What is the solution for 9y^{''}-6y^'+8y=0 ?

The solution for 9y^{''}-6y^'+8y=0 is y=e^{t/3}(c_{1}cos((sqrt(7)t)/3)+c_{2}sin((sqrt(7)t)/3))

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9y^{''}-6y^'+8y=0

9 y^{^{''}} - 6 y^{^{'}} + 8 y = 0

y = e^{\frac{t}{3}} (c_{1} cos (\frac{7 t}{3}) + c_{2} sin (\frac{7 t}{3}))

Solution steps

9 y^{''} (t) - 6 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = e^{\frac{t}{3}} (c_{1} cos (\frac{7 t}{3}) + c_{2} sin (\frac{7 t}{3}))

y = e^{\frac{t}{3}} (c_{1} cos (\frac{7 t}{3}) + c_{2} sin (\frac{7 t}{3}))

\frac{d y}{d x} = 3 x^{2} (y - 1)^{3}

\frac{d}{d x} (2 (x - 2)^{2})

sin^{2} (2 t)

\int ln (x - 3) d x

f (x) = \frac{2}{x - 5}, at x = 7

What is the solution for 9y^{''}-6y^'+8y=0 ?
The solution for 9y^{''}-6y^'+8y=0 is y=e^{t/3}(c_{1}cos((sqrt(7)t)/3)+c_{2}sin((sqrt(7)t)/3))