What is the solution for y^{''}+y=e^x+x^3,y(0)=2,y^'(0)=0 ?

The solution for y^{''}+y=e^x+x^3,y(0)=2,y^'(0)=0 is y= 3/2 cos(x)+11/2 sin(x)+1/2 e^x+x^3-6x

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y^{''}+y=e^x+x^3,y(0)=2,y^'(0)=0

y^{^{''}} + y = e^{x} + x^{3}, y (0) = 2, y^{^{'}} (0) = 0

y = \frac{3}{2} cos (x) + \frac{11}{2} sin (x) + \frac{1}{2} e^{x} + x^{3} - 6 x

Solution steps

y^{''} (x) + y = e^{x} + x^{3}

Solve linear ODE:

y = \frac{3}{2} cos (x) + \frac{11}{2} sin (x) + \frac{1}{2} e^{x} + x^{3} - 6 x

y = \frac{3}{2} cos (x) + \frac{11}{2} sin (x) + \frac{1}{2} e^{x} + x^{3} - 6 x

(D^{3} + 4 D) y = sin (2 x)

y^{^{''}} - 4 y = x^{- 1} e^{2 x}

y^{^{''}} + 2 y^{^{'}} + 4 y = cos (4 x)

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} 4 + 8 y = 8 x^{2} + 8 x + 1

y^{^{'''}} + y^{^{''}} - y^{^{'}} - y = \frac{e ^{x} - e ^{- x}}{2}

What is the solution for y^{''}+y=e^x+x^3,y(0)=2,y^'(0)=0 ?
The solution for y^{''}+y=e^x+x^3,y(0)=2,y^'(0)=0 is y= 3/2 cos(x)+11/2 sin(x)+1/2 e^x+x^3-6x