# PPIISol - Solutions to Practice Problems II Written by...

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Solutions to Practice Problems II Written by Victoria Kala [email protected] SH 6432u Office Hours: T 12:45 - 1:45pm Last updated 7/12/2016 Answers This page contains answers only. Detailed solutions are on the following pages. 1. (a) y = e - 2 x ( c 1 cos 3 x + c 2 sin 3 x ) (b) y = c 1 + c 2 e x + c 3 xe x (c) u = c 1 e - 1 2 t + c 2 e 3 t (d) r = c 1 e s + c 2 e - s + c 3 cos s + c 4 sin s 2. W = 2 e 6 x 3. (a) y 2 = x 2 ln x (b) W = x 3 (c) y = c 1 x 2 + c 2 x 2 ln x 4. y 2 = - t 1 / 2 5. (a) y p = At 2 + Bt + C (b) Answers will vary (c) y p = At 2 + Bt + C + Ee - 2 t (d) y p = Ae 3 t sin 4 t + Be 3 t cos 4 t (e) Answers will vary (f) y p = A + ( Bt 2 + Ct + E ) e - t (g) y p = A cos 2 t + B sin 2 t (h) Answers will vary (i) y p = ( At + B ) + ( Ct + E ) sin t + ( Ft + G ) cos t 6. y = c 1 + c 2 e - 2 t + 3 2 t - 1 2 sin 2 t - 1 2 cos 2 t 7. y = 2 e - 2 t + 9 te - 2 t + ( 1 6 t 3 + 3 2 t 2 ) e - 2 t 8. y = c 1 cos t + c 2 sin t - cos t ln | sec t + tan t | . 9. y = c 1 cos(ln t ) + c 2 sin(ln t ) + ln | cos(ln t ) | (cos(ln t )) + ln | t | (sin(ln t )) 1

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Detailed Solutions 1. Find the general solution of the following homogeneous higher-order differential equations: (a) y 00 + 4 y 0 + 7 y = 0 (use x as the independent variable) Solution. The characteristic equation is r 2 + 4 r + 7 = 0. Using the quadratic equation, we see the roots are r = - 4 ± p 4 2 - 4(1)(7) 2 r = - 4 ± i 2 3 2 = - 2 ± i 3 . We have two complex roots, hence the solution to the differential equation is y = e - 2 x ( c 1 cos 3 x + c 2 sin 3 x ) (b) y (3) + 2 y 00 + y 0 = 0 (use x as the independent variable) Solution. The characteristic equation is r 3 + 2 r 2 + r = 0 which factors as r ( r 2 + 2 r + 1) = 0 r ( r + 1) 2 = 0 . Therefore the roots are r = 0 multiplicity 1, r = - 1 multiplicity 2. The solution to the differential equation is therefore y = c 1 e 0 x + c 2 e 1 x + c 3 xe 1 x or y = c 1 + c 2 e x + c 3 xe x . (c) 2 d 2 u dt 2 - 5 du dt - 3 u = 0 Solution. The characteristic equation is 2 r 2 - 5 r - 3 = 0, which factors as (2 r + 1)( r - 3) = 0 . The roots are r = - 1 2 , r = 3 each with multiplicity 1. The solution to the differential equation is therefore u = c 1 e - 1 2 t + c 2 e 3 t . (d) d 4 r ds 4 - r = 0 Solution. The characteristic equation is m 4 - 1 = 0 (we cannot use r as our characteristic equation variable since it is being used in the differential equation). This factors as ( m 2 - 1)( m 2 + 1) = 0 ( m - 1)( m + 1)( m - i )( m + i ) = 0 . The roots are m = 1 , - 1 , ± i . The solution to the differential equation is r = c 1 e s + c 2 e - s + e 0 s ( c 3 cos s + c 4 sin s ) r = c 1 e s + c 2 e - s + c 3 cos s + c 4 sin s. 2. Calculate W ( y 1 , y 2 , y 3 ) where y 1 = e x , y 2 = e 2 x , y 3 = e 3 x . 2
Solution. W ( e x , e 2 x , e 3 x ) = e x e 2 x e 3 x ( e x ) 0 ( e 2 x ) 0 ( e 3 x ) 0 ( e x ) 00 ( e 2 x ) 00 ( e 3 x ) 00 = e x e 2 x e 3 x e x 2 e 2 x 3 e 3 x e x 4 e 2 x 9 e 3 x = e x 2 e 2 x 3 e 3 x 4 e 2 x 9 e 3 x - e x e 2 x e 3 x 4 e 2 x 9 e 3 x + e x e 2 x e 3 x 2 e 2 x 3 e 3 x = e x (18 e 5 x - 12 e 5 x ) - e x (9 e 5 x - 4 e 5 x ) + e x (3 e 5 x - 2 e 3 x ) = e x (6 e 5 x - 5 e 5 x + e 5 x ) = 2 e 6 x 3. (a) The function y 1 = x 2 is a solution of x 2 y 00 - 3 xy 0 + 4 y = 0. Use the method of reduction of order to find a second solution y 2 to the differential equation on the interval (0 , ).

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