What is the solution for 4y^{''}+51y^'+36y=0,y(0)=-5,y^'(0)=-20 ?

The solution for 4y^{''}+51y^'+36y=0,y(0)=-5,y^'(0)=-20 is y=(-5-19/9)e^{-(3t)/4}+19/9 e^{-12t}

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4y^{''}+51y^'+36y=0,y(0)=-5,y^'(0)=-20

4 y^{^{''}} + 51 y^{^{'}} + 36 y = 0, y (0) = - 5, y^{^{'}} (0) = - 20

y = (- 5 - \frac{19}{9}) e^{- \frac{3 t}{4}} + \frac{19}{9} e^{- 12 t}

Solution steps

4 y^{''} (t) + 51 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = (- 5 - \frac{19}{9}) e^{- \frac{3 t}{4}} + \frac{19}{9} e^{- 12 t}

y = (- 5 - \frac{19}{9}) e^{- \frac{3 t}{4}} + \frac{19}{9} e^{- 12 t}

y^{^{''}} + 9 y = 0, y (0) = - 1, y^{^{'}} (0) = 3

4 y^{^{''}} - 4 y^{^{'}} + 17 y = 0

4 y^{^{''}} - 2 y^{^{'}} + 9 y = 0

y^{^{''}} + 19 y^{^{'}} + 90 y = 0

y^{^{'''}} - 4 y^{^{''}} - 41 y^{^{'}} - 36 y = 0

What is the solution for 4y^{''}+51y^'+36y=0,y(0)=-5,y^'(0)=-20 ?
The solution for 4y^{''}+51y^'+36y=0,y(0)=-5,y^'(0)=-20 is y=(-5-19/9)e^{-(3t)/4}+19/9 e^{-12t}