Question

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

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

Answer 1

The correct option is A $(\frac{\pi }{2}-1)$ sq. units

$C_1:2xy\frac{dy}{dx}=y^2-x^2$
$\Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}$
Put $y=vx$
$\Rightarrow v+x\frac{dv}{dx}=\frac{v^2-1}{2v}$
$\Rightarrow \frac{2v}{v^2+1}dv=-\frac{dx}{x}$
Integrating both sides, we get
$\ln |v^2+1|=-\ln |x|+\ln |c|$
$\Rightarrow |x^2+y^2|=|cx|$
It passes through $(1,1)$
$\therefore |c|=2$
$\therefore C_1:x^2+y^2-2x=0$

$C_2:\frac{2xy}{x^2-y^2}=\frac{dy}{dx}$
Put $y=vx$
$\Rightarrow v+x\frac{dv}{dx}=\frac{2v}{1-v^2}$
$\Rightarrow \frac{1-v^2}{v(1+v^2)}dv=\frac{dx}{x}$
$\Rightarrow \frac{1}{v}dv-\frac{2v}{1+v^2}dv=\frac{dx}{x}$
Integrating both sides, we get
$\ln \left|\frac{v}{1+v^2}\right|=\ln |x|+\ln c$
$\Rightarrow \left|\frac{v}{1+v^2}\right|=|xc|$
$\Rightarrow |x^2+y^2|=\frac{|y|}{|c|}$
It passes through $(1,1)$
$\therefore |c|=\frac{1}{2}$
$\therefore C_2:x^2+y^2-2y=0$

Now area between $C_1$ and $C_2$

As figure is symmetric about $y=x$ line
Area of Arc$\left(OBA\right)=\frac{1}{4}$ Area of $C_1-$ Area of $△OAB$
Area of Arc$\left(OBA\right)=\frac{1}{4}\pi \cdot 1^2-\frac{1}{2}\cdot 1\cdot 1$
$\therefore$ Area enclosed by $C_1$ and $C_2=2\times$Area of Arc$\left(OBA\right)$
$\therefore$ Area enclosed by $C_1$ and $C_2=(\frac{\pi }{2}-1)$ sq. units