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

The solution for y^{''}+6y^'+8y=0,y(0)=0,y^'(0)=-8 is y=-4e^{-2t}+4e^{-4t}

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

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

y = - 4 e^{- 2 t} + 4 e^{- 4 t}

Solution steps

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

Solve linear ODE:

y = - 4 e^{- 2 t} + 4 e^{- 4 t}

y = - 4 e^{- 2 t} + 4 e^{- 4 t}

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (- 0) = 0, y (0) = 3

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 8 y = 0

y^{^{''}} - k \cdot y = 0

(6 d^{3} + 5 d^{2}) y = 0

y^{^{'''}} + 7 y^{^{''}} + 14 y^{^{'}} + 8 y = 0

What is the solution for y^{''}+6y^'+8y=0,y(0)=0,y^'(0)=-8 ?
The solution for y^{''}+6y^'+8y=0,y(0)=0,y^'(0)=-8 is y=-4e^{-2t}+4e^{-4t}