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Using "separation of variables" the PDE $x \cdot u(x,t)_x = u(x,t)_t$

has solutions of the form

$u(x,t) = C\cdot e^{\lambda\cdot t} x^\lambda $

A more general solution could be found by superposition of this, taking different values for k:

$u(x,t) = \sum_i C_i\cdot e^{\lambda_i\cdot t} x^\lambda_i $

However, another solution would be: $u(x,t) = C(t+\ln x)$

I wonder now, how those different types of solutions correspond to each other.

Neither of them can be a complete solution, because of the other one.

If I find a solution by separation of variables, how can I trust to have the most general solution at all and not to exclude something due to the special form of the separation $u(x,t)=v(x)\cdot w(t)$?

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Thanks to the method of characteristics one can find the general solution : $$u(x,t)=F(x\:e^t)$$ where $F$ is an arbitrary function.

For example a particular solution in case of $F(X)=C\ln(X)$ is $u=C(t+\ln(x))$.

This is also consistant with the particular cases of $F(X)=CX^\lambda$ thus $u=Cx^\lambda e^{\lambda t}$.

The above general solution includes the cases of fonctions $F(X)$ expressed on form of series such as the series that you found thanks to the method of separation of variables.

Of course, infinity many other forms of particular solutions can be derived, for example with $F(X)=\sin(X)$ thus $u=\sin(xe^t)$.

If some valid condition is added to the wording of the problem, possibly the function $F(X)$ can be determined and the solution can be a unique particular solution.

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  • $\begingroup$ Nice - I understand now. And I learned how to apply method of characteristics. $\endgroup$ – MichaelW Dec 6 '20 at 19:31

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