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

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

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

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

y = - e^{- 2 t} - 2 e^{- 2 t} t

Solution steps

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

Solve linear ODE:

y = - e^{- 2 t} - 2 e^{- 2 t} t

y = - e^{- 2 t} - 2 e^{- 2 t} t

\frac{\partial}{\partial x} (\frac{4 x - 2 y}{2 x ^{2} + 3 y})

\int_{0}^{4} \frac{1}{x ^{2} + x - 6} d x

\int_{1}^{4} (4 x^{2} + x) d x

\int \frac{1}{1 - 5 t} d t

\frac{d}{d x} (sin (x) + cos (2 x))

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