My problem is that I don't know how to completely solve the following differtential equation.
\begin{equation} \begin{aligned} t \cdot \frac{\partial^2v(x,t)}{\partial x^2}+x \cdot \frac{\partial v(x,t)}{\partial t}=0 &&\text{where $v(x,t):\mathbb{R^2}\rightarrow \mathbb{R}$} \end{aligned} \end{equation}
We can use the method of separation of variable to try to get the solution of this equation.
Let's make the following substitution: $v(x,t)=A(x)B(t)$.
Where $A(x):\mathbb{R}\rightarrow \mathbb{R} \quad B(t):\mathbb{R}\rightarrow \mathbb{R}$
From this substitution we get: \begin{equation} t \cdot B(t)\cdot \frac{\partial^2A(x)}{\partial x^2}+x \cdot A(x) \cdot \frac{\partial B(t)}{\partial t}=0 \Rightarrow \frac{-1}{xA(x)} \cdot \frac{\partial^2A(x)}{\partial x^2} = \frac{1}{tB(t)}\cdot \frac{\partial B(t)}{\partial t} \end{equation} Since the two part of the equality do not depend on the same variable, we can suppose that they should be equal to a constant $c\in\mathbb{R}$. \begin{equation} \begin{cases} \frac{-1}{xA(x)} \cdot \frac{\partial^2A(x)}{\partial x^2}=c &\Leftrightarrow \color{red}{\int \frac{1}{A(x)} d^2(A(x))=-\int cx d^2x} \\ &\Leftrightarrow \int ln(|A(x)|) d(A(x))=-c\frac{x^3}{6}+Lx+M& &\text{where $L,M\in\mathbb{R}$}\\ \\ \frac{1}{tB(t)}\cdot \frac{\partial B(t)}{\partial t}=c &\Leftrightarrow \int \frac{1}{B(t)} \cdot d(B(t))=\int ct \cdot dt \\ &\Leftrightarrow B(t)=N \cdot \exp(c\frac{t^2}{2})& &\text{where $N\in \mathbb{R}$} \\ \end{cases} \end{equation}
The point at which I'm getting in trouble is marked in red. It's hard for me to find a function A from this relation.
What is the easiest and simplest path to find the general solution of this differential equation?
Thank you in advance for your support.
Your equation marked in red in not correct. $d^2x$ is non-sens.
$\frac{-1}{xA(x)} \cdot \frac{\partial^2A(x)}{\partial x^2}=c\quad $or$ \quad\frac{\partial^2A(x)}{\partial x^2}+cxA(x)=0\quad $is an Airy ODE. The solutions are lineaer combinations of the Airy functions : http://mathworld.wolfram.com/AiryFunctions.html
In order to make it on standard form, without loss of generality since $c$ is arbitrary, let $c=-C^3$
\begin{equation} \begin{cases} \frac{-1}{xA(x)} \cdot \frac{\partial^2A(x)}{\partial x^2}=-C^3 &\Leftrightarrow \frac{\partial^2A(x)}{\partial x^2}-C^3xA(x)=0\\ &\Leftrightarrow A(x)=\alpha\,\text{Ai}(Cx)+\beta\,\text{Bi}(Cx)\\ \frac{1}{tB(t)}\cdot \frac{\partial B(t)}{\partial t}=-C^3 &\Leftrightarrow B(t)=\gamma\, \exp(-C^3\frac{t^2}{2}) \\ \end{cases} \end{equation} $\alpha\,,\,\beta\,,\,\gamma$ are arbitrary constants. Ai and Bi are the Airy functions (See above reference). That way some particular solutions of the equation $(1)$ are obtained :
$$t \cdot \frac{\partial^2v(x,t)}{\partial x^2}+x \cdot \frac{\partial v(x,t)}{\partial t}=0 \tag 1$$ $$v(x,t)=\left(\gamma\alpha\,\text{Ai}(Cx)+\gamma\beta\,\text{Bi}(Cx) \right) \exp(-C^3\frac{t^2}{2}) $$ One can combine $\gamma\alpha$ into only one constant $f=\gamma\alpha$. And $g=\gamma\beta$ as well. $$v(x,t)=\left(f\,\text{Ai}(Cx)+g\,\text{Bi}(Cx) \right) \exp(-C^3\frac{t^2}{2}) $$ For each arbitrary values of $C\,,\,f\,,\,g\,,$ we have a particular solution of Eq.$(1)$. All linear combinations of those particular solutions are solutions of $(1)$. Thus a more general solution on discret form is : $$v(x,t)=\sum_{\text{any} C}\left(f_{(C)}\,\text{Ai}(Cx)+g_{(C)}\,\text{Bi}(Cx) \right) \exp(-C^3\frac{t^2}{2}) $$ $f_{(C)}$ and $g_{(C)}$ are arbitrary constants.
An infinity of discret values $C$ can be considered in order to express the general solution : $$v(x,t)=\int\left(f_{(\mu)}\,\text{Ai}(\mu x)+g_{(\mu)}\,\text{Bi}(\mu x) \right) \exp(-\mu^3\frac{t^2}{2}) d\mu$$ For the dummy variable the symbol $\mu$ is introduced instead of $C$ in order to make more clear the change from discret to continuous.
$f(\mu)$ and $g(\mu)$ are arbitrary functions, to be determined according to some boundary conditions. Since no boundary condition is specified in the wording of the question one cannot go further.
Note that, for some kind of boundary conditions eventually the solution might be simpler than the above general form.