The particular solution to the differential equation 4 y ″ + 11 y ′ − 3 y = − 2 t e − 3 t . The particular solution to the differential equation 4 y ″ + 11 y ′ − 3 y = − 2 t e − 3 t is, y ( t ) = ( t 2 13 + 8 t 169 ) e − 3 t _ . Procedure used: For finding a particular solution to the differential equation a y ″ + b y ′ + c y = C t m e r t , where m is a nonnegative integer, use the form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t with: (i) s = 0 , if r is not a root of the associated auxiliary equation; (ii) s = 1 , if r is a simple root of the associated auxiliary equation; (iii) s = 2 , if r is a double root of the associated auxiliary equation. Calculation: The given differential equation is 4 y ″ + 11 y ′ − 3 y = − 2 t e − 3 t . Comparing 4 y ″ + 11 y ′ − 3 y = − 2 t e − 3 t with a y ″ + b y ′ + c y = C t m e r t , it is observed that m = 1 and r = − 3 . The associated auxiliary equation is 4 r 2 + 11 r − 3 = 0 . Simplify 4 r 2 + 11 r − 3 = 0 as, 4 r 2 − r + 12 r − 3 = 0 r ( 4 r − 1 ) + 3 ( 4 r − 1 ) = 0 ( r + 3 ) ( 4 r − 1 ) = 0 r + 3 = 0 , 4 r − 1 = 0 r = − 3 , r = 1 4 That is, the roots are r 1 = − 3 and r 2 = 1 4 . Since r = − 3 is a root of the associated auxiliary equation, s = 1 . The form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t with m = 1 , s = 1 and r = − 3 will be, y p ( t ) = t 1 ( A 1 t + A 0 ) e − 3 t = ( A 1 t 2 + A 0 t ) e − 3 t Differentiate y p ( t ) = ( A 1 t 2 + A 0 t ) e − 3 t with respect to t as, y ′ p ( t ) = ( A 1 t 2 + A 0 t ) ( − 3 e − 3 t ) + ( 2 A 1 t + A 0 ) e − 3 t = ( − 3 A 1 t 2 − 3 A 0 t + 2 A 1 t + A 0 ) e − 3 t Differentiate y ′ p ( t ) = ( − 3 A 1 t 2 − 3 A 0 t + 2 A 1 t + A 0 ) e − 3 t with respect to t as, y ″ p ( t ) = ( − 3 A 1 t 2 − 3 A 0 t + 2 A 1 t + A 0 ) ( − 3 e − 3 t ) + ( − 6 A 1 t − 3 A 0 + 2 A 1 ) e − 3 t = ( 9 A 1 t 2 + 9 A 0 t − 6 A 1 t − 3 A 0 − 6 A 1 t − 3 A 0 + 2 A 1 ) e − 3 t = ( 9 A 1 t 2 + 9 A 0 t − 12 A 1 t − 6 A 0 + 2 A 1 ) e − 3 t Thus, 4 y ″ + 11 y ′ − 3 y for y p will be, 4 y ″ p + 11 y ′ p − 3 y p = [ 4 ( 9 A 1 t 2 + 9 A 0 t − 12 A 1 t − 6 A 0 + 2 A 1 ) e − 3 t + 11 ( − 3 A 1 t 2 − 3 A 0 t + 2 A 1 t + A 0 ) e − 3 t − 3 ( A 1 t 2 + A 0 t ) e − 3 t ] = ( 36 A 1 t 2 + 36 A 0 t − 48 A 1 t − 32 A 0 + 8 A 1 − 33 A 1 t 2 − 33 A 0 t + 22 A 1 t + 11 A 0 − 3 A 1 t 2 − 3 A 0 t ) e − 3 t = ( − 26 A 1 t − 21 A 0 + 8 A 1 ) e − 3 t Setting ( − 26 A 1 t − 13 A 0 + 8 A 1 ) e − 3 t = − 2 t e − 3 t and comparing the coefficients result in − 26 A 1 = − 2 and − 13 A 0 + 8 A 1 = 0 . On simplification of − 26 A 1 = − 2 gives A 1 = 1 13 . Substitute A 1 = 1 13 in − 13 A 0 + 8 A 1 = 0 , − 13 A 0 + 8 ( 1 13 ) = 0 13 A 0 = 8 13 A 0 = 8 13 × 13 A 0 = 8 169 That is, A 0 = 8 169 and A 1 = 1 13 . Substitute A 0 = 8 169 and A 1 = 1 13 in y p ( t ) = ( A 1 t 2 + A 0 t ) e − 3 t to obtain y p ( t ) = ( t 2 13 + 8 t 169 ) e − 3 t . Therefore, the particular solution to the differential equation 4 y ″ + 11 y ′ − 3 y = − 2 t e − 3 t is, y ( t ) = ( t 2 13 + 8 t 169 ) e − 3 t _ .

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