Solve the differential equation $$y'=\frac{y}{2y\ln y+y+x}$$

This doesn't look hard, but the problem is how to spot the right substitution. This seems like a linear differential equation, but I couldn't transform it to the right form.

This is what I did instead:

Firstly, I transformed the equation to the following form: $$y-\left(2y\ln y+y+x\right)y'=0$$ Then I noticed that if I divide it with $y^2$ and recombine the terms in the numerator, I can end with the something like this: $$\frac{y-xy'-y(1+\ln y)y'-y\ln yy'}{y^2}=0$$ which is indeed $$\left(\frac{x-y\ln y\left(1+\ln y\right)}{y}\right)'=0$$ and thus the solution is $$x-y\ln y\left(1+\ln y\right)-c_1y=0$$ My approach works in this particular case. I spent a lot of time trying to figure out how to recombine the terms in the fraction in order to artificially create a derivative of a fraction. Of course, this is not how the general approach should look like.

My question is how to deal with equations like this? Is there a more general algorithm which can be used to lead us to the solution with less effort? What substitution would be in this particular equation and how to spot the right substitution?

  • $\begingroup$ you can solve your equation for $x$ $\endgroup$ – Dr. Sonnhard Graubner Apr 27 '17 at 14:22
  • $\begingroup$ @Dr.SonnhardGraubner. Isn't that what I've did? But I already explained why is my approach bad. $\endgroup$ – duvajtegasvi Apr 27 '17 at 14:24
  • $\begingroup$ your Approach isn't bad and you have the right solution $\endgroup$ – Dr. Sonnhard Graubner Apr 27 '17 at 14:31
  • $\begingroup$ @Dr.SonnhardGraubner. Correct, it leads to the right solution. But, on the test I don't have enough time to recombine the terms. I am looking for more general method. $\endgroup$ – duvajtegasvi Apr 27 '17 at 14:35

You can use homogenous DE techniques after replacing $y'$ with $\frac{1}{x'}$.

First rewrite it as:

$$x'=\frac{2y\ln y+y +x}{y}=2\ln y +1 +\frac{x}{y}$$

Note on the right hand side everywhere you see $x$ it is in the form of $\frac{x}{y}$. This classifies as a homogenous DE.

So let $v=\frac{x}{y}$ so that $x=v\cdot y$ and $x'=v'\cdot y+v$

Substituting in gives:

$$v'\cdot y + v = 2\ln y + 1 + v$$

$$v'\cdot y = 2\ln y + 1$$

$$v' = \frac{2\ln y}{y} + \frac{1}{y}$$

Then simply integrate to get:

$$v=(\ln y)^2+\ln y+c$$

Substitute back in for $v$ and you get:

$$\frac{x}{y} = (\ln y)^2+\ln y+c$$

which is equivalent to your answer.

  • $\begingroup$ Yeah, it works. But I'm afraid that it works only in this particular case. Is this the only approach? I mean, I have a lot of similar equations where your approach doesn't work. $\endgroup$ – duvajtegasvi Apr 27 '17 at 14:33
  • $\begingroup$ Do you have a similar example? I'm not sure exactly what you consider close enough to be similar. $\endgroup$ – Ian Miller Apr 27 '17 at 14:38
  • $\begingroup$ Also there isn't going to be one technique which works for all DEs. Experience doing DEs helps you to narrow down which techniques work in which areas. To me this technique seemed very obvious to try. A different DE may call out a different technique when I see it. $\endgroup$ – Ian Miller Apr 27 '17 at 14:40
  • $\begingroup$ "Also there isn't going to be one technique which works for all DEs" - Yes, but it is proved that there is a technique to prove if the equation can be solved using available methods or not. In this example it can be solved, so there should be some general, universal method which works for all solvable DEs (like Wolfram's Mathematica can solve any DE or prove it cannot be solved). $\endgroup$ – duvajtegasvi Apr 27 '17 at 14:44
  • $\begingroup$ You also asked for similar example. This is the next example from my book where your method doesn't work: $y'\left(x^3\sin y-x\right)=-2y$ $\endgroup$ – duvajtegasvi Apr 27 '17 at 14:50

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