What is the solution for y^{''}+4y^'+4y=0,y(0)=1,y^'(0)=3 ?

The solution for y^{''}+4y^'+4y=0,y(0)=1,y^'(0)=3 is y=e^{-2t}+5e^{-2t}t

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + 5 x^{^{'}} + 6 x = 0

y^{^{''}} + 5 y^{^{'}} = 0, y (0) = 1

y^{^{''''}} + 14 y^{^{''}} + 49 y = 0

y^{^{''}} + 16 y = 0, y (π^{2}) = - 10, y^{^{'}} (π^{2}) = 3

4 y^{^{''}} - 12 y^{^{'}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = 5

What is the solution for y^{''}+4y^'+4y=0,y(0)=1,y^'(0)=3 ?
The solution for y^{''}+4y^'+4y=0,y(0)=1,y^'(0)=3 is y=e^{-2t}+5e^{-2t}t