Solve PDE in 2D Problem How should I go about solving this PDE:
$$ \phi_x+\phi_y=x+y-3c  $$
Where $\phi = \phi(x,y)$, $c$ is a constant, and $\phi$ is specified on the circle
$$ x^2+y^2=1  $$
My Attempt to solve it I would like to use the method of characteristics, but then I get stuck because of the given initial condition. In fact, so far I have the characteristics equations
$$ \dot{{z}}(s)=x+y-3c $$
$$ \dot{{x}}(s)= 1 $$
$$ \dot{{y}}(s)= 1 $$
The last two are easy to solve but then I am not sure how to use the initial condition.
If you know of a different/easier method to solve this PDE, feel free to let me know, thanks!

This problem can also be solved by a substitution:
$$
   x = s+t,\;\;\; y = s-t,\;\;\; \psi(s,t)=\phi(s+t,s-t).
$$
Then
$$
   \psi_{s} = \phi_{x}+\phi_{y} = x+y-3c = 2s-3c
$$
Then there is a function $d(t)$ such that
$$
    \psi = s^{2}-3cs+d(t).
$$
The function $d$ is determined from $\psi(s,t)$, which is assumed to be known on $1=x^{2}+y^{2}=2s^{2}+2t^{2}$.
