**Example:** Find an explicit solution to the initial value problem
$$\dfrac{y'(t)}{\sin(t)} = y^2+1; y \left( \dfrac{\pi}{2} \right)=1.$$
**Solution:** This equation is separable, yielding
$$\displaystyle\int \dfrac{1}{y^2+1} dy = \displaystyle\int \sin(t) dt,$$
or
$$\arctan(y) = -\cos(t)+C.$$
Now the initial condition $y \left( \dfrac{\pi}{2} \right)=1$ and the fact that $\arctan(1)=\dfrac{\pi}{4}$ and $\cos \left( \dfrac{\pi}{2} \right)=0$ yields
$$\dfrac{\pi}{4} = C.$$
Thus we have derived the implicit solution
$$\arctan(y) = -\cos(t) + \dfrac{\pi}{4}.$$
To make this explicit, take the tangent of both sides to get
$$y(t) = \tan \left( - \cos(t) + \dfrac{\pi}{4} \right).$$
Note: Wolfram alpha suggests the solution looks like $\cot \left( \dfrac{\pi}{4} + \cos(t) \right)$. Of course this is an equivalent way to write the same thing (look under "alternate forms").