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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y = - 4 x + 10

y^{^{''''}} + 2 y^{^{''}} + y = 9 + cos (2 t)

y^{^{''}} - 6 y^{^{'}} + 5 = 0

y^{^{''}} + 5 y = 5 sin (t), y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 4 y = arccos (2 x)

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