JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Differential Equations

Rewrite the equation as $xy\frac{dy}{dx}=y^2-x^3$, which on dividing by $x$ becomes $\frac{dy}{dx}=\frac{y}{x}-\frac{x^2}{y}$. The substitution $u=y^2$, with $\frac{du}{dx}=2y\frac{dy}{dx}$, converts it into the linear equation $\frac{du}{dx}-\frac{2}{x}u=-2x^2$. Its integrating factor is $x^{-2}$, and multiplying through gives $\frac{d}{dx}\left(\frac{u}{x^2}\right)=-2$, so that $\frac{u}{x^2}=-2x+c^{\prime }$. Restoring $u=y^2$ yields $y^2=-2x^3+c^{\prime }{x}^2$, which rearranges to $y^2+2x^3+c{x}^2=0$.