MATH 3310 Class Notes 2

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1 MATH 330 Class Notes 2 S. F. Ellermeyer August 2, 200 The differential equation = ky () (where k is a given constant) is extremely important in applications and in the general theory of differential equations. Since we will encounter equation () repeatedly throughout this course, we should become comfortable with its family of solutions. We can find this family of solutions by using the method of separation of variables. Solvingthedifferential equation () will be our first illustration of this method. General Solution of / = ky Our first observation about the differential equation () is that the constant function y =0is a solution. This is easily checked, for if y (t) =0for all t then = d (0) = 0 = k 0=ky. Next, we would like to find all solutions, y, of the differential equation () such that y (t) 6= 0for any t. Ify is a solution of () and y (t) 6= 0for any t, then we can divide both sides of () by y to obtain y = k. (2) Next, we integrate both sides of equation (2) with respect to the independent variable t to obtain µ y = k. (3)

2 To evaluate the indefinite integral on the left hand side of equation (3), we make the substitution u = y. This substitution implies that du =. We thus have µ y = du =ln u + A =ln y + A (4) u (where A is an arbitrary constant). Evaluation of the integral on the right hand side of equation (3) is straightforward. We obtain k= kt + B (5) (where B is an arbitrary constant). Substituting the results of equations (4) and (5) into equation (3), we obtain ln y + A = kt + B or ln y = kt +(B A). Since A and B are both arbitrary constants, we can simply set C = B A and write ln y = kt + C (6) (where C is an arbitrary constant). The only thing that remains is to solve the equation (6) for y. Totothis, we exponentiate both sides of equation (6) to obtain e ln y = e (kt+c) which gives us y = e (kt+c). Using a property of exponents, we can write the above equation as y = e C e kt or, by setting D = e C, we can simply write y = De kt. 2

3 Since C is an arbitrary constant, D = e C is an arbitrary positive constant. (Recall that e C is always positive, no matter what the sign of C.) Finally, recalling that p = q implies that p = ±q, weobtain y = ±De kt. (7) Since D is an arbitrary positive constant, we can set E = ±D and rewrite equation (7) as y = Ee kt (8) where E is an arbitrary positive or negative constant. To check that each member of the family of functions (8) is a solution of the differential equation (), we note that for y = Ee kt,wehave = d Ee kt = Eke kt = kee kt = ky. Finally,notethatifE =0, then equation (8) simply reduces to y =0,which we know is also a solution of the differential equation (). Our conclusion is that the family of solutions of the differential equation () is where C is an arbitrary constant. y = Ce kt Example Find the family of solutions of the differential equation =3y. (9) and draw a picture of some members of this family. Solution: By the work done above, the family of solutions of the differential equation (9) is y = Ce 3t (where C is an arbitrary constant). Some members of this family are pictured in Figure. Example 2 Find the family of solutions of the differential equation = 2y. (0) 3

4 4 3 y t Figure : Family of Solutions of / =3y and draw a picture of some members of this family. Solution: By the work done above, the family of solutions of the differential equation (0) is y = Ce 2t (where C is an arbitrary constant). Some members of this family are pictured in Figure y t Figure 2: Family of Solutions of / = 2y In-class Exercise : Find the family of solutions of the differential 4

5 equation =0.5y and sketch a picture of this family. In-class Exercise 2: Find the family of solutions of the differential equation = 5y and sketch a picture of this family. Remark 3 If k is a positive constant, then lim t e kt = and lim t e kt = 0. Thus,ifk>0, then the family of solutions of / = ky will be similar (qualitatively) to the family shown in Figure of Example ; whereas, if k is a negative constant, then lim t e kt =0and lim t e kt =. Thus,ifk<0, then the family of solutions of / = ky will be similar (qualitatively) to thefamilyshowninfigure2ofexample2. What about the case k =0? This question is addressed in the exercises.. Solution of Initial Value Problem / = ky, y (t 0 )= y 0 Since we know that the general solution of = ky is y = Ce kt, we can easily solve initial value problems for / = ky. Example 4 Find the solution of the initial value problem =3y y (0) = 0.5. Solution: Every solution of the differential equation / =3y has the form y = Ce 3t. In order that the initial condition y (0) = 0.5 also be satisfied, we must choose the constant C such that Ce 3 0 =0.5; i.e., we must choose C =0.5. Thus, the solution of the above initial value problem is y =0.5e 3t. This solution is pictured in Figure 3. 5

6 4 3 y t Figure 3: Solution of / =3y, y (0) = 0.5 Example 5 Find the particular solution of the differential equation = 2y that satisfies the condition y ( ) = 2. Solution: The general solution of / = 2y is y = Ce 2t. In order to satisfy the condition y ( ) = 2, we must choose C such that Ce 2( ) =2. Solving Ce 2 =2for C gives us C =2e 2. Thus, the particular solution we are seeking is y =2e 2 e 2t which can also be written (by using a property of exponents) as or as y =2e 2t 2 y =2e 2(t+). This particular solution of / = 2y is pictured in Figure 4. 6

7 4 3 y t Figure 4: Solution of / = 2y, y ( ) = 2 In-class Exercise 3: Find the solution of the initial value problem =2y y () = and sketch its graph. To be general, let us solve the initial value problem = ky () y (t 0 )=y 0 (2) where k, t 0,an 0 are given constants: The general solution of the differential equation () is y = Ce kt and satisfaction of condition (2) requires that Ce kt 0 = y 0. Solving for C gives C = y 0 e kt 0. Thus, the solution of the initial value problem (,2) is y = y 0 e kt 0 e kt 7

8 which can also be written as y = y 0 e k(t t 0). (3) Example 6 Use the formula (3) to solve the initial value problem = 2y y ( ) = 2 Solution: Using formula (3), since k = 2, t 0 =, an 0 =2,we obtain y =2e 2(t ( )) =2e 2(t+). Exercises. If k =0, then the differential equation () simply reduces to =0. (4) Find the family of solutions of this differential equation and sketch a picture of this family. 2. Suppose that k = Then it would seem that the family of solutions of the differential equation = ky (5) should resemble the family of solutions of the differential equation (4) of exercise (because k is very close to 0). On the other hand, since k>0, the family of solutions of the differential equation (5) should also resemble the family of solutions pictured in Figure (because of Remark 3). Draw a picture of the family of solutions of (5) and explain why both of the above mentioned resemblances are in fact true. 3. Suppose that k = Draw a picture of the family of solutions of = ky. 8

9 4. Find the solution of the initial value problem =4y y (0) = 2 and sketch the graph of this solution. 5. Find the solution of the initial value problem =4y y (2) = 0 and sketch the graph of this solution. 6. Find the solution of the initial value problem = 4y y () = 2 and sketch the graph of this solution. 7. Find the solution of the initial value problem =0 y (0) = 2 and sketch the graph of this solution. 8. Find the solution of the initial value problem = 3y y (2) = 2 and sketch the graph of this solution. 9

10 2 Variable Separable Differential Equations A variable separable (also called separable) differential equation is one which has the form = g (t) h (y). (6) Some examples of separable differential equations are: = t (7) = t2 y 3 (8) = t2 y t 2 + y (9) = 5y. (20) Equation (7) is a separable differential equation because we can set g (t) =t and h (y) = and observe that equation (7) has the form of equation (6). Equation (8) is separable because it has the form of equation (6) with g (t) = t 2 and h (y) = y 3. That equation (9) is separable is not as immediately obvious. We need to recognize that t 2 y t 2 + y = (t 2 +)(y ) which means that equation (9) can be written as = t 2 + (y ). This shows that equation (9) is separable with g (t) =t 2 + and h (y) =y. Finally, equation (20) is separable because it has the form of equation (6) with g (t) = 5 and h (y) =y or with g (t) =and h (y) = 5y. (Note that thechoiceofg (t) and h (y) is not unique in any of the above examples.) An example of a differential equation that is not separable is = t2 + y 3 because there is no way to write t 2 + y 3 in the form g (t) h (y). 2. Solving Separable Differential Equations The method used in solving the separable differential equation () can be attempted in trying to solve any separable differential equation. The procedure is outlined below. 0

11 Procedure for Solving Separable Differential Equations. Formally separate the variables y and t by rewriting the differential equation (6) as = g (t). (2) h (y) 2. Integrate both sides of the above equation to obtain h (y) = g (t). (22) 3. If both of the integrals in step 2 can be computed using integration techniques from calculus, then an algebraic equation in y and t is obtained. (This equation will also contain an arbitrary integration constant because the integrals are indefinite integrals.) One can then attempt to solve this algebraic equation for y and hence express y as a function of t. We now illustrate this procedure with an example. Example 7 Find the general solution of the differential equation = y 3. (23) Solution: First, we note that the differential equation (23) is separable. (It has the form of equation (6) with g (t) =and h (y) =/ (y 3). Separation of variables (step of the outlined procedure) gives us (y 3) = and integration of both sides of this equation (step 2 of the procedure) gives us (y 3) =. Both of these integrals are easily evaluated: (y 3) = 2 y2 3y + A

12 and = t + B. This gives us 2 y2 3y + A = t + B which can be written as 2 y2 3y = t + C (where C is an arbitrary constant). Now, we must solve this equation for y in terms of t (step 3 of the procedure). To do this, we rewrite the above equation as 2 y2 3y (t + C) =0, multiply both sides by 2 (in order to clear away fractions) to obtain y 2 6y (2t + D) =0 and then observe that this equation is a quadratic equation in y. Using the quadratic formula, we obtain y = 6 ± p (2t + D) 2 which can be written more simply as 8t + E y =3± 2 or even more simply as y =3± 2t + F (24) (where F is an arbitrary constant). The family of functions (24) is the general solution of the differential equation (23). Let us check, for example, that the function y =3 2t +57is a solution of the differential equation (23): Setting y =3 (2t +57) /2,weobtain = 2 (2t +57) /2 2= (2t + 57) /2 and y 3 = ³3 /2 = (2t +57) /2. (2t +57) 3 This shows that for y =3 2t +57,wehave/ =/ (y 3). 2

13 2.2 A Formal Justification of the Method of Separation of Variables In using the method of separation of variables, we treat the symbol / as though it is a fraction (in multiplying both sides of equation (6) by ). In Calculus, you learned that / denotes the derivative of the function y with respect to its independent variable t and that the symbol / is not really a fraction but a single entity. Hence, it apparently makes no sense (and is even illegal!) to multiply by. Nonetheless,theexampleswehave seen so far seem to indicate that the method of separation of variables works! Here, we show what is really going on with this method: If the function y = y (t) is a solution of the differential equation = g (t) h (y), then = g (t) h (y (t)) for all values of t in some interval. If h (y (t)) 6= 0for any values of t in this interval, we divide both sides of the above equation by h (y (t)) to obtain h (y (t)) = g (t) and then integrate both sides with respect to t to obtain µ h (y (t)) = g (t). For the integral on the left hand side of the above equation, we make the substitution and this gives us u = y du = h (u) du = g (t) 3

14 which, since u is a dummy variable, we can write as h (y) = g (t). This reasoning shows that equation (22) of the outlined procedure for solving separable differential equation is in fact correct. The preceding equation (2) in step of that procedure is just a shortcut (although not formally correct) that allows us to bypass all of the reasoning supplied in the more rigorous justification given above. Exercises. Explain why any differential equation of the form = f (t) (where the right hand side depends only on the independent variable t) isseparable. 2. Explain why any differential equation of the form = f (y) (where the right hand side depends only on the dependent variable y) is separable. 3. Which of the following differential equations are separable? (a) / =cost (b) / = p t 2 + y 2 (c) / =4 (d) / = ty + y (e) / = ty 4. Find the general solutions of the following differential equations (all of which are separable). 4

15 (a) / = ty (b) / =(y +)/ (t +) (c) / =2y + (d) / =/ (2y +) (e) / = y ( y) Hint: Note that y ( y) = y + y. 5. Find the solutions of the following initial value problems. (a) / = ty, y () = 2 (b) / =(y +)/ (t +), y (0) = 0 (c) / =2y +, y (0) = /2 (Warning: This one is easy but tricky!) (d) / =/ (2y +), y (0) = (e) / = y ( y), y (0) = /2 3 Analytic, Qualitative, and Numerical Techniques The method of separation of variables which we have been considering here is an example of an analytic technique for stuing solutions of differential equations. It involves using calculus to actually find explicit formulas for solutions. Often, it is not possible to find explicit formulas for solutions. Even for separable differential equations, it may be impossible to evaluate the integrals which are involved or impossible to solve the algebraic equations which result from the integrations. Two other approaches to stuing differential equations are the qualitative and numerical approaches. These approaches can often yield useful information about the solutions of differential equations in cases where the solutions cannot be obtained explicitly (as a formula). We will learn about qualitative and numerical approaches in what lies ahead. 5