Question

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

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

Answer 1

Answer: (A), (C)

$y^2=x(1+x)-1$
a system of hyperbola

Rewriting, the given equation as
$2xy\frac{dy}{dx}-y^2=1+x^2$

$\Rightarrow 2y\frac{dy}{dx}-\frac{1}{x}{y}^2=\frac{1}{x}+x$.

Substitute $y^2=u$, we have $\frac{du}{dx}-\frac{1}{x}u=\frac{1}{x}+x$. The I.F. of this equation is $1/x$, so

$u.\frac{1}{x}=\int (\frac{1}{x^2}+1)dx=-\frac{1}{x}+x+C$

$\Rightarrow y^2=(x^2-1)+Cx$. Since $y\left(1\right)=1$ so $C=1$

Hence $y^2=x(1+x)-1$ which represents a system of hyperbola.