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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!

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  • $\begingroup$ @user170231 Thank you!!! $\endgroup$
    – L Garcia
    Jan 6, 2021 at 22:17
  • $\begingroup$ @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 $k_1$. $\endgroup$
    – Aaratrick
    Jan 7, 2021 at 0:13
  • $\begingroup$ @Aaratrick Yes you're right, I don't know what I was thinking... $\endgroup$
    – user170231
    Jan 7, 2021 at 5:44
  • $\begingroup$ @user170231 No problem, it just happens sometimes. However, do you have any idea on how to evaluate the left hand integral analytically? $\endgroup$
    – Aaratrick
    Jan 7, 2021 at 7:50
  • $\begingroup$ I don't think there is an elementary antiderivative, unfortunately. $\endgroup$
    – user170231
    Jan 7, 2021 at 16:14

1 Answer 1

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A particular solution can be obtained of the form:

$$y=kx$$

Substituting this into the ODE gives:

$$k = - \frac{1}{k} \ln{k}$$

Which is satisfied when $k\approx0.653$.

I plotted a few numerical simulations and it appears that they asymptotically approach this particular solution.

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