OK, so, I'm supposed to solve the differential equation $$\frac{dy}{dx} = \frac{y+2x}{y-2x}$$ by making the substitution $y = ux$, to make the equation separable. Then $$\frac{dy}{dx} = u + x\frac{du}{dx},$$ which changes the equation to $$\frac{1}{x} dx = \frac{2-u}{u^2-3u-2}du,$$ and then by integrating I get $$\ln|x|+C = \frac{1}{2\sqrt{17}} \left[ (1-\sqrt{17}) \ln\,\left| u-\frac{3+\sqrt{17}}{2}\right| - (1+\sqrt{17}) \ln\, \left|u-\frac{3-\sqrt{17}}{2} \right| \right].$$

Assuming I did that correctly, my question then becomes: if I resubstitute in for $u \ldots$ is that it? I don't think I can isolate $y$ from this, and Wolfram|Alpha seems to agree with me. I mean, is this implicit solution acceptable?

This isn't the whole story, however; I've found two linear solutions, $$y = \left(\frac{3 \pm \sqrt{17}}{2}\right)x,$$ to the differential equation essentially by showing that $d^2y/dx^2 \rightarrow 0$ as $x\rightarrow\pm\infty$ and then subtituting $y=Ax$. Notwithstanding the fact that it yielded correct solutions, though, I can't vouch for the validity of this method, as I already knew that those were correct and invented some fanciful method for deriving them... If this has any value to it, some basic instruction would be enormously appreciated.

  • $\begingroup$ The linear solutions can be demonstrated by hypothesizing $y=ax$ and then solving $a=(a+2)/(a-2)$. Could you show more work on how you separated the original DE? $\endgroup$ – anon Jul 29 '11 at 11:39
  • $\begingroup$ OK, given $dy/dx = (y+2x)/(y-2x)$, $y = ux$ and $dy/dx = u+x du/dx$, $$u+x\frac{du}{dx} = \frac{ux+2x}{ux-2x},$$ so then $$x \frac{du}{dx} = \frac{-u^2+3u+2}{u-2}$$ which leads to my separated equation. $\endgroup$ – Hargrove Jul 29 '11 at 11:48
  • $\begingroup$ @Hargrove: It is very common in separable equations to end up with an implicit function that cannot be made explicit. So in this case, the solution you gave should be viewed as essentially complete. As to the two linear solutions, they can be found by taking limits in your general solution. So whether they are "covered" by the general solution is a matter of interpretation. $\endgroup$ – André Nicolas Jul 29 '11 at 12:57
  • $\begingroup$ @Hargrove: I've merged several of your duplicate accounts into your current account. Please consider registering your account; this way the system can better keep track of your questions and answers when you log in using different computers or different IP addresses. $\endgroup$ – Willie Wong Jul 29 '11 at 13:05

These solutions are the same as ${\displaystyle u = {3 \pm \sqrt{17} \over 2}}$, which are the constant solutions to ${\displaystyle x{du \over dx} = {-u^2 + 3u + 2 \over u - 2}}$. They dissappear when you write ${\displaystyle {1 \over x} dx = {2 - u \over u^2 - 3u - 2}du}$ because they correspond to when the denominator of the right-hand side is identically zero.

In this sort of situation you have to consider such constant values of $u$ separately and check to see if they correspond to solutions of the original differential equation, and they often do.


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