Form a partial differential equation from the relation Z=f(y)+φ(x+y)

Question

Pls show the steps to form a partial differential equation . The sum is related to elimination method.

Answer 1

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)