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

The solution for y^{''}+4y^'+4y=0,y^'(0)=1,y(0)=1 is y=e^{-2t}+3e^{-2t}t

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

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

y = e^{- 2 t} + 3 e^{- 2 t} t

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} + 3 e^{- 2 t} t

y = e^{- 2 t} + 3 e^{- 2 t} t

y^{^{'''}} + 3 y^{^{''}} - 4 y = 0, y (1) = \frac{1}{2}

9 y^{^{''}} - 12 y^{^{'}} + 40 y = 0, y (0) = 8, y^{^{'}} (0) = 3

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = 10

\frac{1}{y} y^{^{''}} = a, y (0) = 0

y^{^{''''}} - 3 y^{^{''}} + 2 y = 0

What is the solution for y^{''}+4y^'+4y=0,y^'(0)=1,y(0)=1 ?
The solution for y^{''}+4y^'+4y=0,y^'(0)=1,y(0)=1 is y=e^{-2t}+3e^{-2t}t