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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 12 y = 80 sin (2 x)

y^{^{''}} - y^{^{'}} + y = 3 sin (t) - 27 e^{2 t}

\frac{d ^{2} y}{d t ^{2}} + \frac{3}{100} \frac{d y}{d t} = 12

y^{^{''}} = - y + cos (2 t)

y^{^{''}} + 5 y^{^{'}} + 2 = 0

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