Question

Resolva a EDO separável $(ay x^{2} +by)y' -x = 0$

Answer 1

                 $(a x^{2} +b)yy' -x = 0\\ \\ yy'=\dfrac{x}{a x^{2} +b}\\ \\ \\ y~dy=\dfrac{x}{a x^{2} +b}dx\\ \\ \\ \displaystyle \int y~dy=\int\dfrac{x}{a x^{2} +b}dx\\ \\ \\ \dfrac{y^2}{2}=\dfrac{1}{2a}\int\dfrac{2ax}{a x^{2} +b}dx\\ \\ \\ y^2=\dfrac{1}{a}\int\dfrac{d(a x^{2} +b)}{a x^{2} +b}\\ \\ \\ \boxed{\boxed{y^2=\dfrac{1}{a}\ln\left|a x^{2} +b\right|+C}}$