What is the solution for y^{'''}+4y^'=t^2+2t ?

The solution for y^{'''}+4y^'=t^2+2t is y=c_{1}+c_{2}cos(2t)+c_{3}sin(2t)+(t^3}{12}+\frac{t^2)/4-t/8

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y^{'''}+4y^'=t^2+2t

y^{^{'''}} + 4 y^{^{'}} = t^{2} + 2 t

y = c_{1} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{t ^{3}}{12} + \frac{t ^{2}}{4} - \frac{t}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{t ^{3}}{12} + \frac{t ^{2}}{4} - \frac{t}{8}

y = c_{1} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{t ^{3}}{12} + \frac{t ^{2}}{4} - \frac{t}{8}

What is the solution for y^{'''}+4y^'=t^2+2t ?
The solution for y^{'''}+4y^'=t^2+2t is y=c_{1}+c_{2}cos(2t)+c_{3}sin(2t)+(t^3}{12}+\frac{t^2)/4-t/8