The curve satisfying the differential equation, (x^2-y^2 ) d x+2 x y d y=0 and passing through the point (1,1) is

$\left(x^2-y^2\right) d x+2 x y d y=0$(x2−y2)dx+2xydy=0 $$\Rightarrow \frac{d y}{d x}=\frac{y^2-x^2}{2 x y}$$⇒dxdy​=2xyy2−x2​ Let $y=v x$y=vx $$\begin{aligned} & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ \Rightarrow & v+x \frac{d v}{d x}=\frac{v^2 x^2-x^2}{2 v x^2} \\ \Rightarrow & v+x \frac{d v}{d x}=\frac{v^2-1}{2 v} \\ \Rightarrow & x \frac{d v}{d x}=\frac{-v^2-1}{2 v} \\ \Rightarrow & \frac{2 v d v}{v^2+1}=-\frac{d x}{x} \end{aligned}$$⇒⇒⇒⇒​dxdy​=v+xdxdv​v+xdxdv​=2vx2v2x2−x2​v+xdxdv​=2vv2−1​xdxdv​=2v−v2−1​v2+12vdv​=−xdx​​ After integrating, we get $$\begin{aligned} &\ln \left|v^2+1\right|=-\ln |x|+\ln c \\ &\frac{y^2}{x^2}+1=\frac{c}{x} \end{aligned}$$​ln​v2+1​=−ln∣x∣+lncx2y2​+1=xc​​ As curve passes through the point $(1,1)$(1,1), so $1+1=c \Rightarrow c=2$1+1=c⇒c=2 $x^2+y^2-2 x=0$x2+y2−2x=0, which is a circle of radius one.