Question

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

• A. a system of parabolas
• B. a system of circles
• C. $y^2=x(1+x)-1$
• D. $(x-2{)}^2+(y-3{)}^2=5$

Answer 1

Answer: (C)

Rewriting the given equation as
$2xy\frac{dy}{dx}-y^2=1+x^2$
$\Rightarrow 2y\frac{dy}{dx}-\frac{1}{x}{y}^2=\frac{1}{x}+x$.
Putting $y^2=u,$ we have
$\frac{du}{dx}-\frac{1}{x}u=\frac{1}{x}+x.$
The $I.F$. of this equation is $e^{-\int \frac{1}{x}dx}=\frac{1}{x}$
$u.\frac{1}{x}=\int \left(\frac{1}{x^2}+1\right)dx=-\frac{1}{x}+x+C$
$\Rightarrow y^2=\left(x^2-1\right)+Cx$.
Since $y(1)=1$ so $C=1,y^2=x(1+x)-1$
which represents a system of hyperbola.