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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 14 y^{^{'}} + 49 y = 21 x + 6

y^{^{''}} + 2 y^{^{'}} + 2 y = x + cos (x)

D (t) = 30 e^{- 0.1 (3)}

2 y^{^{''}} - y^{^{'}} - 6 y = - 7 e^{2 x}

y^{^{''}} - 8 y^{^{'}} + 16 = 0, y (0) = 4, y (1) = 5

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