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

\int_{0}^{1} (3 x + 1)^{4} d x

f (x) = \frac{1}{1 + x ^{2}}, at x = 2

\frac{d x}{d t} = x^{2} + \frac{1}{49}, x (0) = 1

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0, y (0) = 6

ln (x + 1)

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