What is the solution for 9y^{''}-30y^'+25y=0 ?

The solution for 9y^{''}-30y^'+25y=0 is y=c_{1}e^{(5t)/3}+c_{2}e^{(5t)/3}t

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9y^{''}-30y^'+25y=0

9 y^{^{''}} - 30 y^{^{'}} + 25 y = 0

y = c_{1} e^{\frac{5 t}{3}} + c_{2} e^{\frac{5 t}{3}} t

Solution steps

9 y^{''} (t) - 30 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{5 t}{3}} + c_{2} e^{\frac{5 t}{3}} t

y = c_{1} e^{\frac{5 t}{3}} + c_{2} e^{\frac{5 t}{3}} t

\int \frac{2}{5} x^{\frac{5}{2}} - \frac{54}{5} d x

\frac{\partial}{\partial x} (y + \frac{1}{x})

\int e^{- y x} d x

\frac{d}{d x} (\frac{6 x}{x - 1})

\int y e^{x} + sin (y) d x

What is the solution for 9y^{''}-30y^'+25y=0 ?
The solution for 9y^{''}-30y^'+25y=0 is y=c_{1}e^{(5t)/3}+c_{2}e^{(5t)/3}t