What is the solution for y^{''}+4y^'+4y=0,y(-1)=3,y^'(-1)=6 ?

The solution for y^{''}+4y^'+4y=0,y(-1)=3,y^'(-1)=6 is y=15e^{-2t-2}+12e^{-2t-2}t

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y^{''}+4y^'+4y=0,y(-1)=3,y^'(-1)=6

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 3, y^{^{'}} (- 1) = 6

y = 15 e^{- 2 t - 2} + 12 e^{- 2 t - 2} t

Solution steps

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

Solve linear ODE:

y = 15 e^{- 2 t - 2} + 12 e^{- 2 t - 2} t

y = 15 e^{- 2 t - 2} + 12 e^{- 2 t - 2} t

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 3, y^{^{'}} (- 1) = 5

y^{^{''}} + y^{^{'}} + 11 y = 0, y^{^{'}} (0) = 1, y (0) = 4

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = 3

S^{^{''}} + 16 S = 0, S (0) = 4, S^{^{'}} (0) = 2

y^{^{''}} - y^{^{'}} - 20 y = 0, y (0) = α, y^{^{'}} (0) = 2

What is the solution for y^{''}+4y^'+4y=0,y(-1)=3,y^'(-1)=6 ?
The solution for y^{''}+4y^'+4y=0,y(-1)=3,y^'(-1)=6 is y=15e^{-2t-2}+12e^{-2t-2}t