# Let C1 be the curve obtained by the solution of differential

Question:

Let $C_{1}$ be the curve obtained by the solution of differential equation $2 x y \frac{d y}{d x}=y^{2}-x^{2}, x>0 \&$ Let the curve $C_{2}$ be the solution of $\frac{2 x y}{x^{2}-y^{2}}=\frac{d y}{d x}$

If both the curves pass through $(1,1)$, then the area enclosed by the curves $C_{1}$ and $\mathrm{C}_{2}$ is equal to :

1. (1) $\pi-1$

2. (2) $\frac{\pi}{2}-1$

3. (3) $\pi+1$

4. (4) $\frac{\pi}{4}+1$

Correct Option: , 2

Solution:

$\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}, \quad x \in(0, \infty)$

put $y=v x$

$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v}=\frac{\mathrm{v}^{-}-1}{2 \mathrm{v}}$

$\frac{2 v}{v^{2}+1} d v=-\frac{d x}{x}$

Integrate,

$\ln \left(v^{2}+1\right)=-\ln x+C$

$\ln \left(\frac{y^{2}}{x^{2}}+1\right)=-\ln x+C$

put $x=1, y=1, C=\ln 2$

$\ln \left(\frac{y^{2}}{x^{2}}+1\right)=-\ln x+\ln 2$

$\Rightarrow x^{2}+y^{2}-2 x=0 \quad\left(\right.$ Curve $\left.C_{1}\right)$

Similarly,

$\frac{d y}{d x}=\frac{2 x y}{x^{2}-y^{2}}$

Put $y=v x$

$x^{2}+y^{2}-2 y=0$