Question

If the curve satisfying $(1+e^{x/y})dx+e^{x/y}(1-x/y)dy=0$ passes through $(1,1)$, then $9+y\left(2\right)e^{2/y\left(2\right)}-e$ is equal to

• A. 9
• B. 2
• C. 4
• D. 5

Answer 1

The correct option is D 5

$(1+e^{x/y})dx+e^{x/y}(1-x/y)dy=0$
Substituting $x=vy\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}$
$(1+e^v)(v+y\frac{dv}{dy})+e^v(1-v)=0\Rightarrow \frac{dy}{y}=-\frac{1+e^v}{v+e^v}dv$
Integrating
$\log y(\frac{x}{y}+e^{x/y})=c\Rightarrow x+y{e}^{x/y}=c$
For $y\left(1\right)=1\Rightarrow c=1+e$
Thus $9+y\left(2\right)e^{2/y\left(2\right)}-e=5$