Is it possible to solve the following equation?
\begin{align} \frac{\mathrm{d}^\beta}{\mathrm{d}\alpha^\beta} \frac{\mathrm{d}^\alpha}{\mathrm{d}x^\alpha} f(x)+ \frac{\mathrm{d}^\alpha}{\mathrm{d}x^\alpha} f(x)=\beta\\ \end{align} with \begin{align} \alpha,\beta\in\mathbb{R}_+ \end{align}
Thank you very much
I'm not sure about \begin{align} \alpha,\beta\in\mathbb{R}_+ \end{align} but can imagine it working for
\begin{align} \alpha,\beta\in\mathbb{Z}_+ \end{align}
I wonder if this can be considered more easily as:
\begin{align} \frac{\mathrm{d}^\beta}{\mathrm{d}\alpha^\beta} g(\alpha)+ g(\alpha)=\beta\\ \end{align}
where
\begin{align} g(\alpha)=\frac{\mathrm{d}^\alpha}{\mathrm{d}x^\alpha} f(x) \end{align}
Once we have found $g(\alpha)$ we can just integrate repeatedly to find $f(x)$...
The differential equation has auxiliary equation $\lambda^\beta+1=0$, so $\lambda=(-1)^\frac{1}{\beta}$
For $\beta=1$, we have $\lambda=-1$
General solution of homogeneous differential equation is $g(\alpha)=Ae^{-\alpha}$
But we have the non-homogeneous differential equation $\frac {dg}{d \alpha} +g=1$.
Trial function $g(\alpha)=p \Rightarrow \frac {dg}{d\alpha}=0$
So $0+p=1\Rightarrow p=1$
$g(\alpha)=Ae^{-\alpha}+1$