What is the solution for y^{''}-2y^'-3y=xe^x ?

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

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y^{''}-2y^'-3y=xe^x

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

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

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = x e^{x}

Solve linear ODE:

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

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

e^{\frac{4}{5} x}

\frac{d}{d x} (x + 1 - 2)

\frac{d}{d x} (x^{2} - 32 x)

\frac{d}{d x} (\frac{e ^{4 x^{2 + 1}}}{8})

\frac{d}{d x} (f (x) (x)^{- 1})

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