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

The solution for y^{''}+4y^'+4y=0,y(0)=1,y^'(0)=4 is y=e^{-2t}+6e^{-2t}t

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

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

y = e^{- 2 t} + 6 e^{- 2 t} t

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} + 6 e^{- 2 t} t

y = e^{- 2 t} + 6 e^{- 2 t} t

y^{^{''}} - 3 y^{^{'}} - 40 y = 0

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

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

(D^{2} - 1) y = 0

y^{^{''}} + 16 y = 0, y^{^{'}} (\frac{π}{4}) = - 3, y^{^{'}} (0) = 2

What is the solution for y^{''}+4y^'+4y=0,y(0)=1,y^'(0)=4 ?
The solution for y^{''}+4y^'+4y=0,y(0)=1,y^'(0)=4 is y=e^{-2t}+6e^{-2t}t