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}))

\frac{d}{d x} ((1 - x) e^{- x})

x \to 0 lim (\frac{4 sin ( 3 x )}{2 tan ( 5 x )})

x \to (\infty) lim (\frac{ln ( x )}{x})

x \to 1 lim (\frac{( x ^{2} )}{x - 1})

\int_{0}^{1} \frac{3 x ^{2}}{1 + x ^{3}} d x

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))))