Solve the equation. dy dx = ay + b cy + d, where a, b, c, and d are constants. (assume a ≠ 0 and ay + b ≠ 0.)

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