## Calculus 10th Edition

The particular solution of the given differential equation that passes through the point $(0,2)$ is $$y=\frac{-x^2+\sqrt{x^4+16}}{2}.$$
The solution is in its' implicit form $y(x^2+y)=C$ and contains an arbitrary constant $C$ and we will find it by demanding that the solution contains the point $(0,2)$ i.e. when we put $x=0$ we want to get $y=2$: $$2(0^2+2) = C\Rightarrow C=4.$$ Now we have $$y(x^2+y)=4\Rightarrow y^2+x^2y-4=0.$$ This is the quadratic equation in $y$ and its' solutions are $$y=\frac{-x^2\pm\sqrt{(x^2)^2-4\times(-4)\times1}}{2} = \frac{-x^2\pm\sqrt{x^4+16}}{2}.$$ The solution with the minus sign must be thrown away since when $x=0$ we get $$y=\frac{-0^2-\sqrt{0^4+16}}{2}=-\frac{4}{2}=-2$$ which we didn't want. If we take the plus sign then for $x=0$: $$y=\frac{-0^2+\sqrt{0^4+16}}{2}=\frac{4}{2}=2$$ which is exactly what is required in the problem. So the particular solution of the given differential equation that passes through the point $(0,2)$ is $$y=\frac{-x^2+\sqrt{x^4+16}}{2}.$$