I solved the PDE shown below and I got the solution. I am new to this tool. In the solution, I have a constant multiplied by a vector like function (please refer second term of the solution). What does it mean ? Appreciate your help. Thanks

pde = c1*D[u[x, y, z], x] + c2*D[u[x, y, z], y] + c3*D[u[x, y, z], z] +
a*u[x, y, z] - p == 0

soln = DSolve[pde, u[x, y, z], {x, y, z}]

1 Answer 1


Before we go to the PDE you solved, consider the following pde straight from the from the documentation.

DSolve[3 D[y[x1, x2], x1] + 5 D[y[x1, x2], x2] == x1, y[x1, x2], {x1, x2}]
{{y[x1, x2] -> 1/6 (x1^2 + 6 C[1][1/3 (-5 x1 + 3 x2)])}}

DSolve can solve both ODEs and PDEs with/without boundary conditions. When we don't give boundary conditions, results have arbitrary constants (for ODEs) and arbitrary functions (for PDEs). In the example above, we have a PDE with two independent variables (x1 & x2) and no boundary conditions are given; therefore, the solution has an arbitrary function C[1] (of one argument). This function is C[1][1/3 (-5 x1 + 3 x2)].

Now, in the PDE you solved you have three independent variables (x, y & z) and, therefore, the arbitrary function in the solution has two arguments. C[1][(-c2 x + c1 y)/c1, (-c3 x + c1 z)/c1].


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