A First Course in Elementary Differential Equations: Problems and Solutions. Marcel B. Finan Arkansas Tech University c All Rights Reserved

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1 A First Course in Elementary Differential Equations: Problems and Solutions Marcel B. Finan Arkansas Tech University c All Rights Reserved 1

2 4 The Method of Variation of Parameters Problem 4.1 Solve y + y = sec t by variation of parameters. The characteristic equation r + 1 = 0 has roots r = ±i and y h (t) =c 1 cos t + c sin t Also, y 1 (t) = cos t and y (t) = sin t so that W (t) = cos t + sin t =1. Now, d(cos t) u 1 = sin t sec t = =ln cos t cos t and u = cos t sec t = Hence, the particular solution is given by and the general solution is = t y p (t) =ln cos t cos t + t sin t y(t) =c 1 cos t + c sin t +ln cos t cos t + t sin t Problem 4. Solve y y = e t by undetermined coefficients and by variation of parameters. Explain any differences in the answers. The characteristic equation r 1 = 0 for y y = 0 has roots r = ±1. The homogeneous solution is y h (t) =c 1 e t + c e t. Undetermined Coefficients Summary. The basic trial solution method gives initial trial solution y p (t) =d 1 te t since 1 is a root of the characteristic equation. Substitution into y y = e t gives d 1 e t +d 1 te t d 1 te t = e t. Cancel e t and equate coefficients of like powers of t to find d 1 =1/. Then y p = tet. Variation of Parameters Summary. The homogeneous solution y h (t) = 5

3 c 1 e t + c e t found above implies y 1 = e t, y = e t is a suitable independent pair of solutions. Their Wronskian is W =. The variation of parameters formula applies: e y p (t) =e t t e et e t t et Integration, followed by setting all constants of integration to zero, gives y p (t) = tet et. 4 Differences. The two methods give respectively y p (t) = tet and y p(t) = tet e t. The solutions y 4 p(t) = tet and y p(t) = tet et differ by the homogeneous 4 solution et. In both cases, the general solution is 4 y(t) =c 1 e t + c e t + 1 tet because terms of the homogeneous solution can be absorbed into the arbitrary constants c 1,c Problem 4.3 Solve the following nd order equation using the variation of parameters method: y +4y = t + 8 cos t. The characterisitc equation r +4=0hasrootsr = ±i so that y h (t) = c 1 cos t + c sin t. Hence, y 1 (t) =cost, y (t) = sin t, and W (t) =. Thus, sin t(t + 8 cos t) cos t(t + 8 cos t) y p = cos t + sin t = cos t( 1 4 t sin t cos t 1 4 t cos t cos t) + sin t( 1 4 t cos t 1 8 sin t t sin t +t + 1 sin 4t) = t + cos t cos t +tsin t + 1 sin 4t sin t The general solution is y(t) =c 1 cos t + c sin t t +t sin t 6

4 Problem 4.4 Find a particular solution by the variation of parameters to the equation y +y + y = e t ln t. The characteristic equation r +r +1=0 has roots r 1 = r = 1, so the fundamental solutions of the reduced equation are y 1 (t) =e t,y (t) =te t Compute the Wronskian. W (t) = e t te t e t e t te t =e t (e t te t )+e t te t Compute u 1 (t). =e t te t + te t =e t y (t)g(t) u 1 (t) = W (t) te t e t ln t = e t = t ln t = t t ln t + 1 t = t ln t + t 4 Compute u (t). y1 (t)g(t) u (t) = W (t) e t e t ln t = e t = ln t = t ln t =t ln t t t 1 t 7

5 Note. We used integration by parts to compute the integrals t ln t and ln t. The particular solution to our complete equation is y p (t) =u 1 (t)y 1 (t)+u (t)y (t) ( ) = t ln t + t e t +(tln t t)te t 4 = t ln te t 3t 4 e t =( t ln t 3t 4 )e t Problem 4.5 Solve the following initial value problem by using variation of parameters: y +y 3y = te t,y(0) = 1 64,y (0) = From the characteristic equation, we obtain y 1 (t) = e t,y (t) = e 3t and W (t) = 4e t. Integration then yields e 3t te t u 1 (t) = 4e = t t 8 e t te t u (t) = 4e = 1 t 16 te4t + e4t 64 Thus. y p (t) = et 64 (8t 4t + 1) and the general solution is Initial conditions: y(t) =c 1 e t + c e 3t + t 8 et 1 16 tet y(0) =c 1 + c = 1 64 y (0) =c 1 3c 4 64 = These are satisfied by c 1 = 15 and c 64 = 1. Finally the solution to the initial 4 value problem is y = et 64 (8t 4t + 15) 1 4 e 3t 8

6 Problem 4.6 (a) Verify that {e t,e t } is a fundamental set for the equation 4ty +y y =0 on the interval (0, ). You may assume that the given functions are solutions to the equation. (b) Use the method of variation of parameters to find one solution to the equation 4ty +y y =4 te t. (a) Usually the first thing to do would be to check that y 1 (t) =e t and y (t) =e t really are solutions to the equation. However, the question says that this can be assumed and so we move on to the next step, which is to check that the Wronskian of the two solutions is non-zero on (0, ). We have y 1 = 1 t e t and y = 1 t e t and so W (t) =y 1 y y 1y = 1 t 1 t = 1 t This is indeed non-zero and so {e t,e t } is a fundamental set for the homogeneous equation. (b) The variation of parameters formula says that y = y 1 y g W (t) + y y1 g W (t) is a solution to the nonhomogeneous equation in the form y + py + qy = g. To get the right g, we have to divide the equation through by 4t and so g = 1 t e t. Thus y = e t e t ( 1 t )e t 1/ + e t e t ( 1 t )e t t 1/ t =e t e t e t =te t e t e t 9

7 To evaluate the integral, we substitute u = t so that = 1 udu. We get e t = 1 ue u du = 1 (u 1)eu =( t 1/)e t. Thus y =(t t +1/)e t is one solution to the equation. You might notice that the 1/ can be dropped (because (1/)e t is a solution to the homogeneous equation) so that would also work y =(t t)e t Problem 4.7 Use the method of variation of parameters to find the general solution to the equation y + y = sin t. The characteristic equation r + 1 = 0 has roots r = ±i so that the solution to the homogeneous equation is y h (t) =c 1 cos t + c sin t. The Wronskian is W (cos t, sin t) =1. Now u 1(t) = sin cos (t) 1 t =. Hence u 1 (t) = 1 ( 1 sin(t) t). Similarly, u (t) = sin t cos t. Hence u (t) = 1 sin t. So y p (t) = 1t cos t + 1 sin t. The general solution is given by Problem 4.8 Consider the differential equation y(t) =c 1 cos t + c sin t 1 t cos t t y +3ty 3y =0,t>0. (a) Determine r so that y = t r is a solution. (b) Use (a) to find a fundamental set of solutions. (c) Use the method of variation of parameters for finding a particular solution to t y +3ty 3y = 1 t 3,t>0. 30

8 (a) Inserting y, y, and y into the equation we find r +r 3=0. Solving for r to obtain r 1 = 1 and r = 3. (b) Let y 1 (t) =t and y (t) =t 3. Since W (t) = t t 3 1 3t 4 = 4t 3 {y 1,y } is a fundamental set of solutions for t>0. (c) Recall that the variation of parameters formula states that if y 1 and y form a fundamental solution set for y + p(t)y + q(t)y =0, then y p (t) = u 1 (t)y 1 (t)+u (t)y (t) is a particular solution to the equation y + p(t)y + q(t)y = g(t), where t 3 t 5 u 1 (t) = 4t = t 4 Thus, u (t) = t t 5 4t 3 = 1 4 ln t y p (t) = 1 16 t t 3 ln t Problem 4.9 Use the method of variation of parameters to find the general solution to the differential equations y + y = sin t. The characterisitc equation r + 1 = 0 has roots r = ±i so that y 1 (t) = cos t, y (t) = sin t, and W (t) =1. Hence, u 1 (t) = sin t sin t = (1 cos t)d(cos t) =cost 1 3 cos3 t u (t) = cos t sin t = 1 3 sin3 t Thus, and y p (t) = cos t 1 3 cos4 t sin4 t y(t) =c 1 cos t + c sin t + cos t 1 3 cos4 t sin4 t 31