What is the solution for y^{''}-25y=xe^x ?

The solution for y^{''}-25y=xe^x is y=c_{1}e^{5x}+c_{2}e^{-5x}-(e^xx)/(24)-1/288 e^x

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y^{''}-25y=xe^x

y^{^{''}} - 25 y = x e^{x}

y = c_{1} e^{5 x} + c_{2} e^{- 5 x} - \frac{e ^{x} x}{24} - \frac{1}{288} e^{x}

Solution steps

y^{''} (x) - 25 y = x e^{x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{- 5 x} - \frac{e ^{x} x}{24} - \frac{1}{288} e^{x}

y = c_{1} e^{5 x} + c_{2} e^{- 5 x} - \frac{e ^{x} x}{24} - \frac{1}{288} e^{x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 8 x

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

y^{^{''}} - y^{^{'}} - 6 y = - e^{x} + 12 x, y^{^{'}} (0) = - 2, y (0) = 1

2 x^{^{''}} + 5 x^{^{'}} + 2 x = 4 e^{- 2 t} - t

y^{^{''}} - 9 y = \frac{9 t}{e ^{3 t}}

What is the solution for y^{''}-25y=xe^x ?
The solution for y^{''}-25y=xe^x is y=c_{1}e^{5x}+c_{2}e^{-5x}-(e^xx)/(24)-1/288 e^x