What is the solution for (dy)/(dt)+t^2y=1 ?

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

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(dy)/(dt)+t^2y=1

\frac{d y}{d t} + t^{2} y = 1

y = \frac{\int e ^{\frac{t ^{3}}{3}} d t}{e ^{\frac{t ^{3}}{3}}}

Solution steps

\frac{d y}{d t} + t^{2} y = 1

Solve linear ODE:

y = \frac{\int e ^{\frac{t ^{3}}{3}} d t}{e ^{\frac{t ^{3}}{3}}}

y = \frac{\int e ^{\frac{t ^{3}}{3}} d t}{e ^{\frac{t ^{3}}{3}}}

\frac{\partial}{\partial x} (e^{x} cos (y))

y^{^{'}} - \frac{2}{x ^{2}} y = 0

f (x) = 2 x^{2} ln (2 x)

(\frac{1 - x}{y}) d x = - x^{2} cos (y) d y

\frac{\partial}{\partial x} (4 - x^{2} - y)

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