Question

Consider the differential equation $y^{2}dx+(x-\frac{1}{y})dy=0$. It $y\left(1\right)=1$, then x is given by?

Answer 1

$\begin{array}{c}{y}^{2}dx+\left(x-\frac{1}{y}\right)dy=0,aty\left(1\right)=1,thenisgivenby\\ \frac{dy}{dx}=\frac{y^2}{\left(x-\frac{1}{y}\right)}\\ \Rightarrow \frac{dy}{dx}=\frac{y^3}{\left(xy-1\right)}\\ \int \left(xy-1\right)dy=\int y^{3}dx\\ x\int ydy-\int dy=\int y^{3}dx\\ \left[\frac{x{y}^2}{2}-y^2\right]=\left[y^3.x+c\right]\\ \left[\frac{1}{2}-1\right]=\left[1\times 1+c\right]\\ \frac{-1}{2}-1=c\\ \therefore c=\frac{-3}{2}\\ Putc=\frac{-3}{2}\\ \frac{x{y}^2}{2}-y=y^{3}x\frac{-3}{2}\\ \frac{x{y}^2}{2}-y-y^{3}x=\frac{-3}{2}\\ y\left(\frac{xy}{2}-y^{2}x-1\int \frac{-3}{2}\right)Ans.\end{array}$