ODE Separable Equation Let $y = Φ(x)$ be a solution to $y' = y(5-y)(8-y)$ subject to $y(0) = 7$. Determine $\lim_{x \to ∞} Φ(x)$.
Workings:
I'm thinking I have to solve the differential equation.
$y' = y(5-y)(8-y) dy$
$y' = y(y-5)(y-8) dy$
$∫\frac{1}{(y-5)(y-8)y} dy = ∫1 dx$
$∫\frac{1}{(y-5)(y-8)y} dy$
$\frac{1}{(y-5)(y-8)y} = \frac{A}{(y-5)} + \frac{B}{(y-8)} + \frac{C}{y}$
Greatest Common Divisor gives:
$1 = Ay(y-8) + By(y-5) + C(y-5)(y-8)$
$y = 0:$
$1 = A(0)(0-8) + B(0)(0-5) + C(0-5)(0-8)$
$1 = 40C$
$C = \frac{1}{40}$
$y= 5:$
$1 = A(5)(5-8) + B(5)(5-5) + C(5-5)(5-8)$
$1 = -15A$
$A = \frac{-1}{15}$
$y = 8:$
$1 = C(8)(8-8) + B(8)(8-5) + C(8-5)(8-8)$
$1 = 24B$
$B = \frac{1}{24}$
$\frac{A}{(y-5)} + \frac{B}{y} + \frac{C}/{(y-8)}$
$= \frac{(-1/15)}{(y-5)} + \frac{(1/24)}{y} + \frac{(1/24)}{(y-8)}$
$= \frac{-1}{(15(y-5))} + \frac{1}{40y} + \frac{1}{(24(y-8))}$
$∫\frac{-1}{(15(y-5))} + \frac{1}{40y} + \frac{1}{(24(y-8))} dy$
$= \frac{-1}{15} \ln|y-5| + \frac{1}{40} \ln|y| + \frac{1}{24} \ln|y-8|$
$∫1 dx$
$= x + c$
$\frac{-1}{15} \ln|y-5| + \frac{1}{40} \ln|y| + \frac{1}{24} \ln|y-8| = x + c$
(I'm not too sure if I should solve for y here or not. There is another question later in the assignment where it states it can be expressed implicitly)
Letting $y(0) = 7$:
$\frac{-1}{15} \ln |(7)-5| + \frac{1}{40} \ln|(7)| + \frac{1}{24} \ln|(7)-8| = (0) + c$
$\frac{-1}{15} \ln|2| + \frac{1}{40} \ln|7| + \frac{1}{24} \ln|-1| = c$
$\frac{-1}{15} \ln|y-5| + \frac{1}{40} \ln|y| + \frac{1}{24} \ln|y-8| = x - 1/15 \ln|2| + 1/40 \ln|7| + 1/24 \ln |-1|$
I'm not sure if I did this correctly. Or what to do next. Any help will be appreciated.

to find the limit there is no need to solve the equation. look at the steady state solutions $y = 0, y = 5 y = 8.$ 
this is called the phase portrait:
(a) on  a number line mark these points. 
(b) on each region bounded by these points, determine the sign of $y^\prime$ by evaluating $y^\prime$ at  a test point.
here is what it looks like for your problem:
sign of $y^\prime = y(5-y)(8-y):$  (-)     (+)        (-)         (+) from left to right.
$-\infty$--------<<--------0-------->>------5------<<---------8------>>---------$\infty$
if the initial $y$ value is $7,$ the solution will approach $5$ as $t \to \infty$ and $y \to 8$ as $t \to -\infty$
by uniqueness theorem for differential equation, the solutions may never reach, let alone cross these special solutions.
p.s. appreciate if anyone can augment this post with a line diagram.
