Solutions to Homewor 1, Introduction to Differential Equations, 3450:335-003, Dr. Montero, Spring 2009 problem 2. The problem says that the function yx = ce 2x + e x solves the ODE y + 2y = e x, and ass for the value of the constant c for which y4 = 7. To find c we set y4 = ce 8 + e 4 = 7, from where we find c = 7 e 4 e 8. roblem 3. We see two values of for which yx = e x is a solution of y 8y + 12y = 0. The way to solve this problem is to plug in the function yx above into the equation. For this we compute y x = e x and y x = 2 e x. From here we deduce 2 e x 8e x + 12e x = 2 8 + 12e x = 0. 0.1 Since the function e x is never zero, the only way the equation 0.1 will hold is if 2 8 + 12 = 0. This is a quadratic polynomial. Its roots are = 2 and = 6. roblem 5. In this problem we now that a sample of radioactive material decays at a rate proportional to the quantity of material present in the sample. We also now that after 940 days the quantity decays by 29%. We must find the half-life of the material, which is defined as the time it taes a sample to decay to half its initial
amount. To do this we first set up our variables. From the information we have we set t = time in days and At = amount of material in the sample. Since the material decays at a rate proportional to the amount of material, the differential equation for At is da dt = A, where is a constant that we must determine. Before finding, however, we must solve the equation. To do this we notice that the equation for A is both first order linear as well as separable. We use either method to integrate. For example we loo at the equation as separable, and write da A = dt. Next, we integrate this last equation to obtain da A = dt + C where C is a constant of integration. This means that lna = t + C, and solving for A we find At = e t+c = e C e t. We rename our constant as λ = e C, to obtain finally At = λe t. Note here that A0 = λ, so the constant λ represents the amount present in the sample at t = 0, that is, the initial amount. To find, we use the information we have. In this case, after 940 days the sample has decayed by 29%, that is, at t = 940 we have a 71% of the initial amount remaining. Since the initial amount is λ, we have A940 =.71 λ. On the other hand, we now that At = λe t, so we also have A940 = λe 940.
We then must have.71 λ = λe 940. We cancel λ in this last equation to obtain.71 = e 940. This last equation is easy to solve for. The result is = ln.71 940. The first part of the problem ass about the half-life. Since λ is the initial size of the sample, by definition we see the value of t for which At = λ 2, that is we see the value of t at which the size of the sample is half the initial size. Since At = λe t, we set λe t = λ =.5 λ, 2 and solve this last equation for t. Note that we already now the value of. Note also that, again, λ drops out of the equation. Solving for t this time we obtain t = ln.5 = 940 ln.5 ln.71, where I used the value of we found before. In part b, the problem ass how ong will it tae a sample of 100[mg] to decay to 83[mg]. This means that λ = 100 the initial amount and we want to find the value of t for which At = 83. This gives us the equation At = 100e t = 83. Solving for t we obtain t = ln.83 = 940 ln.83. ln.71 roblem 6. In this problem we are given the ODE d dt = 6 7, 700
along with the initial condition 0 = 2. The first two questions as us to determine the range of in which t is increasing and decreasing respectively. To do this we loo at the expression on righthand side of the equation and determine when is this expression positive and negative. The expression we have is It is easy to notice that f = 7. f > 0 for 0, 7 andf < 0 for > 7. We conclude that t is increasing when 0 < < 7 and t is decreasing when > 7. For problem 6, part C, we need to solve the differential equation d dt = 6 7. 700 We notice that this is a separable, first order equation. To solve it, we write d 7 = 6 700 dt, and then integrate. To integrate the left-hand side we need to use partial fractions, or just loo at the problem and notice that 1 7 = 1 7 1 + 1 7. This means that we need to integrate 1 7 1 + 1 7 d = 6 700 dt. Integrating, we obtain ln = 6 7 100 t + C = 3 50 t + C. Exponentiating this equation we obtain 7 = e 3t 50 e C. Again, we rename our constant as λ = e C, so we get 7 = λe 3t 50.
To determine the value of λ we recall that 0 = 2, so 2 7 2 = 2 5 = λ. We have so far that 7 = 2 5 e 3t 50. Finally, we solve this last equation for. We do this as follows: 1 + 7 7 = 7 = 2 5 e 3t 50, so that 7 7 = 7 = 1 + 2 5 e 3t 5 + 2e 3t 50 50 =. 5 From here we obtain 35 = 7, 5 + 2e 3t 50 so 35 t = 7. 5 + 2e 3t 50 This gives us t for any t. For example 35 2 = 7 5 + 2e 6 50 = 7 35 5 + 2e 3 25. roblem 7. In this problem we consider the ODE d K dt = c ln. This equation models the growth of a limited population and is called Gompertz equation. Assuming, for example, c =.05, K = 5000, and that the initial population is 0 = 600, we are ased first to solve the ODE. To do this we start by writing which is the same as d ln = c dt, K d lnk ln = c dt, and then integrate. This gives you the following result: ln lnk ln = c t + L,
where L is a constant of integration. At this point we need two things: first, we need to solve for as a function of t. Second, we need to determine the value of the constant L. Let us solve for t first. We exponentiate the equation to obtain ln lnk ln = c t + L 1 lnk ln = ec t e L. As we often do, I will rename the constants according to λ = e L. Hence we have so far From here we can now solve for : lnk ln = λ e c t. ln = lnk λ e c t, so t = K. e λe ct To determine the value of λ we can use this last equation, but it is actually easier to recall that ln = lnk λ e c t. Since we now that c =.05, K = 5000, and 0 = 600, the value of λ comes directly from here: 3 λ = ln600 ln5000 = ln. 25 This determines t completely. For part b we need to find the limit For this we go bac to the solution lim t. t t = K e λe ct. Note that, as t, we have e ct 0, so that e λe ct 1. We conclude that lim t = K = 5000. t Finally we need to determine the value of at which the function t is growing fastest. For this we notice that d K dt = c ln.
means that the value of d dt is given by the right-hand side of this equation. Hence will grow fastest when the right-hand side of this last equation is greatest. Since c =.05 > 0, what we need then is to find the maximum value of the function K ψ = ln = lnk ln. This is of course standard. We compute ψ and set it equal to zero. We obtain K ψ = ln 1 = 0, which means t is growing fastest when = K e. roblem 8. In this problem we have an object of mass m = 5 [g] is thrown up with an initial speed of v 0 = 90 [m/s]. The gravity of the place is g = 9.8 [m/s 2, and there is an air resistance given by the constant. To find a formula for the velocity v as a function of t, we recall Newton s second Law, that says that mass times acceleration equals the sum of all forces acting on an object. Now, the acceleration of an object is the derivative of its velocity, so our equation becomes mv = mg v. This is a first order linear equation, but it is also separable. We can choose either method to solve it. I will thin of it as a separable equation. So, I write the equation as from where I get This is easy to integrate: dv dt = m v + g = m dv v + mg = m dt. v + mg ln v + mg = m t + C, where C is a constant of integration. Solving for v from here is a piece of cae: vt = mg + t e m e C. As always, we rename our constant as λ = e C. This means we obtain vt = mg + t λe m.,
Since we now that v0 = v 0 = 90, we obtain which means λ = v 0 + mg vt = mg + mg + λ = v 0,. We finally arrive at v 0 + mg e t m = v0 e t mg m 1 e t m and we now all the constants that appear in the last expression for vt, so this determines vt completely. The second part of this question ass what happens to vt as 0. In other words, we need to compute lim vt. 0 Since vt is the sum of two terms, we can find the limits of these separately and then add the results. First, it is clear that Next, we need to find mg lim 0 lim v 0e t m = v0. 0 1 e t m m = g lim 0 1 e t m Setting x =, we notice that as 0 we have x 0. Hence we need to find the m limit 1 e xt lim. x 0 x Here we notice that this limit is of the form 0 so we use l Hôpital s rule to obtain 0 1 e xt lim x 0 x = t 1 = t. utting this all together we arrive at the conclusion that lim vt = v 0 gt. 0., roblem 9. First we declare our variables: T = temperature of the object. T m = temperature of the room. t = time. Now we now that the temperature of the object changes at a rate proportional to the difference between the temperature of the object and the temperature of the surroundings. Letting be the proportionality constant we obtain dt dt = T T m.
This is both a first order linear equation and a separable equation. In my version of this problem I have that T m = 78, so to solve this equation, for example, we write and then integrate. This gives us dt T 78 = dt, lnt 78 = t + C, so T = 78 + e t+c = 78 + e C e t. It is easier, as we have done several times in class, to set λ = e C, so the solution to our ODE now loos as follows: Tt = 78 + λe t. We need now to find the values of the constants λ and. To do this we use information given in the problem. My version of the problem says that T0 = 195, and also T3 = 180. The first of these data tells us that T0 = 78 + λ = 195, so λ = 117. On the other hand, the second datum says T3 = 180 = 78 + 117e 3, so we can solve for to obtain ln 102 117 =. 3 Hence we now the function Tt completely. Finally my version of the problem ass me to find the value of t for which Tt = 150. To find t I set up the equation 78 + 117e t = 150 and solve for t. This last equation yields t = ln 72 117. Since we actually now the value of, this solves the problem.
roblem 10. This problem says that a bacteria culture doubles its population every 20[min], and that the initial population is 4 bugs. Then it ass to find the population of the culture after t hours, the population after exactly 8 hours, and determine the time it taes the culture to have 32 bugs. To answer these questions, as always, we first we set up our variables. It seems natural to let t be time in hours and At be the number of bugs present at time t. Since the rate of change of the population is proportional to its size, the differential equation for A is A t = At, or A A = 0, where is a constant that we must find. To solve this equation we multiply the equation by e R pt dt = e t to obtain d dt This means that the quantity e t At is constant, so e R pt dt At = d e t At = 0. dt At = A 0 e t. A 0 in this last equation is the population at time t = 0, in this case 4 bugs. Hence At = 4e t. To find we use the fact that the population doubles every 20[min]. Since at t = 0 the population is 4 bugs, 20[min] later, that is, at time t = 1 hours the population 3 will be 8 bugs. Hence 1 A = 4e 3 = 8. 3 Solving this last equation for we find that = 3 ln2 = ln8. This means that the solution to the equation we see is At = 4e 3ln2t = 4e ln8t = 4 8 t. From here we obtain A8 = 4 8 8 = 2 26 bugs. Finally we want to now the value of t for which At = 32. We can just reason this out as follows: if initially we have 4 bugs, and the population doubles every 20[mins], after 20 minutes we will have 8 bugs, after 40 we will have 16 and after an hour we will have 32 bugs. Since we measure time in hours, the answer is t = 1. We could also set up the equation 32 = At = 4 8 t or 32 = 4 8 t,
and solve for t. The solution to this last equation is clearly t = 1, which agrees with our previous reasoning. roblem 11. This problem ass to solve the differential equation subject to the initial condition y0 = 7. 11x 4y x 2 + 1 dy dx = 0, Solution: First we notice this is a separable equation. We re-write it as Then we integrate y dy dx = 11 4 y dy = 11 4 x x2 + 1. x dx + C. x2 + 1 For the integral on the left it is convenient to use the substitution u = x 2 + 1, and du = 2xdx. Hence x x2 + 1 dx = 1 2 2x x2 + 1 dx = 1 2 1 u du = u = x 2 + 1. This means that y 2 2 = 11 x2 + 1 + C, 4 where C is a constant of integration. To find C we use the fact that y0 = 7 to obtain Our solution is then 49 2 = 11 4 + C so C = 87 4. y = 2 11 x2 + 1 + 87. 4 4 roblem 12. Here we want to solve with y0 = 8. 4 dy dt + y = 24t, Solution: This equation, and those in the remaining two problems, is a first order linear equation. The general method to solve these equations goes lie this: first we consider the equation y + pty = ft,
and multiply the whole thing by e R pt dt. This maes the left hand side into a perfect derivative: d e R pt dt y = e R pt dt y + pte R pt dt y = e R pt dt ft, dt so we obtain d e R pt dt y = e R pt dt ft. dt To solve this equation for y we must integrate. In roblem 12 we have which we re-write as This means that pt = 1 4, so 4 dy dt + y = 24t, dy dt + y 4 = 6t. pt dt = t 4 and then er pt dt = e t 4. Note that in this integral we do not need a constant of integration. Once we have this we write and integrate to obtain d e t 4 y = 6te t 4, dt e t 4 y = 6 te t 4 dt + C. Note that we do need a constant of integration here. To compute the integral te t 4 dt we go by parts. Let u = t, so du = dt, and dv = e4, t so v = 4e4. t Then te t 4 dt = uv v du = 4te t 4 4 e t t t 4 dt = 4te 4 16e 4. So far, then, we have e t 4 y = 6 4te t t 4 16e 4 + C = 32 e t 4 t 4 + C. From here we obtain yt = 24 t 4 + C e t 4.
To determine the constant t we use the condition y0 = 8, which gives us 8 = 24 4 + C, so C = 104. The solution to our problem is yt = 24 t 4 + 104 e t 4. roblem 13. Here we consider the equation and the condition y0 = 1. dy dt 2ty = 3t2 e t2, Solution: The equation is already in the form y + pty = ft. Here pt = 2t so e R pt dt = e t2. This allows us to write Integrating this we obtain d e t2 y = 3t 2. dt e t2 y = t 3 + C, where C is a constant of integration. We have so far yt = t 3 + Ce t2. We use finally the condition y0 = 1 to obtain y0 = C = 1 so C = 1. The solution to the problem is yt = t 3 1e t2. roblem 14. Our last problem is to solve the ODE with the initial condition y0 = 10. 9t + 1 dy dt 8y = 8t,
Solution: Again, we first write the equation in the form y + pty = ft. This means Hence dy dt 8 9t + 1 y = 8t 9t + 1. pt = 8 9t + 1 so e R pt dt 1 =. 1 + t 8 9 The equation then becomes d y 8t =. dt 1 + t 8 9 91 + t 17 9 Here we need to find the integral 8t 91 + t 17 9 dt = 8 9 t 1 + t 17 9 dt. For the last integral we could go by parts, but we can also notice that t 1 + t 1 dt = dt 1 + t 17 9 1 + t 17 9 1 1 = dt dt 1 + t 8 9 1 + t 17 9 = 91 + t 1 9 9 +. 81 + t 8 9 From this we obtain y 1 + t 8 9 = 8 9 91 + t 1 9 + 9 81 + t 8 9 + C, so yt = 81 + t + 1 + C1 + t 8 9. We finally use the condition y0 = 10 to obtain 10 = 8 + 1 + C, so C = 1. The solution we see is yt = 81 + t + 1 + 1 + t 8 9.