What is the solution for 9y^{''}-18y^'+15y=0 ?

The solution for 9y^{''}-18y^'+15y=0 is y=e^t(c_{1}cos(sqrt(2/3)t)+c_{2}sin(sqrt(2/3)t))

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9y^{''}-18y^'+15y=0

9 y^{^{''}} - 18 y^{^{'}} + 15 y = 0

y = e^{t} (c_{1} cos (\frac{2}{3} t) + c_{2} sin (\frac{2}{3} t))

Solution steps

9 y^{''} (t) - 18 y^{'} (t) + 15 y = 0

Solve linear ODE:

y = e^{t} (c_{1} cos (\frac{2}{3} t) + c_{2} sin (\frac{2}{3} t))

y = e^{t} (c_{1} cos (\frac{2}{3} t) + c_{2} sin (\frac{2}{3} t))

x \to 0 lim \frac{x ^{2} + p ^{2} - p}{x ^{2} + q ( x ) ^{2} - q ( x )}

3 sin (x)

\frac{2}{4 - x}

\int - 2 t d t

\int \frac{sec ^{2} ( x )}{tan ^{5} ( x )} d x

What is the solution for 9y^{''}-18y^'+15y=0 ?
The solution for 9y^{''}-18y^'+15y=0 is y=e^t(c_{1}cos(sqrt(2/3)t)+c_{2}sin(sqrt(2/3)t))