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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").