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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 = e^{- 5 x}

(d^{2} - 9) y = 30 sin^{3} (x)

3 y^{^{''}} - 8 y^{^{'}} - 3 y = 6

y^{^{''}} + 4 y^{^{'}} + 5 y = 13 sin (2 t)

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

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