Question

Equation of the curve which passes through the point (1, 1) and whose D.E is $(y-yx)dx+(x+xy)dy=0$ :

• A. $xy=e^{x-y}$
• B. $\frac{x}{y}=e^{x+y}$
• C. $xy=e^{x/}y$
• D. $\frac{x}{y}=e^{x/}y$

Answer 1

Answer: (A)

$xy=e^{x-y}$

$y(1-x)dx+x(1+y)dy=0$
$\Rightarrow \int \left(\frac{1-x}{x}\right)dx=\int -\frac{(1+y)}{y}dy+c$
$\Rightarrow \log x-x=-(y+\log y)+c$
$\Rightarrow \log xy=x-y+c$
$\Rightarrow xy=e^c.e^{x-y}.$
Put $(x,y)=(1,1)$
$1=e^c.1$
$\Rightarrow c=0$
$\Rightarrow xy=e^{x-y}$