How to solve $(y)^{y'}=(y')^{y+c},c \in \mathbb{R}$

In the case when $ c=0 $ this ode will be $(y)^{y'}=(y')^{y}$ , let's assume that $y$ and $y'$ are strictly positive functions so:

$$(y)^{y'}=(y')^{y} \iff e^{y' \log(y)}= e^{y \log(y')} \iff {y' \log(y)}= {y \log(y')} \iff \frac{y'}{y}=\frac{\log(y')}{\log(y)}$$

I have no idea what to do now? Any help please ?

Solution 1:

Let $z=\ln(y)$, we have that $zz'=z+\ln(z')$.

If $z'=1$, then $z(x)=x+k$ where $k\in\Bbb R$ i.e. $y(x)=e^{x+k}$.

If $z'\neq1$, then $$z=\frac{\ln(z')}{z'-1}=f(z')$$ where $$f:\quad t\mapsto\frac{\ln(t)}{t-1}.$$ $f:\Bbb R^*_+\to\Bbb R^*_+$ is a bijective function, thus $f^{-1}$ is well-defined. Hence, $z'=f^{-1}(z)$. Let $z(0)=a$, by Cauchy–Lipschitz theorem, the equation has a unique maximal solution, and we have \begin{align} \frac{z'}{f^{-1}(z)}=1&\implies(\int_a^z\frac{dt}{f^{-1}(t)})'=1\\ &\implies\int_a^z\frac{dt}{f^{-1}(t)}=x\\ &\implies\int_{f^{-1}(a)}^{f^{-1}(z)}\frac{f'(s)}{s}ds=x\tag{1}\\ &\implies I\circ f^{-1}(z)=x+I\circ f^{-1}(a)\tag{2}\\ &\implies z(x)=f\circ I^{-1}(x+I\circ f^{-1}(a))\tag{3}\\ &\implies y(x)=\exp\circ f\circ I^{-1}(x+k)\tag{4} \end{align} $(1)$: $t=f(s)$;

$(2)$: $I(x)=-\int_{x}^{+\infty}\frac{f'(s)}{s}ds$;

$(3)$: $I:\Bbb R^*_+\to\Bbb R^*_+$ is bijective;

$(4)$: $k=I\circ f^{-1}(a)$.

The (maximal) domain of $y$ is therefore $(-k,+\infty)$.

(Before assuming $y>0$, note that $y=0$ is also a trivial solution. Generally, for $c\in\Bbb R$, $y=-c$ is the unique constant solution.)

The function $I$ has a closed-form expression using $\operatorname {Li}_{2}$: $$I(x)=\frac1x+f(x)-\left(\operatorname {Li}_{2}(-x)+\frac12\ln^2 x+\frac{\pi^2}6\right).$$