Solving first order non-linear ODE with logarithm

Question

I want to solve for $t \in \mathbb{R}, u'(t)=-u(t)\ln \lvert u(t) \rvert$.

I defined two cases: $\mathbb{R^*_+}$ and $\mathbb{R^*_-}$.

For $\mathbb{R^*_+}$:

$$\frac{du}{u}=-\ln(u(t))dt$$

And by integrating, it follows that:

$$\ln\lvert u \rvert = -u\ln(u) + u$$

But I'm stuck here, I don't know what to do next.

Answer 1

$$\frac{du}{u}=-\ln(u(t))dt$$ You cannot integrate $\int-\ln(u(t))dt$ directly.

The mistake is : You integrate $\quad \int-\ln(u(t))du= -u\ln(u)+u\quad$ but this is not $\quad \int -\ln(u(t))dt$.

Back to the original ODE : $$\frac{du}{dt}=u\ln(u)$$ This is a separable ODE thus $$\frac{du}{u\ln(u)}=dt$$ $$\int\frac{du}{u\ln(u)}=\int dt$$ $$\ln(\ln(u))=t+c$$ $$u=\exp(\exp(t+c))$$

Answer 2

The above solution is absolutely correct. A slightly quicker way is to substitute $$ u(t) = e^{f(t)} $$ producing the differential equation: $$ f'=-f $$ eventually leading to the same answer.