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

The solution for y^{''}+2y^'+y=0,y(0)=0,y^'(0)=1 is y=e^{-t}t

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

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

y = e^{- t} t

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 0

Solve linear ODE:

y = e^{- t} t

y = e^{- t} t

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

\frac{d}{d x} (c_{1} e^{- x} + c_{2} x e^{- x})

\int cot (d) x d x

\int 5^{u} d u

\int ln (x^{2} + 2) d x

What is the solution for y^{''}+2y^'+y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+2y^'+y=0,y(0)=0,y^'(0)=1 is y=e^{-t}t