Question

If $\frac{dy}{dx}=e^{x+y}$ and at $x=1;y=1$, then for $x=-1;y$ ___.

• A. $-1$
• B. $e$
• C. $1$
• D. $e+\frac{1}{e}$

Answer 1

The correct option is A $-1$

$\frac{dy}{dx}=e^x\cdot e^y$
$\Rightarrow e^{-y}dy=e^{x}dx$
Integrating both sides, we get
$\Rightarrow \frac{e^{-y}}{-1}=e^x-c$
$\Rightarrow e^x+e^{-y}=c$
Substitute $(x,y)=(1,1)$
$\Rightarrow e+\frac{1}{e}=c$
Now, for $x=-1$
$\Rightarrow \frac{1}{e}+e^{-y}=e+\frac{1}{e}$
$\Rightarrow e^{-y}=e$
$\Rightarrow y=-1$