While there is a simple solution to the Euler equation $f(x) = \mu x f'(x) + \lambda x^2 f''(x)$ as pointed out in my previous question (see here), the one discussed here seems much more challenging.
I don't think there is a simple solution for the general case, but the case $\mu = 1$ leads to an interesting result. I was wondering if the result I am presenting here (see below) is correct. I proceeded as follows:
Step 1
Let $f(x) = x\phi(x)$.
Step 2
This leads to $(x+2\lambda) \phi'(x) + \lambda x \phi''(x) = 0$, that is
$$\frac{d \log\phi'(x)}{dx} = -\frac{1}{\lambda}-\frac{2}{x}.$$
Step 3
Simple integration (twice) leads to
$$f(x) = x\phi(x) = Ax + Bx\int \frac{1}{x^2}\cdot\exp\Big(-\frac{x}{\lambda}\Big) dx.$$
So we have the general solution if $\mu =1$. It is trivial to verify that $f(x) = Ax$ is indeed a particular solution.
Step 4
Using WolframAlpha or by other means, the solution can be expressed in terms of the exponential integral (Ei) function:
$$f(x) = Ax - \frac{B}{\lambda} \Big\{ x \mbox{Ei}\Big(-\frac{x}{\lambda}\Big) + \lambda\exp\Big(-\frac{x}{\lambda}\Big)\Big\}.$$
Is this final result correct (if $\mu=1$)? Is this a well known differential equation?
When $b=1$, $f(x)=x$ is the soltion of $$axf''(x)+bxf'(x)=f(x)~~~~(1)$$ If onw solution of (1) is $x$, then other solution can be found as $f(x)=u (x)$, we get $$axu''+(2a+x)u'=0 \implies axv'+(2a+x)v=0 \implies v=u'=C \exp[-2 \ln x -x/a] \implies u(x)=C\int \frac{1}{x^2}e^{-x/a}~dx+D$$ So when $b=1$, the total solution of the second order ODE (1) is $$f(x)=Cx\int \frac{1}{x^2} e^{-x/a} dx+ Dx,$$ where $C$ and $D$ are two constants of integration.