What is the solution for y^{''}+8y^'+16y=0,y(-1)=2,y^'(-1)=4 ?

The solution for y^{''}+8y^'+16y=0,y(-1)=2,y^'(-1)=4 is y=14e^{-4t-4}+12e^{-4t-4}t

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y^{''}+8y^'+16y=0,y(-1)=2,y^'(-1)=4

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 4

y = 14 e^{- 4 t - 4} + 12 e^{- 4 t - 4} t

Solution steps

y^{''} (t) + 8 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = 14 e^{- 4 t - 4} + 12 e^{- 4 t - 4} t

y = 14 e^{- 4 t - 4} + 12 e^{- 4 t - 4} t

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 2

x^{^{''}} + 10 x^{^{'}} + 16 x = 0, x (0) = 1, x^{^{'}} (0) = - 13

4 y^{^{''}} + 8 y^{^{'}} - 21 y = 0

6 y^{^{''}} - 5 y^{^{'}} + y = 0, y (0) = e^{\frac{x}{2}}

y^{^{'''}} + 25 y^{^{'}} = 0

What is the solution for y^{''}+8y^'+16y=0,y(-1)=2,y^'(-1)=4 ?
The solution for y^{''}+8y^'+16y=0,y(-1)=2,y^'(-1)=4 is y=14e^{-4t-4}+12e^{-4t-4}t