# Differential equation $y'=\frac{y}{2y\ln y+y+x}$

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?

• you can solve your equation for $x$ – Dr. Sonnhard Graubner Apr 27 '17 at 14:22
• @Dr.SonnhardGraubner. Isn't that what I've did? But I already explained why is my approach bad. – duvajtegasvi Apr 27 '17 at 14:24
• your Approach isn't bad and you have the right solution – Dr. Sonnhard Graubner Apr 27 '17 at 14:31
• @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. – 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:

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

• You also asked for similar example. This is the next example from my book where your method doesn't work: $y'\left(x^3\sin y-x\right)=-2y$ – duvajtegasvi Apr 27 '17 at 14:50