Answer:
[tex]y=\frac{1}{8}(x^{2} +4Cx+C^{2}ln(x) )+C2[/tex]
Step-by-step explanation:
First we have to find the slope of the family curves (m1=dy/dx)
[tex]y=\frac{2}{x^{2} +C} , \frac{dy}{dx}= \frac{-4x}{(x^{2} +C)^{2}}[/tex]
The slope of the perpendicular family of curves will be:
[tex]m2=-\frac{1}{m1} =\frac{dy2}{dx}=\frac{(x^{2}+C)^{2}}{4x}=\frac{x^{2}+2Cx+C^{2}}{4x}\\ \int{} \, dy =\frac{1}{4} \int{(x+2C+\frac{C^{2}}{x})} \, dx \\y=\frac{1}{8} (x^{2}+4Cx+C^{2}ln(x))+C2[/tex]
Note that the associated D.E to the orthogonal family of curves is solved by using separated variables. C2 is the constant getting out from the integrals.
