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

Answer 1

The correct option is A a circle with centre on the x-axis

$(x^2-y^2)dx+2xydy=0$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\left[\frac{y^2-x^2}{xy}\right]$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\frac{y}{x}[1-{\left(\frac{x}{y}\right)}^2]$
Put $y=vx\therefore \frac{dy}{dx}=x\frac{dv}{dx}+v$
$\Rightarrow x\frac{dv}{dx}+v=\frac{1}{2}v(1-\frac{1}{v^2})$
$\Rightarrow \int \frac{2v}{1+v^2}dv=-\int \frac{dx}{x}$
$\Rightarrow \ln (v^2+1)=-\ln x+\ln C$
$\Rightarrow y^2+x^2=Cx$
Passes through $(1,1)$
So, $C=2$
$\therefore y^2+x^2=2x$
$\Rightarrow x^2+y^2-2x=0$
$\Rightarrow (x-1{)}^2+y^2=1$
Hence, It is a circle with centre on $x-\text{axis}.$