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

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

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

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

x = 2 e^{- 2 t} t

Solution steps

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

Solve linear ODE:

x = 2 e^{- 2 t} t

x = 2 e^{- 2 t} t

15 y^{^{''''}} + 83 y^{^{''}} + 16 y = 0

y^{^{''}} + 64 y = 0, y (0) = 4, y^{^{'}} (0) = 3

12 y^{^{''}} - 5 y^{^{'}} - 3 y = 0, y (0) = 2

y^{^{''}} + 64 y = 0, y (0) = 4, y^{^{'}} (0) = 7

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

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