It solves it in x. the solution for y includes heavy use of the product log function.
dy/dx = ay + b/cy +d
(cy +d/ ay + b) dy = dx
∫ (cy +d/ ay + b) dy = x (t) + C
Into solving the integral, integration by parts followed by u substitution and another integration by parts.
∫ (cy +d/ ay + b) dy
u = cy + d dv = dy/ay + b
du = c dy v = ln I ay + b l / a
Then, use u substitution for the new integral
u = ay + b
du = a dy
∫ ln l ay + b I dy = ∫ ln IuI /a du = 1/a ∫ ln luI du
Integrating the natural log includes thus far another integration by parts
r = ln IuI ds = du
dr = du / u (s) = du
∫ ln IuI / du = u ln IuI - ∫ du = u ln IuI - ∫ a dy = (ay + b) ln Iay +bl – ay
The original integral of expression
∫ (cy +d/ ay + b) dy = cy + d/a ln lay+bl – c/a² [(ay+b) ln lay+bl – ay]
Then simplify
∫ (cy +d/ ay + b) dy = cy + d/a ln lay+bl – c/a²[(ay+b) ln lay+bl – ay]
= a (cy + d)/a² ln lay+bl – c (ay+b)/ a²ln lay+bl + c/a² ay
= cay + ad – cay – cb/ a² ln lay+bl + cay/a²
= ad – cb/a²ln lay+bl + cy/a
The final answer will be
x(t) + C = ad – cb/a² ln lay+bl + cy/a
x(t) = ad – cb/a² ln lay+bl + cy/a + k
