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

The correct option is A hyperbola

$\frac{dy}{dx}=\frac{x^2+y^2+1}{2xy}$

$⟹2xydy=(x^2+1)dx+y^{2}dx$

$⟹\frac{2xydy-y^{2}dx}{x^2}=\left(\frac{x^2+1}{x^2}\right)dx$

$⟹\frac{d\left(x{y}^2\right)}{x^2}=\left(\frac{x^2+1}{x^2}\right)dx$

$⟹\int d\left(\frac{y^2}{x}\right)=\int (1+\frac{1}{x^2})dx$

$⟹\frac{y^2}{x}=x-\frac{1}{x}+C$

$⟹y^2=(x^2-1+Cx)$

When x = 1, y = 0

$⟹0=1-1+C$

$⟹C=0$

$\therefore$ The solution is $x^2-y^2=1$ i.e., hyperbola.