TITLE: To draw integral curves near to an equilibrium point
QUESTION [1 upvotes]: consider the System given by 
\begin{align*}
x'&=e^{y-x}-1\\
y'&= \sin(\pi+x+y)
\end{align*}
The point $(\pi,\pi)$ is an equilibrium point which is asymptotically stable (which can be figured out through a linearization of the vector field). I would like to sketch some integral curves nearby, but I do not know how to do it the best way. 
First I tried to find a first integral for the vector field but I couldn't find one. I am wondering if there is no such first integral or if I just couldn't find one. Is there a general method to find a first integral for such systems or a method by which one can decide whether there exists one or not?
Suppose we can't find a first integral. What is the best way to sketch the integral curves? 
Best wishes

REPLY [2 votes]: Here is a streamplot:  the vectors have components $(x^\prime, y^\prime )$.

If you integrate to find $y(x)$ under the condition $y(0) = 3$, you find:
$y(x) = 2 \cos ^{-1}\left(\frac{\left(\sin \left(\frac{3}{2}\right)+\cos
   \left(\frac{3}{2}\right)\right) \left(2 \cos \left(\frac{3-x}{2}\right)+x \left(\cos
   \left(\frac{x+3}{2}\right)-\sin \left(\frac{3-x}{2}\right)\right)\right)}{\sqrt{x (x+x
   \cos (6)+2 \sin (6)+4 \cos (3))+4 (1+\sin (3))}}\right)$ .
Rather complicated.
I recommend using software for all of this.
Here's my Mathematica code:
StreamPlot[{Exp[y - x] - 1, Sin[\[Pi] + x + y]}, 
    {x, 0, 5}, {y, 0, 5},
 FrameLabel -> {Text[Style["x", 18, Italic]], 
   Text[Style["y", 18, Italic]]},
 Epilog -> {Red, PointSize[0.05], Point[{\[Pi], \[Pi]}]}]
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