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

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.