Solve dy/dx = (x+2y−3)/(2x+y−3)

dy/dx = (x + 2y − 3)/(2x + y − 3)

The numerator and denominator are linear but not homogeneous. The lines x + 2y − 3 = 0 and 2x + y − 3 = 0 meet at a point; solving them gives x = 1, y = 1.

Shift the origin to (1, 1)

Put x = X + 1, y = Y + 1. Then dy/dx = dY/dX and

dY/dX = (X + 2Y)/(2X + Y)

Homogeneous substitution

Put Y = vX, so dY/dX = v + X dv/dX:

v + X dv/dX = (1 + 2v)/(2 + v)

X dv/dX = (1 + 2v − 2v − v²)/(2 + v) = (1 − v²)/(2 + v)

(2 + v)dv/[(1 − v)(1 + v)] = dX/X

Partial fractions

(2 + v)/[(1 − v)(1 + v)] = (3/2)/(1 − v) + (1/2)/(1 + v)

Integrating:

−(3/2) ln|1 − v| + (1/2) ln|1 + v| = ln|X| + const

(1 + v)/(1 − v)³ = C X²

Put v = Y/X: (X + Y)/X · X³/(X − Y)³ = CX², so X + Y = C(X − Y)³.

Back to x and y

X + Y = x + y − 2 and X − Y = x − y:

x + y − 2 = C(x − y)³