Let C1 be the curve obtained by the solution of differential equation 2xydy \over dx = y^2 - x^2,x \gt 0 . Let the curve C2 be the solution of 2xy \over x^2 - y^2 = dy \over dx . If both the curves pass through (1, 1), then the area enclosed by the curves C1 and C2 is equal to :

Differential Equations PYQ — JEE Main 2021 · 16 Mar · Shift 2 · Q31 with Solution

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Correct answer: D

Solve for curve $C_1$

Given $2xy\frac{dy}{dx}=y^2-x^2, \qquad x>0$ So, $\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$ This is homogeneous. Put $y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$ Substitute: $v+x\frac{dv}{dx}=\frac{v^2x^2-x^2}{2x\cdot vx}=\frac{v^2-1}{2v}$ Hence $x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{v^2-1-2v^2}{2v}=-\frac{v^2+1}{2v}$ So $\frac{2v}{v^2+1}\,dv=-\frac{dx}{x}$ Integrating, $\ln(v^2+1)=-\ln x + C$ $x(v^2+1)=C$ Now $v=\frac{y}{x}$ , so

\Rightarrow \frac{x^2+y^2}{x}=C$$ $$x^2+y^2=Cx$$ Since it passes through $(1,1)$, $$1+1=C\cdot 1 \Rightarrow C=2$$ Thus $$C_1: x^2+y^2=2x$$ which is $$(x-1)^2+y^2=1$$ This is a circle centered at $(1,0)$ with radius $1$. --- 2. **Solve for curve $C_2$** Given $$\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$$ Again homogeneous. Put $$y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$$ Then $$v+x\frac{dv}{dx}=\frac{2x\cdot vx}{x^2-v^2x^2}=\frac{2v}{1-v^2}$$ So $$x\frac{dv}{dx}=\frac{2v}{1-v^2}-v =\frac{2v-v(1-v^2)}{1-v^2} =\frac{v+v^3}{1-v^2} =\frac{v(1+v^2)}{1-v^2}$$ Thus $$\frac{1-v^2}{v(1+v^2)}\,dv=\frac{dx}{x}$$ Now $$\frac{1-v^2}{v(1+v^2)}=\frac{1}{v}-\frac{2v}{1+v^2}$$ Therefore $$\int\left(\frac{1}{v}-\frac{2v}{1+v^2}\right)dv=\int \frac{dx}{x}$$ $$\ln|v|-\ln(1+v^2)=\ln x + C$$ $$\ln\left|\frac{v}{1+v^2}\right|=\ln x + C$$ $$\frac{v}{1+v^2}=Cx$$ Substitute $v=\frac{y}{x}$: $$\frac{y/x}{1+y^2/x^2}=Cx \Rightarrow \frac{yx}{x^2+y^2}=Cx$$ $$\frac{y}{x^2+y^2}=C$$ So $$x^2+y^2=ky$$ for some constant $k$. Using $(1,1)$, $$1+1=k\cdot 1 \Rightarrow k=2$$ Hence $$C_2: x^2+y^2=2y$$ which is $$x^2+(y-1)^2=1$$ This is a circle centered at $(0,1)$ with radius $1$. --- 3. **Find intersection points** The two circles are $$x^2+y^2=2x$$ $$x^2+y^2=2y$$ Subtracting, $$2x=2y \Rightarrow x=y$$ Substitute into one circle: $$2x^2=2x \Rightarrow x(x-1)=0$$ So intersections are $$(0,0), \quad (1,1)$$ --- 4. **Identify enclosed region** Both circles have radius $1$. Distance between centers $(1,0)$ and $(0,1)$ is $$d=\sqrt{(1-0)^2+(0-1)^2}=\sqrt{2}$$ The common enclosed region is the lens formed by two equal circles. At each center, the chord joining the intersection points subtends angle $90^\circ$ because triangle formed by the two centers and either intersection point is isosceles right: $$\cos\theta=\frac{d}{2r}=\frac{\sqrt2}{2} \Rightarrow \theta=45^\circ$$ Hence sector angle is $$2\theta=90^\circ=\frac{\pi}{2}$$ Area of one sector: $$\frac{1}{2}r^2\cdot \frac{\pi}{2}=\frac{\pi}{4}$$ Since $r=1$. Area of corresponding triangle: $$\frac{1}{2}\cdot 1\cdot 1=\frac{1}{2}$$ So one segment area is $$\frac{\pi}{4}-\frac{1}{2}$$ There are two such equal segments, so enclosed area is $$2\left(\frac{\pi}{4}-\frac{1}{2}\right)=\frac{\pi}{2}-1$$ --- 5. **Final answer** $$\boxed{\frac{\pi}{2}-1}$$ So the correct option is **D**.

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Correct answer: D

Solve for curve $C_1$

\Rightarrow \frac{x^2+y^2}{x}=C$$ $$x^2+y^2=Cx$$ Since it passes through $(1,1)$, $$1+1=C\cdot 1 \Rightarrow C=2$$ Thus $$C_1: x^2+y^2=2x$$ which is $$(x-1)^2+y^2=1$$ This is a circle centered at $(1,0)$ with radius $1$. --- 2. **Solve for curve $C_2$** Given $$\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$$ Again homogeneous. Put $$y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$$ Then $$v+x\frac{dv}{dx}=\frac{2x\cdot vx}{x^2-v^2x^2}=\frac{2v}{1-v^2}$$ So $$x\frac{dv}{dx}=\frac{2v}{1-v^2}-v =\frac{2v-v(1-v^2)}{1-v^2} =\frac{v+v^3}{1-v^2} =\frac{v(1+v^2)}{1-v^2}$$ Thus $$\frac{1-v^2}{v(1+v^2)}\,dv=\frac{dx}{x}$$ Now $$\frac{1-v^2}{v(1+v^2)}=\frac{1}{v}-\frac{2v}{1+v^2}$$ Therefore $$\int\left(\frac{1}{v}-\frac{2v}{1+v^2}\right)dv=\int \frac{dx}{x}$$ $$\ln|v|-\ln(1+v^2)=\ln x + C$$ $$\ln\left|\frac{v}{1+v^2}\right|=\ln x + C$$ $$\frac{v}{1+v^2}=Cx$$ Substitute $v=\frac{y}{x}$: $$\frac{y/x}{1+y^2/x^2}=Cx \Rightarrow \frac{yx}{x^2+y^2}=Cx$$ $$\frac{y}{x^2+y^2}=C$$ So $$x^2+y^2=ky$$ for some constant $k$. Using $(1,1)$, $$1+1=k\cdot 1 \Rightarrow k=2$$ Hence $$C_2: x^2+y^2=2y$$ which is $$x^2+(y-1)^2=1$$ This is a circle centered at $(0,1)$ with radius $1$. --- 3. **Find intersection points** The two circles are $$x^2+y^2=2x$$ $$x^2+y^2=2y$$ Subtracting, $$2x=2y \Rightarrow x=y$$ Substitute into one circle: $$2x^2=2x \Rightarrow x(x-1)=0$$ So intersections are $$(0,0), \quad (1,1)$$ --- 4. **Identify enclosed region** Both circles have radius $1$. Distance between centers $(1,0)$ and $(0,1)$ is $$d=\sqrt{(1-0)^2+(0-1)^2}=\sqrt{2}$$ The common enclosed region is the lens formed by two equal circles. At each center, the chord joining the intersection points subtends angle $90^\circ$ because triangle formed by the two centers and either intersection point is isosceles right: $$\cos\theta=\frac{d}{2r}=\frac{\sqrt2}{2} \Rightarrow \theta=45^\circ$$ Hence sector angle is $$2\theta=90^\circ=\frac{\pi}{2}$$ Area of one sector: $$\frac{1}{2}r^2\cdot \frac{\pi}{2}=\frac{\pi}{4}$$ Since $r=1$. Area of corresponding triangle: $$\frac{1}{2}\cdot 1\cdot 1=\frac{1}{2}$$ So one segment area is $$\frac{\pi}{4}-\frac{1}{2}$$ There are two such equal segments, so enclosed area is $$2\left(\frac{\pi}{4}-\frac{1}{2}\right)=\frac{\pi}{2}-1$$ --- 5. **Final answer** $$\boxed{\frac{\pi}{2}-1}$$ So the correct option is **D**.

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