What is the solution for y^{''}+8y^'+16y=0,y(-1)=3,y^'(-1)=5 ?

The solution for y^{''}+8y^'+16y=0,y(-1)=3,y^'(-1)=5 is y=20e^{-4t-4}+17e^{-4t-4}t

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y^{''}+8y^'+16y=0,y(-1)=3,y^'(-1)=5

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (- 1) = 3, y^{^{'}} (- 1) = 5

y = 20 e^{- 4 t - 4} + 17 e^{- 4 t - 4} t

Solution steps

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

Solve linear ODE:

y = 20 e^{- 4 t - 4} + 17 e^{- 4 t - 4} t

y = 20 e^{- 4 t - 4} + 17 e^{- 4 t - 4} t

4 y^{^{''}} - y^{^{'}} - 6 y = 0

y^{^{''}} - \frac{2}{3} y^{^{'}} + \frac{y}{9} = 0, y (0) = 17, y^{^{'}} (0) = b

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 1, y^{^{'}} (0) = 7

y^{^{''}} + 10 y + 16 y = 0

8 \frac{d ^{2} y}{d t ^{2}} + 12 \frac{d y}{d t} + 5 y = 0

What is the solution for y^{''}+8y^'+16y=0,y(-1)=3,y^'(-1)=5 ?
The solution for y^{''}+8y^'+16y=0,y(-1)=3,y^'(-1)=5 is y=20e^{-4t-4}+17e^{-4t-4}t