Question

The curve amongst the family of curves, represented by the differential equation, $(x^2-y^2)dx+2xydy=0$ which passes through $(1,1)$ is :

• A. A circle with centre on the y-axis
• B. A circle with centre on the x-axis
• C. An ellipse with major axis along the y-axis
• D. A hyperbola with transverse axis along the x-axis

Answer 1

Answer: (B)

A circle with centre on the x-axis

$(x^2-y^2)dx+2xydy=0$

$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$
Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

Solving we get,

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

Integrating both sides we get
$\ln (v^2+1)=-\ln x+\ln C$

After putting the value of $y$, we get
$(y^2+x^2)=Cx$

$1+1=C\Rightarrow C=2$
$y^2+x^2=2x$

$(x-1{)}^2+y^2=1$