What is the solution for y^{''}+4y^'+4y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+4y^'+4y=0,y(0)=0,y^'(0)=1 is y=e^{-2t}t

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y^{''}+4y^'+4y=0,y(0)=0,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = e^{- 2 t} t

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{- 2 t} t

y = e^{- 2 t} t

\frac{d}{d x} (\frac{- 6 x}{x ^{2} + 1})

9 x^{2} y^{^{'}} = y^{^{'}} + 5 x e^{- y}

\frac{d y}{d x} = x \cdot e^{- 2 x}

\int \frac{x - 1}{( x + 1 ) ( 2 x ^{2} + 3 )} d x

\frac{d ^{2} y}{d x ^{2}} + y = csc (x)

What is the solution for y^{''}+4y^'+4y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+4y^'+4y=0,y(0)=0,y^'(0)=1 is y=e^{-2t}t