I am having trouble isolating the $x$ and $y$ to separate side in the differential equations below. Could someone give me a hint as to how to to this.

Equation 1: $$\frac{dy}{dx} - \frac{x}{y} = \frac{1}{x}$$ Equation 2: $$xy\frac{dy}{dx} = y^2$$

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    $\begingroup$ Were you told they are both separable? $\endgroup$ – Amzoti Oct 27 '13 at 16:20
  • $\begingroup$ How else would I get a general solution to each of the above without separation? $\endgroup$ – user Oct 27 '13 at 16:21

The first is not separable.

For the second, we have:

$$xy\frac{dy}{dx} = y^2$$

Dividing, we have:

$$\dfrac{y~dy}{y^2} = \dfrac{dx}{x}$$

This is separable and we can now integrate each side as:

$$\int \dfrac{1}{y}~ dy = \int \dfrac{1}{x}~ dx$$

I think you can take it from here.

  • $\begingroup$ Could you explain how you divided and how I should solve the first equation? $\endgroup$ – user Oct 27 '13 at 16:30
  • $\begingroup$ Yes I was told it was separable. Could you explain what you divided by in the second equation. $\endgroup$ – user Oct 27 '13 at 16:37
  • $\begingroup$ I added the details in my answer. I do not see an approach for the first as it is a 1st-order nonlinear DEQ and you have to resort to numerical methods. Regards $\endgroup$ – Amzoti Oct 27 '13 at 16:40
  • $\begingroup$ @user2352274: Are you sure you wrote the first one correctly? $\endgroup$ – Amzoti Oct 27 '13 at 16:48
  • $\begingroup$ Yes, and thank you for your help. $\endgroup$ – user Oct 27 '13 at 16:56

The second one is $\dot y = \frac{y}{x}$, and then you should try to put $y=xz$ and solve for $z$.


As far as I can see, only the second one is separable and after simplifying it becomes:

$$\frac{dx}{x}=\frac{dy}{y} $$

Straightforward integration will do the trick.

As a general rule, more often than not it is the multiplicative types of ODE's that are separable.


For $\dfrac{dy}{dx}-\dfrac{x}{y}=\dfrac{1}{x}$ ,


This belongs to an Abel equation of the second kind.

Let $x=e^t$ ,

Then $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{dy}{dt}}{e^t}=e^{-t}\dfrac{dy}{dt}$

$\therefore e^{-t}y\dfrac{dy}{dt}=e^{-t}y+e^t$


This belongs to an Abel equation of the second kind in the canonical form.

Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf

For $xy\dfrac{dy}{dx}=y^2$ , it simply belongs to a separable ODE.



$\ln y=\ln x+c$



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