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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = t e^{- 3 t}

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = cos (2 x)

y^{^{''}} + y = 2 + t, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} - 6 y^{^{'}} + 8 y = 4 e^{e - 2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = - 80 sin (3 x) - 23 cos (3 x)

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