Added January 2, 2019.
Problem 1.3 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for \(w(x,y)\) \[ w_x = w f(x,y) \]
Mathematica ✓
ClearAll["Global`*"]; pde = D[w[x, y], x] == w[x, y]*f[x, y]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, w[x, y], {x, y}], 60*10]];
\[\left \{\left \{w(x,y)\to c_1(y) \exp \left (\int _1^xf(K[1],y)dK[1]\right )\right \}\right \}\]
Maple ✓
restart; pde := diff(w(x,y),x)=w(x,y)*f(x,y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,w(x,y))),output='realtime'));
\[w \left ( x,y \right ) ={\it \_F1} \left ( y \right ) {{\rm e}^{\int \!f \left ( x,y \right ) \,{\rm d}x}}\]
Hand solution
\begin {align*} \frac {\partial w}{\partial x} & =wf\left ( x,y\right ) \\ \frac {1}{w}\frac {\partial w}{\partial x} & =f\left ( x,y\right ) \end {align*}
Integrating both sides w.r.t. \(x\) gives\begin {align*} \ln \left ( w\right ) & =\int _{0}^{x}f\left ( s,y\right ) ds+G\left ( y\right ) \\ w & =e^{\int _{0}^{x}f\left ( s,y\right ) ds+G\left ( y\right ) }\\ & =F\left ( y\right ) e^{\int _{0}^{x}f\left ( s,y\right ) ds} \end {align*}
Where \(F\left ( y\right ) =e^{G\left ( y\right ) }\)
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