What is the solution for 3y^{''}-2y^'+4y=0 ?

The solution for 3y^{''}-2y^'+4y=0 is y=e^{t/3}(c_{1}cos((sqrt(11)t)/3)+c_{2}sin((sqrt(11)t)/3))

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3y^{''}-2y^'+4y=0

3 y^{^{''}} - 2 y^{^{'}} + 4 y = 0

y = e^{\frac{t}{3}} (c_{1} cos (\frac{11 t}{3}) + c_{2} sin (\frac{11 t}{3}))

Solution steps

3 y^{''} (t) - 2 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{\frac{t}{3}} (c_{1} cos (\frac{11 t}{3}) + c_{2} sin (\frac{11 t}{3}))

y = e^{\frac{t}{3}} (c_{1} cos (\frac{11 t}{3}) + c_{2} sin (\frac{11 t}{3}))

\frac{d}{d x} (2 sin (x^{2} - 1))

\int \frac{1}{- x ^{2} + 4 x + 12} d x

\frac{x ^{2} + 5 x}{x + 2}

n = 1 \sum \infty \frac{8}{n ^{5}}

e^{- 2 x}

What is the solution for 3y^{''}-2y^'+4y=0 ?
The solution for 3y^{''}-2y^'+4y=0 is y=e^{t/3}(c_{1}cos((sqrt(11)t)/3)+c_{2}sin((sqrt(11)t)/3))