What is the solution for y^{''}+4y^'+4y=0,y(0)=5 ?

The solution for y^{''}+4y^'+4y=0,y(0)=5 is y=5e^{-2t}+c_{2}te^{-2t}

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y^{''}+4y^'+4y=0,y(0)=5

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 5

y = 5 e^{- 2 t} + c_{2} t e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = 5 e^{- 2 t} + c_{2} t e^{- 2 t}

y = 5 e^{- 2 t} + c_{2} t e^{- 2 t}

y^{^{''}} - 16 y^{^{'}} + 68 y = 0

y^{^{''''}} + 4 y^{^{'''}} + 6 y^{^{''}} + 4 y^{^{'}} + y = 0

11 x^{^{''}} + 112.5 x = 0

\frac{y ^{^{''}}}{y} = a

\frac{d ^{2} y}{d x ^{2}} + 6 \cdot (\frac{d y}{d x}) + 8 \cdot y = 0

What is the solution for y^{''}+4y^'+4y=0,y(0)=5 ?
The solution for y^{''}+4y^'+4y=0,y(0)=5 is y=5e^{-2t}+c_{2}te^{-2t}