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

VITEEEVITEEE 2014Differential Equations

Solution:

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.