What is the solution for 9y^{''}+24y^'+16y=0 ?

The solution for 9y^{''}+24y^'+16y=0 is y=c_{1}e^{-(4t)/3}+c_{2}e^{-(4t)/3}t

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9y^{''}+24y^'+16y=0

9 y^{^{''}} + 24 y^{^{'}} + 16 y = 0

y = c_{1} e^{- \frac{4 t}{3}} + c_{2} e^{- \frac{4 t}{3}} t

Solution steps

9 y^{''} (t) + 24 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{4 t}{3}} + c_{2} e^{- \frac{4 t}{3}} t

y = c_{1} e^{- \frac{4 t}{3}} + c_{2} e^{- \frac{4 t}{3}} t

(\frac{1}{1 + x ^{2}})^{^{'}}

\int_{- 1}^{5} 3 x^{2} + 6 x d x

y = 4 x^{3} - 4 x, (- 1, 0)

x \to - 2 lim (- 3)

\int_{0}^{1} (\frac{1}{x ^{2}}) d x

What is the solution for 9y^{''}+24y^'+16y=0 ?
The solution for 9y^{''}+24y^'+16y=0 is y=c_{1}e^{-(4t)/3}+c_{2}e^{-(4t)/3}t