Solve `(dy)/(dx) = (x+2y - 3)/(2x + y - 3)`

Explanation

Here `a_(1) = 1, b_(1) = 2 " " :. (a_(1))/(b_(1)) = (1)/(2)`
`a_(2) = 2, b_(2) = 1 " " :. (a_(2))/(b_(2)) = 2`
Now, `(a_(1))/(b_(1)) != (a_(2))/(b_(2))`
Thus case - I applies, we have setting,
`{:(x=u+h),(y=v+k):}}`
`(dv)/(du) = ((u+2v) + (h+2k-3))/((2u + v) + (2h + k-3))0`
Choose h, k such that
`{:(h+2k-3=0),(2h + k - 3 =0):}}`
The solution is h = 1, k = 1
`rArr (dv)/(du) = (u+2v)/(2u + v)`
Set `v = tu, rArr (dv)/(du) = t + (dt)/(du)`
Our equation reduces to
`t + u(dt)/(du) = (u+2tu)/(2u + tu) = (1+2t)/(2+t)`
`rArr u(dt)/(du) = (1+2t)/(2+t) - t = (1+2t-2t-t^(2))/(2+t) = (1-t^(2))/(2+t)`
`rArr (2+t)/(1-t^(2)) dt = (du)/(2)`
`rArr (2)/(1-t^(2))dt+(1)/(2).2(t)/(1-t^(2)) dt = (du)/(u)`
On integrating, we get
`ln|(1+t)/(1-t)| - (1)/(2)ln|1-t^(2)| = ln|u|+ ln|k|`, k being the constant of integration.
`rArr ln|((1+(v)/(u))/(1-(v)/(u)))|-(1)/(2)ln(1-(v^(2))/(u^(2)))| = ln(uk)|`
`rArr |((u+v)/(u-v))| - ln|(sqrt(u^(2)-v^(2)))/(u)| = ln|(uk)|`
`rArr ln|((u+v)/(u-v).(u)/(sqrt(u^(2)-v^(2))))| = ln|uk|`
`rArr (usqrt(u+v))/((u-v)^((3)/(2))) = +- uk`
`rArr k(u-v)^((3)/(2)) = +-sqrt(u+v)`
`rArr k(x-1-y+1)^((3)/(2)) = sqrt(x-1+y-1)`
`rArr k(x-y)^((3)/(2)) = sqrt(x+y-2)`
`:. k^(2)(x-y)^(3) = (x+y-2)`
`rArr (x-y)^(3) = lambda(x+y-2), lambda` being another constant.