What is the solution for y^{''}+4y^'+4y=0,y(0)=2,y^'(0)=-9 ?

The solution for y^{''}+4y^'+4y=0,y(0)=2,y^'(0)=-9 is y=2e^{-2t}-5e^{-2t}t

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y^{''}+4y^'+4y=0,y(0)=2,y^'(0)=-9

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 9

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

Solution steps

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

Solve linear ODE:

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

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

y^{2} - 6 x = 1, x - y = 1

\int - e^{- 2 x} x d x

\int (e) d x

x \to 0 lim (\frac{sin ( x )}{x} + 3)

f (x) = arccos (e^{x})

What is the solution for y^{''}+4y^'+4y=0,y(0)=2,y^'(0)=-9 ?
The solution for y^{''}+4y^'+4y=0,y(0)=2,y^'(0)=-9 is y=2e^{-2t}-5e^{-2t}t