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

\int_{\frac{1}{2}}^{1} (x^{- 3} - 25) d x

y^{^{''}} + \frac{1}{x} y^{^{'}} = 0

y^{^{''}} + 2 x = 6 x^{2}, y (0) = 3

\int_{- 3}^{3} π (9 - x^{2})^{2} d x

x \to 5 lim ((4 - x)^{\frac{3}{2}})

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