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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 7 y = 0

4 y^{^{''}} + 4 y^{^{'}} + 4 y = 0

5 y^{^{''}} + 5 y = 0

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

4 y^{^{''}} - 2 y^{^{'}} - 2 y = 0, y (0) = 4, y^{^{'}} (0) = 1

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