What is the solution for y^{''}+4y^'+4y=0,y(0)=3,y^'(0)=2 ?

The solution for y^{''}+4y^'+4y=0,y(0)=3,y^'(0)=2 is y=3e^{-2t}+8te^{-2t}

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y^{''}+4y^'+4y=0,y(0)=3,y^'(0)=2

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = 2

y = 3 e^{- 2 t} + 8 t e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = 3 e^{- 2 t} + 8 t e^{- 2 t}

y = 3 e^{- 2 t} + 8 t e^{- 2 t}

y^{^{'''}} + 4 y^{^{''}} + y^{^{'}} - 26 y = 0, y (0) = e^{2 x}

y^{^{''}} - 2 y^{^{'}} = - 4 y

7 y^{^{''}} - 8 y^{^{'}} + y = 0

60 x^{^{''}} + 240 x = 0, x (0) = 0, x^{^{'}} (0) = - 8

4 y^{^{''}} + 9 y^{^{'}} + 9 y = 0

What is the solution for y^{''}+4y^'+4y=0,y(0)=3,y^'(0)=2 ?
The solution for y^{''}+4y^'+4y=0,y(0)=3,y^'(0)=2 is y=3e^{-2t}+8te^{-2t}