Solve the phase plane equation to obtain the integral curves for the system: Solve the phase plane equation to obtain the integral curves for the system:
$$\begin{align*}\frac{\mathrm dx}{\mathrm dt}&=2y-x\\\frac{\mathrm dy}{\mathrm dt}&=e^x+y\end{align*}$$
It's for a 200 level paper; differential equations. What is important with this question is that it is non-linear. So he can't use linear methods. He will try a Jacobian matrix and see if that gives anything useful. His textbook also says he could maybe change it into polar coordinates, but that doesn't seem to be helping.

The phase plane equation for the integral curves is
$$\def\part#1#2{\frac{\mathrm d#1}{\mathrm d#2}}
\part yx=\frac{\part yt}{\part xt}=\frac{\mathrm e^x+y}{2y-x}\;.$$
To get a grip on this, it helps to expand around the equilibrium. At equilibrium, $\part xt=2y-x=0$, and thus $y=x/2$, so we can write $y=x/2+z$, with $y'=1/2+z'$. Substituting this into the differential equation for $y$ yields
$$
\begin{align}
\frac12+z'
&=\frac{\mathrm e^x+\frac x2+z}{2(\frac x2 + z)-x}\\
&=\frac{\mathrm e^x+\frac x2+z}{2z}\;,\\
z'
&=\frac{\mathrm e^x+\frac x2}{2z}\;,\\
2zz'
&=\mathrm e^x+\frac x2\;,\\
z^2
&=\mathrm e^x+\frac {x^2}4+C\;,\\
z
&=\pm\sqrt{\mathrm e^x+\frac {x^2}4+C}\;,\\
y
&=\frac x2\pm\sqrt{\mathrm e^x+\frac {x^2}4+C}\;.
\end{align}$$
