Answer to Question #205038 in Calculus for Kaustav Borah

Question

(D − 1)^2(D^2 + 1)^2 y = sin^2(x/2) + e^x + x

Answer 1

Solution

For homogeneous solution let’s solve characteristic equation

(k-1)²(k²+1)=0 => k_1,2=1, k_3,4=i, k_5,6=-i

So the homogeneous solution is y_H(x) = (A+Bx)e^x+(C+Dx)e^ix +(E+Fx)e^-ix, where A,B,C,D,E,F are arbitrary constants. 

sin²(x/2) + e^x + x = x+1/2-cos(x)/2+e^x = x+1/2 - e^ix /4 - e^-ix /4 +e^x    

Four parts of partial solutions

y₁(x) =ax+b: (D − 1)²(D² + 1)²y₁(x) = (D − 1)²(D² + 1)(ax+b) = (D − 1)² (ax+b) = (D − 1)(a-ax-b) = -2a+ax+b

ax+b-2a = x+1/2 =>a=1, b=5/2 => y₁(x) =x+5/2

y₂(x) =cx²e^x: (D − 1)²(D² + 1)²y₂(x) = (D² + 1)²(D − 1)² cx²e^x = c(D² + 1)²(D − 1) (2xe^x+x²e^x- x²e^x) = c(D² + 1)²(D − 1) (2xe^x) = c(D² + 1)² (2e^x+2xe^x-2xe^x) = 2c(D² + 1)² e^x = 2c(D² + 1) (2e^x ) = 8ce^x => c=1/8 +> y₂(x) =x²e^x /8

y₃(x) =dx²e^ix: (D − 1)²(D² + 1)²y₃(x) = d(D − 1)²(D² + 1) ( 2e^ix + 2ixe^ix +2ixe^ix - x²e^ix + x²e^ix) = d(D − 1)² (D² + 1) (2e^ix + 4ixe^ix) = d(D − 1)² (-2e^ix - 4e^ix - 4e^ix – 4ixe^ix + 2e^ix + 4ixe^ix) = d(D − 1)² ( - 4e^ix - 4e^ix) = -8d(D − 1)² (e^ix ) = -8d(D − 1) (ie^ix - e^ix ) = -8d(-e^ix - ie^ix - ie^ix + e^ix ) = 16d ie^ix => 16d i =-1/4 => d= i /64 => y₃(x) = ix²e^ix/64

y₄(x) =fx²e^-ix: (D − 1)²(D² + 1)²y₄(x) = f(D − 1)²(D² + 1) (2e^-ix – 2ixe^-ix - 2ixe^-ix - x²e^-ix + x²e^-ix) = f(D − 1)² (D² + 1) (2e^-ix – 4ixe^-ix) = f(D − 1)² (-2e^-ix – 4e^-ix – 4e^-ix + 4ixe^-ix +2e^-ix – 4ixe^-ix) = f(D − 1)² ( - 4e^-ix - 4e^-ix) = -8f(D − 1)² (e^-ix ) = -8f(D − 1) (-ie^-ix - e^-ix ) = -8f(-e^-ix + ie^-ix + ie^-ix + e^-ix ) = -16f ie^-ix => -16f i =-1/4 => f= -i /64 => y₄(x) =-ix²e^-ix/64

Solution of the given equation:

y(x) = y_H(x) + y₁(x) + y₂(x) + y₃(x)+ y₄(x) = (A+Bx)e^x+(C+Dx)e^ix +(E+Fx)e^-ix + x+5/2 + x²e^x /8 + ix²e^ix/64 - ix²e^ix/64 = (A+Bx)e^x+(C+Dx)e^ix +(E+Fx)e^-ix + x+5/2 + x²e^x/8 - x²sin(x)/32

Answer

y(x) = (A+Bx)e^x+(C+Dx)e^ix +(E+Fx)e^-ix + x+5/2 + x²e^x/8 - x²sin(x)/32