JEE Main 2018 (Online) 15th April Evening ShiftQuestion 1
The curve satisfying the differential equation $\left(x^{2}-y^{2}\right)dx+2xy\,dy=0$ and passing through the point $(1,1)$ is
Differential Equation
Solution
Concept. $2y\,dy=dw$ with $w=y^2$; the equation becomes linear in $w$. Formulas used. (1) $\left(\dfrac wx\right)'=\dfrac{xw'-w}{x^2}$Step 1 - Rewrite. $2xy\,y'=y^2-x^2$, i.e. $xw'-w=-x^2$. Step 2 - Integrate. By (1), $\left(\dfrac wx\right)'=-1$, so $w=-x^2+Cx$ and $x^2+y^2=Cx$. Step 3 - Through $(1,1)$. $C=2$: $(x-1)^2+y^2=1$, a circle of radius $1$. Answer: (B)