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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial z} (x^{a} y^{b} z^{c})

\frac{d y}{d x} = \frac{( y - 1 )}{x + 3}

y^{^{'}} + \frac{y}{x} = 7 x^{3} y^{2}

3^{x}

\frac{d}{d x} (e^{3 x} sin (x))

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