The homogeneous solution satisfies y c t c 1 e t c 2 e 2 t Neither solution

The homogeneous solution satisfies y c t c 1 e t c 2

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The homogeneous solution satisfies: y c ( t ) = c 1 e - t + c 2 e - 2 t . Neither solution matches the forcing function, so try a general cubic polynomial: y p ( t ) = At 3 + Bt 2 + Ct + D. Differentiating: y 0 p ( t ) = 3 At 2 + 2 Bt + C and y 00 p ( t ) = 6 At + 2 B. In the original differential equation, we find: (6 At + 2 B ) + 3(3 At 2 + 2 Bt + C ) + 2( At 3 + Bt 2 + Ct + D ) = 8 t 3 . We collect the coefficients of the powers of t : t 3 : 2 A = 8 , or A = 4 , t 2 : 9 A + 2 B = 0 , or B = - 18 , t 1 : 6 A + 6 B + 2 C = 0 , or C = 42 , t 0 : 2 B + 3 C + 2 D = 0 , or D = - 45 . This gives the particular solution: y p ( t ) = 4 t 3 - 18 t 2 + 42 t - 45 . The general solution is: y ( t ) = c 1 e - t + c 2 e - 2 t + 4 t 3 - 18 t 2 + 42 t - 45 . 21. (3 pts) Consider the differential equation given by: x 2 y 00 - 7 xy 0 + 16 y = x 2
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First, we consider the homogeneous case (Euler-Cauchy equation): x 2 y 00 - 7 xy 0 + 16 y = 0 Let y ( x ) = x r , then y 0 ( t ) = rx r - 1 and y 00 ( x ) = r ( r - 1) x r - 2 , so: x 2 ( r ( r - 1) x r - 2 ) - 7 x ( rx r - 1 ) + 16 x r = x r ( r ( r - 1) - 7 r + 16) = 0 . This gives the auxiliary equation: r 2 - 8 r + 16 = ( r - 4) 2 = 0 . This gives equal roots with r = 4. So, the general solution to the homogeneous equation is: y c ( x ) = c 1 x 4 + c 2 x 4 ln( x ) . This problem requires the variation of parameters method, so we compute the Wronskian with the homogeneous solutions y 1 ( x ) = x 4 and y 2 ( x ) = x 4 ln( x ): W [ x 4 , x 4 ln( x )]( x ) = det x 4 x 4 ln( x ) 4 x 3 x 3 (4 ln( x ) + 1) = x 7 To use variation of parameters, we rewrite our differential equation: y 00 - 7 y 0 x + 16
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