Form a partial differential equation from the relation z=f(y)+φ(x+y)
Integrating partially wrt y,
δz/δy = δf/δy + δφ/δy
δz/δy = δf/δy + (δφ/δ(x+y)).(δ(x+y)/δy)
δz/δy = δf/δy + (δφ/δ(x+y)).(1)
δz/δy = f’ + φ’ -------------------- (1)
Integrating partially wrt x,
δz/δx = δf/δx + δφ/δx
δz/δx = (δφ/δ(x+y)).(δ(x+y)/δx)
δz/δx = (δφ/δ(x+y)).(1)
δz/δx = φ’ ----------------------- (2)
Substituting for φ’ from (2) into (1),
δz/δy = δf/δy + δz/δx ------------- (3)
The above DE is a partial differential equation formed from the relation z=f(y)+φ(x+y).
The solution to the DE (3) generates an arbitrary function of (x + y), viz. φ(x+y).
For the sake of interest and clarification, the solution follows.
Rearranging (3),
δz/δy - δf/δy = δz/δx
δ(z – f)/δy = δz/δx ---------------- (4)
Let u(x,y) = z – f
Then δu/δx = δz/δx
Substituting for u and δu/δx into (4),
δu/δy = δu/δx
=> u(x,y) = φ(x + y)
Check:
δu/δx = δφ/δx = (δφ(x+y)/δ(x+y))(δ(x+y)/δx) = φ’
δu/δy = δφ/δy = (δφ(x+y)/δ(x+y))(δ(x+y)/δy) = φ’
i.e. δu/δx = δu/δy
Finally, from u(x,y) = φ(x + y),
z – f = φ(x + y)
z(x,y) = f(y) + φ(x + y)