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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (x^{\frac{1}{3}} y^{3})

\int \frac{5 x - 4}{2 x ^{2} + x - 1} d x

t \to 0 lim (\frac{sin ( 4 t )}{t})

65 x^{\frac{9}{10}}

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

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