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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 6 y = 2 e^{- t}

\frac{d ^{2} y}{d x ^{2}} + y = tan (x)

y^{^{''}} + 10 y^{^{'}} + 25 y = 100 sin (5 x)

(d^{2} + 1) y = sin (x)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2 x} (cos (x) - 5 sin (x))

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