What is the solution for y^{''}+2y^'+y=0,y^1=xe^{-x} ?

The solution for y^{''}+2y^'+y=0,y^1=xe^{-x} is y=c_{1}e^{-x}+c_{2}xe^{-x}

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y^{''}+2y^'+y=0,y^1=xe^{-x}

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{1} (0) = x e^{- x}

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = 0

Solve linear ODE:

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

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

6 y^{^{''''}} + 5 y^{^{'''}} + 25 y^{^{''}} + 20 y^{^{'}} + 4 y = 0

4 \frac{d ^{2} y}{d x ^{2}} - 12 \frac{d y}{d x} + 5 y = 0

4 y^{^{''}} + y^{^{'}} + \frac{1}{16} y = 0

y^{^{'''''}} + 16 y^{^{''''}} + 80 y^{^{'''}} = 0

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 0

What is the solution for y^{''}+2y^'+y=0,y^1=xe^{-x} ?
The solution for y^{''}+2y^'+y=0,y^1=xe^{-x} is y=c_{1}e^{-x}+c_{2}xe^{-x}