What is the solution for y^{''}-16y=xe^x ?

The solution for y^{''}-16y=xe^x is y=c_{1}e^{4x}+c_{2}e^{-4x}-(e^xx)/(15)-2/225 e^x

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y^{''}-16y=xe^x

y^{^{''}} - 16 y = x e^{x}

y = c_{1} e^{4 x} + c_{2} e^{- 4 x} - \frac{e ^{x} x}{15} - \frac{2}{225} e^{x}

Solution steps

y^{''} (x) - 16 y = x e^{x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- 4 x} - \frac{e ^{x} x}{15} - \frac{2}{225} e^{x}

y = c_{1} e^{4 x} + c_{2} e^{- 4 x} - \frac{e ^{x} x}{15} - \frac{2}{225} e^{x}

y^{^{''}} - 2 y^{^{'}} - 8 y = e^{- 2 x} + 2 x^{2}

\frac{d ^{2} y}{d x ^{2}} + c^{2} y = sec^{2} (c x)

y^{^{''}} - 3 y^{^{'}} + 5 y = 4 x^{3} - 2 x

y^{^{''}} - 8 y^{^{'}} + 16 y = 24 x + 2, y (0) = 2, y^{^{'}} (0) = - 1

y^{^{'''}} + y^{^{''}} - 5 y^{^{'}} + 3 y = e^{- t} + sin (t)

What is the solution for y^{''}-16y=xe^x ?
The solution for y^{''}-16y=xe^x is y=c_{1}e^{4x}+c_{2}e^{-4x}-(e^xx)/(15)-2/225 e^x