Taken from Mathematica help pages
Solve for \(u(x,y)\) \[ 3 u_x + 5 u_y = x \]
Mathematica ✓
ClearAll["Global`*"]; sol = AbsoluteTiming[TimeConstrained[DSolve[3*D[u[x, y], x] + 5*D[u[x, y], y] == x, u[x, y], {x, y}], 60*10]];
\[\left \{\left \{u(x,y)\to \frac {x^2}{6}+c_1\left (y-\frac {5 x}{3}\right )\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde :=3*diff(u(x, y), x) + 5*diff(u(x, y), y) = x; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime'));
\[u \left ( x,y \right ) ={\frac {{x}^{2}}{6}}+{\it \_F1} \left ( -{\frac {5\,x}{3}}+y \right ) \]
Hand solution
Solve\begin {align} 3u_{x}+5u_{y} & =x\nonumber \\ u_{x}+\frac {5}{3}u_{y} & =\frac {x}{3}\tag {1} \end {align}
Solution
Let \(u=u\left ( y\left ( x\right ) ,x\right ) \). Then \begin {equation} \frac {du}{dx}=\frac {\partial u}{\partial y}\frac {dy}{dx}+\frac {\partial u}{\partial x}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dx} & =\frac {x}{3}\tag {3}\\ \frac {dy}{dx} & =\frac {5}{3}\tag {4} \end {align}
Solving (3) gives\begin {align*} u & =\frac {x^{2}}{6}+C_{1}\\ C_{1} & =u-\frac {x^{2}}{6} \end {align*}
From (4)\begin {align*} y & =\frac {5}{3}x+C_{2}\\ C_{2} & =y-\frac {5}{3}x \end {align*}
Let \(C_{1}=F\left ( C_{2}\right ) \) where \(F\) is arbutrary function. This gives\begin {align*} u-\frac {x^{2}}{6} & =F\left ( y-\frac {5}{3}x\right ) \\ u\left ( x,y\right ) & =F\left ( y-\frac {5}{3}x\right ) +\frac {x^{2}}{6} \end {align*}
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