What is the solution for y^{''}-4y=x^3 ?

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

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y^{''}-4y=x^3

y^{^{''}} - 4 y = x^{3}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{x ^{3}}{4} - \frac{3 x}{8}

Solution steps

y^{''} (x) - 4 y = x^{3}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{x ^{3}}{4} - \frac{3 x}{8}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{x ^{3}}{4} - \frac{3 x}{8}

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

y^{^{''}} - 2 y^{^{'}} + y = 12 x^{3} e^{x}

2 y^{^{''}} - 2 y^{^{'}} - 4 y = 2 e^{2 t}

y^{^{'''}} - 6 y^{^{''}} - 6 = cos (x)

y^{^{'''}} + 6 y^{^{''}} + 11 y^{^{'}} + 6 y = 3 + 3 x

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