What is the solution for y^{''}+4y^'+4y=0,y(-1)=4,y^'(-1)=6 ?

The solution for y^{''}+4y^'+4y=0,y(-1)=4,y^'(-1)=6 is y=18e^{-2t-2}+14e^{-2t-2}t

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y^{''}+4y^'+4y=0,y(-1)=4,y^'(-1)=6

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 4, y^{^{'}} (- 1) = 6

y = 18 e^{- 2 t - 2} + 14 e^{- 2 t - 2} t

Solution steps

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

Solve linear ODE:

y = 18 e^{- 2 t - 2} + 14 e^{- 2 t - 2} t

y = 18 e^{- 2 t - 2} + 14 e^{- 2 t - 2} t

\frac{d ^{2} y}{d x ^{2}} = a y

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y^{1} (0) = e^{3} x

5 \frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = 0

2 y^{^{'''}} + 9 y^{^{''}} - 6 y^{^{'}} - 5 y = 0

y^{^{''}} + 8 y^{^{'}} + 4 y = 0, y (0) = 0

What is the solution for y^{''}+4y^'+4y=0,y(-1)=4,y^'(-1)=6 ?
The solution for y^{''}+4y^'+4y=0,y(-1)=4,y^'(-1)=6 is y=18e^{-2t-2}+14e^{-2t-2}t