What is the solution for 9y^{''}-18y^'+11y=0 ?

The solution for 9y^{''}-18y^'+11y=0 is y=e^t(c_{1}cos((sqrt(2)t)/3)+c_{2}sin((sqrt(2)t)/3))

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9y^{''}-18y^'+11y=0

9 y^{^{''}} - 18 y^{^{'}} + 11 y = 0

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

Solution steps

9 y^{''} (t) - 18 y^{'} (t) + 11 y = 0

Solve linear ODE:

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

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

x \to - 4 lim (\frac{2 x + 3}{x + 4})

\frac{d}{d x} ((a - x) a + x)

\frac{d}{d x} (e^{2 x} cos (π x))

\int sin (ln (8 x)) d x

arcsin (x)

What is the solution for 9y^{''}-18y^'+11y=0 ?
The solution for 9y^{''}-18y^'+11y=0 is y=e^t(c_{1}cos((sqrt(2)t)/3)+c_{2}sin((sqrt(2)t)/3))