Question

The solution of the differential equation $xdy + ydx = xydx$ when $y(1) = 1$ is

• A. $y=\frac{e^{x}}{x}$
• B. $\frac{e^{x}}{ex}$
• C. $y=\frac{xe^{x}}{e}$
• D. none of these

Answer 1

Answer: (B)

$\frac{e^{x}}{ex}$

Answer 2

From given $\frac{xdy+ydx}{xy}=dx$ On integration, we get $log_{e}\left(xy\right)=x+k$ $\therefore x=1$, $y=1$ $\Rightarrow k=-1$ $\Rightarrow xy=e^{x-1}$ $\Rightarrow xy=\frac{e^{x}}{e}$ $\therefore y=\frac{e^{x}}{ex}$