What is the solution for 9y^{''}+12y^'+5y=0 ?

The solution for 9y^{''}+12y^'+5y=0 is y=e^{-(2t)/3}(c_{1}cos(t/3)+c_{2}sin(t/3))

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9y^{''}+12y^'+5y=0

9 y^{^{''}} + 12 y^{^{'}} + 5 y = 0

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{t}{3}) + c_{2} sin (\frac{t}{3}))

Solution steps

9 y^{''} (t) + 12 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{t}{3}) + c_{2} sin (\frac{t}{3}))

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{t}{3}) + c_{2} sin (\frac{t}{3}))

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

2 y^{^{''}} + 4 y^{^{'}} + 5 y = 0

y^{^{''}} - 10 y^{^{'}} - 24 y = 0

\frac{d ^{2} y}{d x ^{2}} + 100 y = 0

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 12 y = 0

What is the solution for 9y^{''}+12y^'+5y=0 ?
The solution for 9y^{''}+12y^'+5y=0 is y=e^{-(2t)/3}(c_{1}cos(t/3)+c_{2}sin(t/3))