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}-2e^{-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} - 2 e^{- 2 t} t

Solution steps

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

Solve linear ODE:

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

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

\int tan (2 x se) c^{2} 2 x d x

y = 3 x - 2 x^{2}, (- 3, - 27)

x \to 0 + lim (x + 1)

\frac{d y}{d x} = e^{2 y} sin (x)

\int x - 6 cos (x) d x

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}-2e^{-2t}t