What is the solution for 3y^{''}+13y^'+4y=0,y(0)=-2,y^'(0)=-2 ?

The solution for 3y^{''}+13y^'+4y=0,y(0)=-2,y^'(0)=-2 is y=(-2-8/11)e^{-t/3}+8/11 e^{-4t}

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3y^{''}+13y^'+4y=0,y(0)=-2,y^'(0)=-2

3 y^{^{''}} + 13 y^{^{'}} + 4 y = 0, y (0) = - 2, y^{^{'}} (0) = - 2

y = (- 2 - \frac{8}{11}) e^{- \frac{t}{3}} + \frac{8}{11} e^{- 4 t}

Solution steps

3 y^{''} (t) + 13 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = (- 2 - \frac{8}{11}) e^{- \frac{t}{3}} + \frac{8}{11} e^{- 4 t}

y = (- 2 - \frac{8}{11}) e^{- \frac{t}{3}} + \frac{8}{11} e^{- 4 t}

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What is the solution for 3y^{''}+13y^'+4y=0,y(0)=-2,y^'(0)=-2 ?
The solution for 3y^{''}+13y^'+4y=0,y(0)=-2,y^'(0)=-2 is y=(-2-8/11)e^{-t/3}+8/11 e^{-4t}