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

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

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

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

y = c_{1} e^{t} + c_{2} - \frac{t ^{3}}{3} - t^{2} - 2 t

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} - \frac{t ^{3}}{3} - t^{2} - 2 t

y = c_{1} e^{t} + c_{2} - \frac{t ^{3}}{3} - t^{2} - 2 t

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