The PDE is the following :


To be honest, I haven't quite understood the method and it seems to be used in different ways and that confuses me(in some solutions I see parametrizations going on). I managed to apply it in some problems using the following characteristics equations, which in this problem are written like this :


From the first equality we get:

$$\frac{dy}{dx}=2$$ $$=>y=2x+C_1=>C_1=y-2x$$

We can also get this:

$$\frac{du}{dx}=\frac{(x+y)u}{xy}=(1/y+1/x)u$$ Integrating I get something strange (from what I've seen at least).


Am on the right track? Can I use my initial equations to get a better relationship for C2? By cancelling (x+y) perhaps?

  • $\begingroup$ You're fine. The last step is to just take the exponential of both sides to get $$ u = Cxe^{x/y} = xe^{x/y}f(y-2x) $$ $\endgroup$ – Dylan Jun 23 '18 at 14:33
  • $\begingroup$ I did think of that but how does the absolute value disappear? I had the same problem in other examples but I found another way around which didn't require producing an absolute value. I couldn't avoid it on this one. $\endgroup$ – John Katsantas Jun 23 '18 at 14:40
  • $\begingroup$ @Dylan Maybe I've made a mistake, as far as I can tell your solution doesn't satisfy the PDE. The integration at $$u' = \left( \frac{1}{x} + \frac{1}{y} \right)u$$ is incorrect as $y = y(x)$. $\endgroup$ – mattos Jun 23 '18 at 14:58
  • $\begingroup$ Ok, I made a mistake. The integration should have $dy=2dx$. But then @Mattos your solution is also not correct. It should be $$ u = x\sqrt{y}f(y-2x) $$ $\endgroup$ – Dylan Jun 23 '18 at 20:15
  • 1
    $\begingroup$ @JohnKatsantas I wouldn't worry too much about the absolute value. $|x|$ produces either a positive or negative sign which gets absorbed into the arbitrary function $\endgroup$ – Dylan Jun 23 '18 at 20:17

I should clarify that I'm also not an expert in the method, but I can tell you what went wrong.

Your first part up to $c_1 = y-2x$ is correct

For the second integral, you have $dy=2dx$, therefore

$$ \frac{du}{u} = \left(\frac{1}{y} + \frac{1}{x}\right)dx = \frac{1}{2y}dy + \frac{1}{x}dx $$

Integrating the above

$$ \ln |u| = \frac12 \ln |y| + \ln|x| + c_2 $$

Taking the exponential

$$ |u| = e^{c_2}|x|\sqrt{|y|} \implies u = \pm e^{c_2}x\sqrt{|y|} = Cx\sqrt{|y|} $$

The last step is to take $C = f(c_1)$ so that

$$ u = x\sqrt{y} f(y-2x) $$

You can usually drop the absolute value on $|y|$ for convenience, unless a boundary condition is given.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.