Question

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

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

Answer 1

Answer: (A), (C)

The correct options are
A a pair of straight lines
C $y^2=x^2$

$\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$ satisfying $y\left(1\right)=1$
$y=vx$
$x\frac{dv}{dx}+v=\frac{1+v^2}{2v}$
$\frac{2vdv}{1-v^2}=\frac{dx}{x}$
Integrating both sides
$-\ln (1-v^2)=\ln x+c$
$-\ln (1-\frac{x^2}{y^2})=\ln x+c$
Put $y\left(1\right)=1$
$c=0$
$x^2=y^2$
$x=y$ and $x=-y$