Okay I think I missed the fact that this is a linear ODE…Here’s a solution I think is right:
dy/dx = x + y
GENERAL FORM:
dy/dx + H(x) y + Z(x) = 0; H(x) = -1 and Z(x) = -x
THEREFORE:
dy/dx – y – x = 0
MULTIPLY THROUGH INTEGRATING FACTOR e^∫ [ H(x) dx ]=e^ (-∫dx)=e^(-x+C)=Ce^(-x):
(dy/dx)Ce^(-x) – yCe^(-x) – xCe^(-x) = 0
YOU CAN PROVE d/dx [ yCe^(-x) ] = (dy/dx)Ce^(-x) – yCe^(-x) VIA CHAIN RULE:
d/dx ( yCe^(-x) ) = -yCe^(-x) + Ce(-x) dy/dx
THEREFORE SUBSTITUTING THE FORMER FOR THE LATTER:
d/dx ( yCe^(-x) ) – xCe^(-x) = 0
INTEGRATE BOTH SIDES:
∫ [d/dx ( yCe^(-x) ) dx – ∫ [xCe^(-x)] dx = 0
SIMPLIFY:
yCe^(-x) – C∫ [xe^(-x)] dx = 0
REARRANGE:
yCe^(-x) = C∫ [xe^(-x)] dx
SIMPLIFY:
ye^(-x) = ∫ [xe^(-x)] dx
y = e^x ∫ [xe^(-x)] dx
y = e^x [ – e^(-x) (x+1) + D ]; where D is a constant
y = e^x [ -e^(-x)x -e^(-x) + D ]
y = – x e^x e^(-x) – e^x e^(-x) + D e^x
y = -x -1 + De^x
y = De^x – x – 1
Can anyone verify as it has been a long time since I’ve done anything like this…And what’s with the reference in the question to u = x + y?