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I have the partial differential equation:

$\qquad U_{x} - 4 x^{2} U_{t} = x U, \qquad U(x,0)=f(x), \qquad x \in (- \infty, \infty), \qquad t \in (0, \infty)$

Where $U_{x} = \frac{\partial U}{\partial x}$, and the solution I arrived at using Method of Characteristics is:

$\qquad U(x,t) = f\Big( (x^{3}+\frac{3}{4}t)^{\frac{1}{3}} \Big) * e^{\frac{x^{2}}{2} - \frac{1}{2} (x^{3} + \frac{3}{4}t)^{\frac{2}{3}}}$

I am unsure how to define the arbitrary function $f(x,t)$ in that way when trying to plug the solution back into the differential equation to check to see if it is a solution. (This solution may not be correct)

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    $\begingroup$ Have you read the relevant part in the documentation? $\endgroup$ – yohbs Sep 23 '15 at 21:09
  • $\begingroup$ @yohbs Yeah I did but I tried it and it didn't work. I think I screwed up a bracket or something. Thank you though. $\endgroup$ – Kvothealar Sep 25 '15 at 14:10
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In case reading the documentation, as recommended by yohbs, did not fully answer your question, try

U[x_, t] = f[(x^3 + 3 t/4)^(1/3)] Exp[x^2/2 - (x^3 + 3 t/4)^(2/3)/2];
Simplify[D[U[x, t], x] - 4 x^2 D[U[x, t], t] - x U[x, t]]
(* 0 *)

By the way, Exp[ - (x^3 + 3 t/4)^(2/3)/2] can be absorbed into f without loss of generality.

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  • $\begingroup$ Ah! Thank you. I tried something and it didn't work. I must have just screwed up the brackets or something. This worked perfectly. $\endgroup$ – Kvothealar Sep 25 '15 at 14:09

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