What is the solution for y^{''}+6y^'+8y=0,y(0)=-6,y^'(0)=18 ?

The solution for y^{''}+6y^'+8y=0,y(0)=-6,y^'(0)=18 is y=-3e^{-2t}-3e^{-4t}

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y^{''}+6y^'+8y=0,y(0)=-6,y^'(0)=18

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = - 6, y^{^{'}} (0) = 18

y = - 3 e^{- 2 t} - 3 e^{- 4 t}

Solution steps

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

Solve linear ODE:

y = - 3 e^{- 2 t} - 3 e^{- 4 t}

y = - 3 e^{- 2 t} - 3 e^{- 4 t}

y 4 + 3 y^{^{''}} + y = 0

y^{^{''}} + \frac{8}{100} y^{^{'}} + \frac{12}{10000} y = 0

y^{^{''}} - y = 0, y (ln (2)) = 5, y (0) = 1

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 2 x = 0, x (0) = 0

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

What is the solution for y^{''}+6y^'+8y=0,y(0)=-6,y^'(0)=18 ?
The solution for y^{''}+6y^'+8y=0,y(0)=-6,y^'(0)=18 is y=-3e^{-2t}-3e^{-4t}