What is the solution for 9y^{''}+48y^'+64y=0 ?

The solution for 9y^{''}+48y^'+64y=0 is y=c_{1}e^{-(8t)/3}+c_{2}e^{-(8t)/3}t

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9y^{''}+48y^'+64y=0

9 y^{^{''}} + 48 y^{^{'}} + 64 y = 0

y = c_{1} e^{- \frac{8 t}{3}} + c_{2} e^{- \frac{8 t}{3}} t

Solution steps

9 y^{''} (t) + 48 y^{'} (t) + 64 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{8 t}{3}} + c_{2} e^{- \frac{8 t}{3}} t

y = c_{1} e^{- \frac{8 t}{3}} + c_{2} e^{- \frac{8 t}{3}} t

f^{^{'}} (x) = - x

\frac{d}{d x} ((arctan (9 x))^{2})

\frac{d}{d t} (a cos (10 t) + b sin (10 t))

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

\frac{d}{d x} (89 x)

What is the solution for 9y^{''}+48y^'+64y=0 ?
The solution for 9y^{''}+48y^'+64y=0 is y=c_{1}e^{-(8t)/3}+c_{2}e^{-(8t)/3}t