Question

The solution of $\frac{dy}{dx}=\frac{x^2+y^2+1}{2xy}$, satisfying $y\left(1\right)=0$ is given by

• A. hyperbola
• B. circle
• C. ellipse
• D. parabola

Answer 1

Answer: (A)

Given differential equation is
$\frac{dy}{dx}=\frac{x^2+y^2+1}{2xy}$
$\Rightarrow 2xydy=\left(x^2+1\right)dx+y^{2}dx$
$\Rightarrow \frac{xd\left(y^2\right)-y^{2}dx}{x^2}$
$=\left(\frac{x^2+1}{x^2}\right)dx$
$\Rightarrow \int d\left(\frac{y^2}{x}\right)=\int \left(1+\frac{1}{x^2}\right)dx$
$\Rightarrow \frac{y^2}{x}=x-\frac{1}{x}C$
$\Rightarrow y^2=\left(x^2-1+Cx\right)$
When $x=1,y=0$
Then, $0=1-1+C$
$\Rightarrow C=0$
$\therefore$ The solution is $x^2-y^2=1$ i.e., hyperbola.