DSolve and coupled linear first order PDEs

Does any one know a trick to make DSolve find solution to this coupled linear first order PDE system: (these are Cauchy-Riemann PDE equations, but with one of them having one of the dependent variables as well).

ClearAll[F1,F2,x,y];

ode1  = D[F1[x,y],y]-D[F2[x,y],x] == 0
ode2  = D[F1[x,y],x]+D[F2[x,y],y] == y  (*y here causes the problem*)

DSolve[{ode1,ode2},{F1[x,y],F2[x,y]},{x,y}] This can be solved in Maple:

restart;
#infolevel[pdsolve]:=3;
eq1:= diff(F1(x,y),y)-diff(F2(x,y),x) = 0;
eq2:= diff(F1(x,y),x)+diff(F2(x,y),y) = y;
pdsolve({eq1,eq2},{F1(x,y),F2(x,y)});

Solution it gives is

F1(x, y) = _F1(y-I*x)+_F2(y+I*x)
F2(x, y) = I*_F1(y-I*x)-I*_F2(y+I*x)+(1/2)*y^2+_C1

Screen shot: If the RHS of the second equation is not y but a constant or some other parameter, then Mathematica can now solve it:

ClearAll[F1,F2,x,y,m];
ode1  =  D[F1[x,y],y]-D[F2[x,y],x]  == 0
ode2  = D[F1[x,y],x]+D[F2[x,y],y]   == m
DSolve[{ode1,ode2},{F1[x,y],F2[x,y]},{x,y}] Is this a known limitation of DSolve or is there a trick or some other method to get the same solution as in Maple?

Using version 11.2 on windows 7.

• Would you view Simplify[Unevaluated[D[F1[x, y], x] + D[F2[x, y], y] == y] /. F2[x, y] -> G2[x, y] + y^2/2] as cheating? Jan 11 '18 at 4:39
• @bbgodfrey thanks, but I am not sure how to use the above trick. I trired ClearAll[F1,F2,G2,x,y,m]; ode1=D[F1[x,y],y]-D[F2[x,y],x]==0; ode2=Simplify[Unevaluated[D[F1[x,y],x]+D[F2[x,y],y]==y]/.F2[x,y]->G2[x,y]+y^2/2]; DSolve[{ode1,ode2},{F1[x,y],G2[x,y]},{x,y}] but it still does not solve it. Jan 11 '18 at 4:51
• Below is what I had in mind. Jan 11 '18 at 4:56