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

The solution for y^{''}-4y^'+4y=e^x-x^3 is y=c_{1}e^{2x}+c_{2}xe^{2x}+e^x+(-x^3-3x^2-3)/4-(9x)/8

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{x} + \frac{- x ^{3} - 3 x ^{2} - 3}{4} - \frac{9 x}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{x} + \frac{- x ^{3} - 3 x ^{2} - 3}{4} - \frac{9 x}{8}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{x} + \frac{- x ^{3} - 3 x ^{2} - 3}{4} - \frac{9 x}{8}

9 y^{^{''}} + 4 y = 3 x + 12 e^{- x}, y (0) = 1, y^{^{'}} (0) = 0

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

4 y^{^{''}} - 3 y^{^{'}} = 4 x - 6

y^{^{''}} - 9 y^{^{'}} + 8 y = x, y (0) = e^{x}

y^{^{''}} + 2 y^{^{'}} + 4 y = t + cos (2 t)

What is the solution for y^{''}-4y^'+4y=e^x-x^3 ?
The solution for y^{''}-4y^'+4y=e^x-x^3 is y=c_{1}e^{2x}+c_{2}xe^{2x}+e^x+(-x^3-3x^2-3)/4-(9x)/8