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)

hyperbola

Answer 2

Given differential equation is
$\frac{dy}{dx} = \frac{x^{2} + y^{2} + 1}{2xy}$
$\Rightarrow 2xy dy = \left(x^{2} + 1\right)dx +y^{2} dx$
$\Rightarrow \frac{xd\left(y^{2}\right) -y^{2} dx}{x^{2}}$
$= \left(\frac{x^{2} + 1 }{x^{2}} \right) dx$
$\Rightarrow \int d\left(\frac{y^{2}}{x}\right) = \int\left(1+ \frac{1}{x^{2}}\right)dx$
$\Rightarrow \frac{y^{2}}{x} = x - \frac{1}{x}C$
$\Rightarrow y^{2} = \left(x^{2} - 1 + Cx\right)$
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.