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
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