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

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

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

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

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

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

y^{^{''}} + y = e^{t^{2}}

y^{^{''}} - 3 y + 2 y = e^{t} sin (t)

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = x^{3} + 4 x^{2} + 2 x + 1

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