What is the solution for y^{'''}+y^'=t^2+1 ?

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

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y^{'''}+y^'=t^2+1

y^{^{'''}} + y^{^{'}} = t^{2} + 1

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

Solution steps

y^{'''} (t) + y^{'} (t) = t^{2} + 1

Solve linear ODE:

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

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

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 25

y^{^{'''}} + 4 y^{^{''}} + 8 y^{^{'}} = 11 + 5 e^{2 x}

y^{^{''}} - 9 y^{^{'}} = 81 x^{2} + 7, y (0) = - 1, y^{^{'}} (0) = 2

y^{^{''}} - y^{^{'}} - 6 y = 14 sin (3)

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