What is the solution for 9y^{''}+6y^'+4y=0 ?

The solution for 9y^{''}+6y^'+4y=0 is y=e^{-t/3}(c_{1}cos(t/(sqrt(3)))+c_{2}sin(t/(sqrt(3))))

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9y^{''}+6y^'+4y=0

9 y^{^{''}} + 6 y^{^{'}} + 4 y = 0

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

Solution steps

9 y^{''} (t) + 6 y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

\int 2 e^{t} d t

7 cos (9 x), 7 - 7 cos (9 x), 0, \frac{π}{9}

x y^{^{'}} + 2 y = x^{2}

\int_{0}^{1} \frac{9}{x ( 1 + x )} d x

\frac{d y}{d x} = \frac{x}{y}, y (0) = - 5

What is the solution for 9y^{''}+6y^'+4y=0 ?
The solution for 9y^{''}+6y^'+4y=0 is y=e^{-t/3}(c_{1}cos(t/(sqrt(3)))+c_{2}sin(t/(sqrt(3))))