Question

Let $C_1$ be the curve obtained by solution of differential equation $2xy\frac{dy}{dx}=y^2-x^2,$ $x>0.$ Let the curve $C_2$ be the solution of $\frac{2xy}{x^2-y^2}=\frac{dy}{dx}.$ If both the curves pass through $(1,1),$ then the area enclosed by the curves $C_1$ and $C_2$ is equal to :

• A. $\frac{\pi }{2}-1$
• B. $\frac{\pi }{4}+1$
• C. $\pi -1$
• D. $\pi +1$

Answer 1

The correct option is A $\frac{\pi }{2}-1$

$2xy\frac{dy}{dx}=y^2-x^2$
$\Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}$
Put $y=vx$
$v+x\frac{dv}{dx}=\frac{v^2{x}^2-x^2}{2v{x}^2}=\frac{v^2-1}{2v}$
$\Rightarrow x\frac{dv}{dx}=\frac{v^2-1-2v^2}{2v}=-\frac{(v^2+1)}{2v}$
$\Rightarrow \frac{2v}{v^2+1}dv=-\frac{dx}{x}$
Integrating both sides,
$\ln (v^2+1)=-\ln x+\ln c$
$\Rightarrow v^2+1=\frac{c}{x}$
$\Rightarrow \frac{y^2}{x^2}+1=\frac{c}{x}$
$\Rightarrow x^2+y^2=cx$
It passes through $(1,1)$
$\therefore x^2+y^2-2x=0$

Similarly, for second differential equation $\frac{dx}{dy}=\frac{x^2-y^2}{2xy},$
equation of curve is $x^2+y^2-2y=0$


Required area
$=2\times (\frac{1}{4}\times \pi \times 1^2-\frac{1}{2}\times 1\times 1)$
$=(\frac{\pi }{2}-1)$ sq. units