Added May 23, 2019.
From Math 5587 midterm I, Fall 2016, practice exam, problem 6.
Solve for \(u(x,t)\) with \(x>0,t>0\) and \(u(x,0)=x^2\) \begin {align*} x u_t - t u_x = 0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = x D[u[x, t], t] - t*D[u[x, t], x] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to c_1\left (\frac {1}{2} \left (t^2+x^2\right )\right )\right \}\right \}\]
Maple ✓
restart; pde :=x*diff(u(x,t),t)-t*diff(u(x,t),x)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,t))),output='realtime'));
\[u \left (x , t\right ) = \textit {\_F1} \left (t^{2}+x^{2}\right )\]
Hand solution
Solve \(xu_{t}-tu_{x}=0\). Using the method of characteristics\[ \frac {dt}{x}=-\frac {dx}{t}=\frac {du}{0}\] From the first pair of equations we obtain \(tdt=-xdx\) or \(\frac {t^{2}}{2}=-\frac {x^{2}}{2}+C_{1}\) or \(C_{1}=\frac {t^{2}}{2}+\frac {x^{2}}{2}\). From \(du=0\) then \(u=C_{2}\). But \(C_{2}=F\left ( C_{1}\right ) \) where \(F\) is arbitrary function. Hence \[ u=F\left ( \frac {t^{2}}{2}+\frac {x^{2}}{2}\right ) \]
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