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

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

d er i v a t i v e 65 x^{\frac{9}{10}}

\int cos (x) csc^{6} (x) d x

\int \frac{1}{3 + y} d y

\int_{0}^{1} \frac{1}{1 + 9 x ^{2}} d 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