What is the solution for 9y^{''}+12y^'+4y=0 ?

The solution for 9y^{''}+12y^'+4y=0 is y=c_{1}e^{-(2t)/3}+c_{2}e^{-(2t)/3}t

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9y^{''}+12y^'+4y=0

9 y^{^{''}} + 12 y^{^{'}} + 4 y = 0

y = c_{1} e^{- \frac{2 t}{3}} + c_{2} e^{- \frac{2 t}{3}} t

Solution steps

9 y^{''} (t) + 12 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{2 t}{3}} + c_{2} e^{- \frac{2 t}{3}} t

y = c_{1} e^{- \frac{2 t}{3}} + c_{2} e^{- \frac{2 t}{3}} t

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''''}} - 14 y^{^{''}} - 32 y = 0

y^{^{''}} - 23 y^{^{'}} + 3 y = 0

y^{^{'''}} - 2 y^{^{''}} - 5 y^{^{'}} + 6 y = 0

2 y^{^{''}} + 20 y^{^{'}} + 51 y = 0, y (0) = 2, y^{^{'}} (0) = 0

What is the solution for 9y^{''}+12y^'+4y=0 ?
The solution for 9y^{''}+12y^'+4y=0 is y=c_{1}e^{-(2t)/3}+c_{2}e^{-(2t)/3}t