JEE Main · 2004 · 24 Apr 2004 · Offline · Differential Equations

Treat $x$ as the dependent variable. Dividing $ydx+(x+x^{2}y)dy=0$ by $ydy$ and rearranging gives $\frac{dx}{dy}+\frac{x}{y}=-x^2$, a Bernoulli equation. Dividing by $x^2$ and substituting $v=\frac{1}{x}$, so that $\frac{dv}{dy}=-\frac{1}{x^2}\frac{dx}{dy}$, turns it into the linear equation $\frac{dv}{dy}-\frac{v}{y}=1$. The integrating factor $\frac{1}{y}$ gives $\frac{d}{dy}\left(\frac{v}{y}\right)=\frac{1}{y}$, so $\frac{v}{y}=\log y+C$, that is $v=y\log y+Cy$. Replacing $v=\frac{1}{x}$ and dividing by $y$ gives $\frac{1}{xy}=\log y+C$, which rearranges to $-\frac{1}{xy}+\log y=C$.