Question

If the solution of the differential equation $\left(1+e^{\frac{x}{y}}\right)dx+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)dy=0$ is $x+ky{e}^{\frac{x}{y}}=C$ (where, $C$ is an arbitrary constant), then the value of $k$ is equal to

Answer 1

Answer: 1

The given equation is
$e^{\frac{x}{y}}\left(dx-\frac{x}{y}dy\right)+e^{\frac{x}{y}}dy+dx=0$
or $e^{\frac{x}{y}}yd\left(\frac{x}{y}\right)+e^{\frac{x}{y}}dy+dx=0$
$\Rightarrow d\left(e^{\frac{x}{y}}y\right)+dx=0$
On integrating, we get,
$e^{\frac{x}{y}}y+x=C$
$\Rightarrow k=1$