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

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

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

y^{^{''}} + \frac{64}{5} y^{^{'}} + 64 y = 0

25 y^{^{''}} - 25 y^{^{'}} - 6 y = 0

y^{^{''}} + 6 y^{^{'}} + 34 y = 0, y (0) = 8

\frac{1}{8} y^{^{''}} + y^{^{'}} + 2 y = 0

y^{^{''}} = - 6 y

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