Example 1

Consider

\[y' = x^2 e^y.\]

To solve this we rewrite as

\[\begin{split} \frac{dy}{dx} &= x^2 e^y\\ e^{-y} dy &= x^2\,dx\\ \int e^{-y}\,dy &= \int x^2\,dx\\ -e^{-y} &= \frac{1}{3} x^3 + c\\ -y &= \ln \left(-\frac{1}{3}x^3 + C \right)\\ y &= -\ln \left( -\frac{1}{3}x^3 + C \right). \end{split}\]

Example 2

Consider $y' - t = ty^2.$

This can be solved by

\[\begin{split} \frac{dy}{dt} &= t + ty^2\\ \frac{dy}{dt}&= t(1+y^2)\\ \frac{dy}{1+y^2} &= t\,dt\\ \int \frac{dy}{1+y^2} &= \int t\,dt\\ \arctan(y)&= \frac{1}{2}t^2 + c\\ y &= \tan\left( \frac{1}{2}t^2 +C \right). \end{split}\]

Example 3

Consider

$\frac{dy}{dt} = \frac{1}{\pi}\left(3+ \sin t \right) y$ We will write $\exp(x) = e^x$.

This can be solved by

\[\begin{split} \frac{dy}{y} &= \frac{3+\sin t}{\pi}\,dt\\ \int \frac{dy}{y} &= \int \frac{3+\sin t}{\pi} \,dt\\ \ln |y| &= \frac{3}{\pi}t - \frac{1}{\pi} \cos t + C\\ |y| &= \exp\left(\frac{3}{\pi}t -\frac{1}{\pi}\cos t + C \right)\\ y &= \pm e^C\exp\left(\frac{3}{\pi}t -\frac{1}{\pi}\cos t \right)\\ y&= A\exp\left(\frac{3}{\pi}t -\frac{1}{\pi}\cos t \right). \end{split}\]