What is the solution for y^{''}+2y^'=8x^3 ?

The solution for y^{''}+2y^'=8x^3 is y=c_{1}+c_{2}e^{-2x}+x^4-2x^3+3x^2-3x

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y^{''}+2y^'=8x^3

y^{^{''}} + 2 y^{^{'}} = 8 x^{3}

y = c_{1} + c_{2} e^{- 2 x} + x^{4} - 2 x^{3} + 3 x^{2} - 3 x

Solution steps

y^{''} (x) + 2 y^{'} (x) = 8 x^{3}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + x^{4} - 2 x^{3} + 3 x^{2} - 3 x

y = c_{1} + c_{2} e^{- 2 x} + x^{4} - 2 x^{3} + 3 x^{2} - 3 x

y^{^{''}} + 6 y^{^{'}} + 8 y = 4 t^{2} - 3 t + 2

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 x} \cdot sin (x)

\frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} = 6 t

y^{^{''}} + 9 y = 9 cos^{3} (3 x)

y^{^{''}} + 2 y^{^{'}} = 49 e^{x} sin (2 x)

What is the solution for y^{''}+2y^'=8x^3 ?
The solution for y^{''}+2y^'=8x^3 is y=c_{1}+c_{2}e^{-2x}+x^4-2x^3+3x^2-3x