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

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

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

y^{^{''}} + 2 y^{^{'}} + y = 0, y (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}

y^{^{''}} + 5 y^{^{'}} - 4 y = 0

16 y^{^{''}} + 56 y^{^{'}} + 49 y = 0

y^{^{''}} - 25 y^{^{'}} = 0

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

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

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