dy/dx = x - yUse substitution:v = x - yDifferentiate with respect to xdv/dx = 1 - dy/dxdy/dx = 1 - dv/dxNow we use above substitutions in differential equationsdy/dx = x - y1 - dv/dx = vdv/dx = 1 - vdv/(1-v) = dxNow integrate both sides:

intg dv/(1-v) = intg dx

-ln (1-v) = x+c

ln(1-v) = -x - c

1 - v = e^(-x-c) = [e^(-x)][e^(-c)]

1 - v = (C)e^(-x) , where C = e^(-Cc)

1 - (x - y) = (C)e^(-x)

y - x + 1 = (C)e^(-x)

y = (C)e^(-x) + x - 1

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