Question

Show that the differential equation $2y{e}^{x/y}dx+\left(y2x{e}^{x/y}\right)dy=0$ is homogeneous.
Find the particular solution of this differential equation, given that $x=0$ when $y=1$.

• A. $y+\log x=0$
• B. $2y+\log x=0$
• C. $2+2\log x=y$
• D. None of these

Answer 1

Answer: (A)

$y+\log x=0$

Given, $2y{e}^{x/y}dx+y2x{e}^{x/y}dy=0$
$\Rightarrow 2y{e}^{x/y}dx=-y2x{e}^{x/y}dy$
$\Rightarrow \frac{dy}{dx}=\frac{-2y{e}^{x/y}}{y2x{e}^{x/y}}$
$\Rightarrow \frac{dy}{dx}=\frac{-1}{x}$
$\Rightarrow x\frac{dy}{dx}+1=0$
Therefore, the given differential equation is
$dy=\frac{-dx}{x}$
Integrate both sides
$\int dy=\int \frac{-dx}{x}$
$y=-\log x+c$
when $y=0,x=1$
$0=-\log 1+C$
$0=0+C$
$C=0$
$\therefore y=-\log x$
$y+\log x=0$