Right answer is (b) (x+y-2)=c(x-y)^3
Explanation: \(\frac{dy}{dx} = \frac{x+2y-3}{2x+y-3}\) –>x=X+h, y=Y+k
h+2k-3=0 & 2h+k-3=0 solving we get h=1=k
new differential equation is \(\frac{dY}{dX} = \frac{X+2Y}{2X+Y}\), put Y=vX
\(v + X\frac{dv}{dX} = \frac{1+2v}{2+v} \rightarrow X \frac{dv}{dX} = \frac{1-v^2}{2+v}\)
or
\( \int(\frac{2+v}{1-v^2}) dv = \int \frac{1}{X} \,dX \rightarrow \int(\frac{2}{1-v^2} + \frac{v}{1-v^2}) \,dv = log X + log c\)
\(log\left(\frac{1+v}{1-v}\right) – 0.5 log (1-v^2) = log X + log c\)
\(log \left(\frac{X+Y}{X-Y}\right) – 0.5 log (X^2-Y^2) + log X = log X + log c\)
\(log \left(\frac{X+Y}{X-Y}.\frac{1}{\sqrt{X^2-Y^2}}\right) = log c\)
\(\sqrt{X+Y}\) = (X-Y)^1.5 c or (X+Y)=(X-Y)^3c
(x+y-2)=c(x-y)^3 is the solution.