# How do I solve de ODE $y' = \frac xy\ln\left(\frac xy\right)$?

I need some help finishing the ODE, please. This is what I have so far:

$$w=\frac xy$$, or $$y=\frac xw$$. Then we have

$$y’=\frac1w-\frac{xw'}{w^2}.$$

Then the DE becomes

$$y’=\frac1w-\frac{xw'}{w^2}=w\log w$$

$$\frac1w-w\log w=\frac{xw'}{w^2}$$

and clear $$w^2$$ out of denominator to get

$$w-w^3\log w=x\frac{dw}{dx}.$$

The equation is now separable as

$$\frac{dw}{w-w^3\log w}=\frac{dx}x$$

so integrate both sides to yield

$$\int\frac{dw}{w-w^3\log w}=\log x+k_1$$

and exponentiate to get

$$\exp\left(\int\frac{dw}{w-w^3\log w}\right)=k_2x.$$

However, I can't go any further than this.

My professor said I might be ignoring other solutions. He also said there is another way to solve the ODE that is way easier.

I appreciate your help, thank you!

• @user170231 Thank you!!! Jan 6, 2021 at 22:17
• @user170231 If we take $k_1 = \log k_2$, then we get, $\exp(\log x + k _1) = \exp(\log(k_2 x)) = k_2 x$. Also, the constant from the integral on the left can be absorbed into $k_1$. Jan 7, 2021 at 0:13
• @Aaratrick Yes you're right, I don't know what I was thinking... Jan 7, 2021 at 5:44
• @user170231 No problem, it just happens sometimes. However, do you have any idea on how to evaluate the left hand integral analytically? Jan 7, 2021 at 7:50
• I don't think there is an elementary antiderivative, unfortunately. Jan 7, 2021 at 16:14

$$y=kx$$
$$k = - \frac{1}{k} \ln{k}$$
Which is satisfied when $$k\approx0.653$$.