Question

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

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

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{2}-1$

Find the area enclosed by the curves $C_1$ and $C_2$

Step 1: Given data,

$2xy\left(\frac{dy}{dx}\right)=y^2-x^2$

Simplify the equation.

$\left(\frac{dy}{dx}\right)=\frac{y^2-x^2}{2xy}$

Let,$\frac{y}{x}=v\mathrm{or}y=vx$

Differentiating with respect to $x$, we get:

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Then, on substituting, we have:

$$\begin{gathered} v+x\frac{dv}{dx}=\frac{v^2-1}{2v} \\ x\frac{dv}{dx}=-\left(\frac{v^2+1}{2v}\right) \\ \frac{2vdv}{v^2+1}=-\frac{dx}{x}...\left(1\right) \end{gathered}$$

Step 2: Integrate the Equation $\left(1\right)$

$\begin{array}{rcl}\int \frac{2vdv}{v^2+1}& =& -\int \frac{dx}{x}\\ & \Rightarrow & \log \left|v^2+1\right|=-\log x+\log c\\ & \Rightarrow & \log \left|\frac{x^2+y^2}{x^2}\right|=\log \left|\frac{c}{x}\right|\end{array}$

Put $y=1,x=1$

$\begin{array}{rcl}\left|1^2+1^2\right|& =& \left|c\right|\\ \left|c\right|& =& 2\end{array}$

So, the equation of $C_1$ will be $x^2+y^2-2x=0$.

Similarly,

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

We know that,

$$\begin{gathered} {\left(x-1\right)}^2+y^2=1 \\ {y}^2+x^2=2x \end{gathered}$$

Then,

$$\begin{gathered} \frac{dy}{dx}=\frac{2xy}{x^2-y^2} \\ lety=tx \\ \Rightarrow t+x\frac{dt}{dx}=\frac{2t}{1-t^2} \\ \Rightarrow \frac{1-t^2}{t\left(1+t^2\right)}dt=\frac{dx}{x} \end{gathered}$$

Integrating both sides.

$$\begin{gathered} \ln \left|\frac{t}{1+t^2}\right|=\ln \left|x\right|+\ln c \\ \Rightarrow \left|\frac{t}{1+t^2}\right|=\left|xc\right| \\ \Rightarrow \left|x^2+y^2\right|=\left|\frac{y}{c}\right| \end{gathered}$$

Put $y=1,x=1$

$\begin{array}{rcl}\left|1^2+1^2\right|& =& \left|\frac{1}{c}\right|\\ \left|c\right|& =& \frac{1}{2}\end{array}$

We get

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

Step-3 Bounded area :

From the above Equation draw diagram to get the bounded area

Thus, the bounded area is Equal to

As the figure is symmetric about $y=x$ line

$$\begin{gathered} \text{Bounded area}=2·\left(\frac{1}{4}\mathrm{Ar}.of{C}_1-Ar.∆OAB\right) \\ =2·\left(\frac{1}{4}\times \mathrm{\pi }\times 1\right)-\frac{1}{2}·1·1 \\ =\frac{\mathrm{\pi }}{2}-1 \end{gathered}$$

Hence, the correct answer is option (A).