What is the solution for y^{''}+8y^'+6y=0,y(-1)=6,y^'(-1)=5 ?

The solution for y^{''}+8y^'+6y=0,y(-1)=6,y^'(-1)=5 is

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y^{''}+8y^'+6y=0,y(-1)=6,y^'(-1)=5

y^{^{''}} + 8 y^{^{'}} + 6 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 5

y = (6 e^{- 4 + 10} + \frac{29 - 6 10}{2 10 e ^{4 - 10}}) e^{(- 4 + 10) t} - \frac{( 29 - 6 10 ) e ^{- 4 t - 10 t - 4 - 10}}{2 10}

Solution steps

y^{''} (t) + 8 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = (6 e^{- 4 + 10} + \frac{29 - 6 10}{2 10 e ^{4 - 10}}) e^{(- 4 + 10) t} - \frac{( 29 - 6 10 ) e ^{- 4 t - 10 t - 4 - 10}}{2 10}

y = (6 e^{- 4 + 10} + \frac{29 - 6 10}{2 10 e ^{4 - 10}}) e^{(- 4 + 10) t} - \frac{( 29 - 6 10 ) e ^{- 4 t - 10 t - 4 - 10}}{2 10}

2 y^{^{''}} + 4 y^{^{'}} + 20 y = 0

y^{^{''}} - 4 y^{^{'}} - 96 y = 0, y (0) = 13, y^{^{'}} (0) = 36

y^{^{''}} + 25 y^{^{'}} + \frac{1}{0.046} y = 0

y^{^{'''''}} - 6 y^{^{''''}} + 11 y^{^{'''}} - 6 y^{^{''}} = 0

y^{^{'''}} + 8 y^{^{'}} = 0

What is the solution for y^{''}+8y^'+6y=0,y(-1)=6,y^'(-1)=5 ?
The solution for y^{''}+8y^'+6y=0,y(-1)=6,y^'(-1)=5 is