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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.

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The references to "him" no longer make sense after you edited out your boyfriend. –  joriki Oct 7 '11 at 7:11
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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

$$ \begin{align} \frac12+z' &=\frac{\mathrm e^x+\frac x2+z}{2(\frac x2 + z)-x}\\ &=\frac{\mathrm e^x+\frac x2+z}{2z}\;,\\ z' &=\frac{\mathrm e^x+\frac x2}{2z}\;,\\ 2zz' &=\mathrm e^x+\frac x2\;,\\ z^2 &=\mathrm e^x+\frac {x^2}4+C\;,\\ z &=\pm\sqrt{\mathrm e^x+\frac {x^2}4+C}\;,\\ y &=\frac x2\pm\sqrt{\mathrm e^x+\frac {x^2}4+C}\;. \end{align}$$

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I follow until your last sentence. What is the resulting differential equation for $z$? (More specifically, how do we see $e^x+y$ solely as a function of $z$?) –  anon Oct 7 '11 at 7:37
    
@anon: I'd written this out but then removed it because I saw the question is on homework; I'll write it out again. –  joriki Oct 7 '11 at 8:01
    
Oh, of course. For some reason I was thinking we had to get it autonomous in $z$, not sure why I thought that. Moment of brain seepage. –  anon Oct 7 '11 at 8:11
    
Thanks. This was posted by my gf on my behalf. This is exactly what i needed. Knew it would be something simple! –  user17256 Oct 7 '11 at 8:49
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