<s>[INST] 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.
 [/INST] 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.
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