What is the solution for y^{''}+6y^'+8y=0,y(0)=-1,y^'(0)=2 ?

The solution for y^{''}+6y^'+8y=0,y(0)=-1,y^'(0)=2 is y=-e^{-2t}

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y^{''}+6y^'+8y=0,y(0)=-1,y^'(0)=2

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

y = - e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = - e^{- 2 t}

y = - e^{- 2 t}

\frac{d ^{2} y}{d t ^{2}} - 10 \frac{d y}{d t} = 0

y^{^{''}} - 4 y + 13 y = 0

49 y^{^{'''}} + 56 y^{^{''}} + 16 y^{^{'}} = 0

\frac{d ^{3} y}{d x ^{3}} + 8 \frac{d y}{d x} = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 5

What is the solution for y^{''}+6y^'+8y=0,y(0)=-1,y^'(0)=2 ?
The solution for y^{''}+6y^'+8y=0,y(0)=-1,y^'(0)=2 is y=-e^{-2t}