Added May 23, 2019.
From Math 5587 midterm I, Fall 2016, practice exam, problem 5.
Solve for \(u(x,y)\) \begin {align*} 2 u_x + 3 u_y = 1 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = 2 D[u[x, y], x] + 3*D[u[x, y], y] == 1; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y], {x, y}], 60*10]];
\[\left \{\left \{u(x,y)\to \frac {x}{2}+c_1\left (y-\frac {3 x}{2}\right )\right \}\right \}\]
Maple ✓
restart; pde :=2*diff(u(x,y),x)+3*diff(u(x,y),y)=1; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime'));
\[u \left (x , y\right ) = \frac {x}{2}+\textit {\_F1} \left (-\frac {3 x}{2}+y \right )\]
Hand solution
Solve \(2u_{x}+3u_{y}=1\). Using the method of characteristics\[ \frac {dx}{2}=\frac {dy}{3}=\frac {du}{1}\]
From the first pair of equations we obtain \(\frac {1}{2}x=\frac {1}{3}y+C_{1}\) or \(C_{1}=\frac {1}{2}x-\frac {1}{3}y\). From the pair \(\frac {dx}{2}=\frac {du}{1}\) we obtain
\(\frac {1}{2}x=u+C_{2}\) or \(C_{2}=\frac {1}{2}x-u\). But \(C_{2}=F\left ( C_{1}\right ) \) where \(F\) is arbitrary function. Hence \begin {align*} \frac {1}{2}x-u & =F\left ( \frac {1}{2}x-\frac {1}{3}y\right ) \\ u & =\frac {1}{2}x-F\left ( \frac {1}{2}x-\frac {1}{3}y\right ) \end {align*}
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