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We are looking for solutions to the following differential equation $$ tf'(t)-\mu f(\frac{t}{\mu})+\mu f(0)=0\;\; ;\;\;t\in \mu I, $$ where $I$ is an interval containing $0$, $\mu\in (0,1)$ is a constant, $\mu I=\{ \mu x: x\in I\}$, and $f:I\rightarrow \mathbb{R}$ is a differentiable function.

Note. The functions $f(t)=kt$ satisfy the equation, and the related integral equation is $$ \int^x\frac{f(\frac{1}{\mu}t)}{t}dt=\frac{1}{\mu}f(x)+f(0)\log|x|\; ; \; x\neq 0. $$

Now,

(1) Can one obtain its general solution?, if no,

(2) Are there some other infinite classes for its solutions?

(3) What about uniqueness conditions for a special solution?

Thanks in advance

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    $\begingroup$ This is a functional differential equation, setting $t=e^{-x}$ you can transform it into a delay-differential equation. Usually the solution depends on a history function on an interval, here it looks like $f|_{[\mu,1]}$ gives this initial history, not just an initial condition in one point. $\endgroup$ Feb 9 '21 at 10:07
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Let $u(x)=f(e^{-x})$. Then $$ u'(x)=-e^{-x}f'(e^{-x})=-tf'(t)=−μf(t/μ)+μf(0)=−μ(u(x-\delta)-A) $$δ where $δ=-\lnμ$ and $A=\lim_{x\to\infty}u(x)$.

The solution formula is $$ u(x)=u(0)−μ\int_0^{x-δ}(u(s)-A)\,ds,~~x>δ. $$ The solution is completely determined by the values of $u$ on $[0,δ]$, or of $f$ on $[μ,1]$.

Using an exponential trial one finds a solution of $u(x)=A+Be^{-x}\implies f(t)=A+Bt$. This is of course only one special solution class among many more.

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