ONLY thing that we can consider is birth rate. Nothing else can enter into the picture and clearl we have other influences in the differential equation. 007 Paul Dawkins 64

66 So, just how does this tripling come into pla? Well, we should also note that without knowing r we will have a difficult time solving the IVP completel. We will use the fact that the population triples in two week time to help us find r. In the absence of outside factors the differential equation would become. P = rp P 0 = 00 P 4 = 300 Note that since we used das as the time frame in the actual IVP I needed to convert the two weeks to 4 das. We could have just as easil converted the original IVP to weeks as the time frame, in which case there would have been a net change of 56 per week instead of the 8 per da that we are currentl using in the original differential equation. Oka back to the differential equation that ignores all the outside factors. This differential equation is separable and linear and is a simple differential equation to solve. I ll leave the detail to ou to get the general solution. P t = ce rt Appling the initial condition gives c = 00. Now appl the second condition. r r 300= P 4 = 00e 300= 00e 4 4 We need to solve this for r. First divide both sides b 00, then take the natural log of both sides. 4r 3 = e We made use of the fact that ln g ln3= ln e ln3= 4r r = ln3 4 4r e = g here to simplif the problem. Now, that we have r we can go back and solve the original differential equation. We ll rewrite it a little for the solution process. ln3 P - P=- 8 P( 0) = 00 4 This is a fairl simple linear differential equation, but that coefficient of P alwas get people bent out of shape, so we ll go through at least some of the details here. m () t Û ln3 - dt ln3 ı 4 - t 4 = e = e t Now, don t get ecited about the integrating factor here. It s just like e onl this time the constant is a little more complicated than just a, but it is a constant! Now, solve the differential equation. 007 Paul Dawkins 65

67 Û Ù ı Pe ln3t ( e ) - - ln3t 4 4 P dt = -8e dt - ln3t - ln3t 4 4 () P t Ú Ê 4 ˆ =-8Á- e + c Ë ln3 ln3t 4 = + ce ln3 Again, do not get ecited about doing the right hand integral, it s just like integrating Appling the initial condition gives the following. Ê ˆ ln3t ln3t 4 4 P() t = + Á00- = ln3 Ë ln3 e ln3 e Now, the eponential has a positive eponent and so will go to plus infinit as t increases. Its coefficient, however, is negative and so the whole population will go negative eventuall. Clearl, population can t be negative, but in order for the population to go negative it must pass through zero. In other words, eventuall all the insects must die. So, the don t survive and we can solve the following to determine when the die out. ln3t 4 0= e fi t = das ln3 t e! So, the insects will survive for around 7. weeks. Here is a graph of the population during the time in which the survive. As with the miing problems, we could make the population problems more complicated b changing the circumstances at some point in time. For instance, if at some point in time the local bird population saw a decrease due to disease the wouldn t eat as much after that point and a second differential equation to govern the time after this point. Let s now take a look at the final tpe of problem that we ll be modeling in this section. 007 Paul Dawkins 66

68 Falling Bod This will not be the first time that we ve looked into falling bodies. If ou recall, we looked at one of these when we were looking at Direction Fields. In that section we saw that the basic equation that we ll use is Newton s Second Law of Motion. mv = F tv, The two forces that we ll be looking at here are gravit and air resistance. The main issue with these problems is to correctl define conventions and then remember to keep those conventions. B this we mean define which direction will be termed the positive direction and then make sure that all our forces match that convention. This is especiall important for air resistance as this is usuall dependent on the velocit and so the sign of the velocit can and does affect the sign of the air resistance force. Let s take a look at an eample. Eample 4 A 50 kg mass is shot from a cannon straight up with an initial velocit of 0m/s off a bridge that is 00 meters above the ground. If air resistance is given b 5v determine the velocit of the mass when it hits the ground. Solution First, notice that when we sa straight up, we reall mean straight up, but in such a wa that it will miss the bridge on the wa back down. Here is a sketch of the situation. Notice the conventions that we set up for this problem. Since the vast majorit of the motion will be in the downward direction we decided to assume that everthing acting in the downward direction should be positive. Note that we also defined the zero position as the bridge, which makes the ground have a position of 00. Oka, if ou think about it we actuall have two situations here. The initial phase in which the mass is rising in the air and the second phase when the mass is on its wa down. We will need to eamine both situations and set up an IVP for each. We will do this simultaneousl. Here are the forces that are acting on the object on the wa up and on the wa down. 007 Paul Dawkins 67

69 Notice that the air resistance force needs a negative in both cases in order to get the correct sign or direction on the force. When the mass is moving upwards the velocit (and hence v) is negative, et the force must be acting in a downward direction. Therefore, the - must be part of the force to make sure that, overall, the force is positive and hence acting in the downward direction. Likewise, when the mass is moving downward the velocit (and so v) is positive. Therefore, the air resistance must also have a - in order to make sure that it s negative and hence acting in the upward direction. So, the IVP for each of these situations are. Up Down mv = mg- 5v mv = mg-5v v( t ) v 0 =- 0 = 0 In the second IVP, the t 0 is the time when the object is at the highest point and is read to start on the wa down. Note that at this time the velocit would be zero. Also note that the initial condition of the first differential equation will have to be negative since the initial velocit is upward. In this case, the differential equation for both of the situations is identical. This won t alwas happen, but in those cases where it does, we can ignore the second IVP and just let the first govern the whole process. So, let s actuall plug in for the mass and gravit (we ll be using g = 9.8 m/s here). We ll go ahead and divide out the mass while we re at it since we ll need to do that eventuall anwa. 5v v v = 9.8- = 9.8- v( 0) = This is a simple linear differential equation to solve so we ll leave the details to ou. Upon solving we arrive at the following equation for the velocit of the object at an time t. - vt = 98-08e t () 0 Oka, we want the velocit of the ball when it hits the ground. Of course we need to know when it hits the ground before we can ask this. In order to find this we will need to find the position function. This is eas enough to do () () Ú Ú e e s t = v t dt = - dt = t+ + c We can now use the fact that I took the convention that s(0) = 0 to find that c = The position at an time is then. t 0 t 007 Paul Dawkins 68

70 - t 0 () t e s t = To determine when the mass hits the ground we just need to solve. - t , = t+ e - t =- We ve got two solutions here, but since we are starting things at t = 0, the negative is clearl the incorrect value. Therefore the mass hits the ground at t = The velocit of the object upon hitting the ground is then. v ( ) = This last eample gave us an eample of a situation where the two differential equations needed for the problem ended up being identical and so we didn t need the second one after all. Be careful however to not alwas epect this. We could ver easil change this problem so that it required two different differential equations. For instance we could have had a parachute on the mass open at the top of its arc changing its air resistance. This would have completel changed the second differential equation and forced us to use it as well. Or, we could have put a river under the bridge so that before it actuall hit the ground it would have first had to go through some water which would have a different air resistance for that phase necessitating a new differential equation for that portion. Or, we could be reall craz and have both the parachute and the river which would then require three IVP s to be solved before we determined the velocit of the mass before it actuall hits the solid ground. Before leaving this section let s work a couple eamples illustrating the importance of remembering the conventions that ou set up for the positive direction in these problems. Awhile back I gave m students a problem in which a sk diver jumps out of a plane. Most of m students are engineering majors and following the standard convention from most of their engineering classes the defined the positive direction as upward, despite the fact that all the motion in the problem was downward. There is nothing wrong with this assumption, however, because the forgot the convention that up was positive the did not correctl deal with the air resistance which caused them to get the incorrect answer. So, let s take a look at the problem and set up the IVP that will give the sk diver s velocit at an time t. Eample 5 Set up the IVP that will give the velocit of a 60 kg sk diver that jumps out of a plane with no initial velocit and an air resistance of 0.8 v. For this eample assume that the positive direction is upward. Solution Here are the forces that are acting on the sk diver 007 Paul Dawkins 69

71 Because of the conventions the force due to gravit is negative and the force due to air resistance is positive. As set up, these forces have the correct sign and so the IVP is mv =- mg+ 0.8 v v 0 = 0 The problem arises when ou go to remove the absolute value bars. In order to do the problem the do need to be removed. This is where most of the students made their mistake. Because the had forgotten about the convention and the direction of motion the just dropped the absolute value bars to get. mv =- mg+ 0.8v v 0 = 0 incorrect IVP!! So, wh is this incorrect? Well remember that the convention is that positive is upward. However in this case the object is moving downward and so v is negative! Upon dropping the absolute value bars the air resistance became a negative force and hence was acting in the downward direction! To get the correct IVP recall that because v is negative then v = -v. Using this, the air resistance becomes F A = -0.8v and despite appearances this is a positive force since the - cancels out against the velocit (which is negative) to get a positive force. The correct IVP is then Plugging in the mass gives mv =-mg- 0.8v v 0 = 0 v v =-9.8- v( 0) = 0 75 For the sake of completeness the velocit of the sk diver, at least until the parachute opens, which I didn t include in this problem is. - vt = e t () 75 This mistake was made in part because the students were in a hurr and weren t paing attention, but also because the simpl forgot about their convention and the direction of motion! Don t fall into this mistake. Alwas pa attention to our conventions and what is happening in the problems. Just to show ou the difference here is the problem worked b assuming that down is positive. 007 Paul Dawkins 70

72 Eample 6 Set up the IVP that will give the velocit of a 60 kg sk diver that jumps out of a plane with no initial velocit and an air resistance of 0.8 v. For this eample assume that the positive direction is downward. Solution Here are the forces that are acting on the sk diver In this case the force due to gravit is positive since it s a downward force and air resistance is an upward force and so needs to be negative. In this case since the motion is downward the velocit is positive so v = v. The air resistance is then F A = -0.8v. The IVP for this case is mv = mg- 0.8v v 0 = 0 Plugging in the mass gives Solving this gives v v = 9.8- v( 0) = () = - e vt t This is the same solution as the previous eample, ecept that it s got the opposite sign. This is to be epected since the conventions have been switched between the two eamples. 007 Paul Dawkins 7

73 Equilibrium Solutions In the previous section we modeled a population based on the assumption that the growth rate would be a constant. However, in realit this doesn t make much sense. Clearl a population cannot be allowed to grow forever at the same rate. The growth rate of a population needs to depend on the population itself. Once a population reaches a certain point the growth rate will start reduce, often drasticall. A much more realistic model of a population growth is given b the logistic growth equation. Here is the logistic growth equation. P Ê r P ˆ = Á - P Ë K In the logistic growth equation r is the intrinsic growth rate and is the same r as in the last section. In other words, it is the growth rate that will occur in the absence of an limiting factors. K is called either the saturation level or the carring capacit. Now, we claimed that this was a more realistic model for a population. Let s see if that in fact is correct. To allow us to sketch a direction field let s pick a couple of numbers for r and K. We ll use r = and K = 0. For these values the logistics equation is. P P Ê ˆ = Á - P Ë 0 If ou need a refresher on sketching direction fields go back and take a look at that section. First notice that the derivative will be zero at P = 0 and P = 0. Also notice that these are in fact solutions to the differential equation. These two values are called equilibrium solutions since the are constant solutions to the differential equation. We ll leave the rest of the details on sketching the direction field to ou. Here is the direction field as well as a couple of solutions sketched in as well. Note, that we included a small portion of negative P s in here even though the reall don t make an sense for a population problem. The reason for this will be apparent down the road. Also, notice that a population of sa 8 doesn t make all that much sense so let s assume that population is in thousands or millions so that 8 actuall represents 8,000 or 8,000,000 individuals in a population. 007 Paul Dawkins 7