What is the solution for (dy)/(dt)+t^2y=t^2 ?

The solution for (dy)/(dt)+t^2y=t^2 is y=e^{-(t^3)/3+c_{1}}+1

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(dy)/(dt)+t^2y=t^2

\frac{d y}{d t} + t^{2} y = t^{2}

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

Solution steps

\frac{d y}{d t} + t^{2} y = t^{2}

Solve separable ODE:

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

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

d y = - y d x

x^{^{'}} = (\frac{2}{t} - t) x

\frac{2 x - 6}{25} + \frac{2 ( y - 5 ) ( \frac{d y}{d x} )}{9} = 1

(x^{2} + 6 x^{2} y^{2}) d x + e^{x^{3}} y d y = 0

y^{\frac{9}{2}} \frac{d y}{d x} - y^{4} = - y^{\frac{11}{2}}

What is the solution for (dy)/(dt)+t^2y=t^2 ?
The solution for (dy)/(dt)+t^2y=t^2 is y=e^{-(t^3)/3+c_{1}}+1