6.10 Further example

Solution. Dividing both sides by $(y^{2}+1)$, we obtain

and hence $\tan^{-1}y=\frac{x^{2}}{2}-x+c$. To get this into the form $y=f(x)$ we apply $\tan$ to both sides to get the general solution $y={\tan(\frac{x^{2}}{2}-x+c).}$ For the particular solution, we equate $y(0)={\tan c}=1$, so we have $c={\frac{\pi}{4}.}$ (Adding an integer multiple of $\pi$ to $c$ has no effect on the function, so we may assume $c\in[0,\pi)$.)