<s>[INST] Differential Equation $y' = \frac{ty(4-y)}{(1+y)}$ $y' = \frac{ty(4-y)}{(1+y)}$ 
given $ y_0 = 2 $
At what time $t$  when the solution will first be 3.9
I tried solving this but it didnt work out so well.
What I did was:


*

*separate dy/dx and move all $y$ terms to one side with $dy$

*Some how I don't know how to deal with the $dx$, so I couldn't come up with the solution , $y$.

 [/INST] Separate the $y$ and $t$ terms on either side of the equal side and get
$$ t \, dt = \frac{1+y}{y (4-y)} dy $$
The right-hand side may be broken up as
$$\begin{align} \frac{1+y}{y (4-y)} &= \frac{1}{y (4-y)} + \frac{1}{4-y} \\ &=\frac{1}{4} \left ( \frac{1}{y} + \frac{1}{4-y} \right ) + \frac{1}{4-y} \\ &= \frac{1}{4 y} + \frac{5}{4} \frac{1}{4-y} \end{align}$$
Now integrate both sides:
$$ \frac{1}{2} t^2 + C = \frac{1}{4} [\log{y} - 5 \log{(4-y)}] $$
or
$$ 2 t^2 + C = \log{\left [ \frac{y}{(4-y)^5} \right ]}$$
Using $y(0)=2$ implies that $C=-\log{16}$.  Then
$$t^2 = \frac{1}{2} \log{\left [ \frac{16 y}{(4-y)^5} \right ]}$$
When $y=3.9$, $t$ will be
$$t^2 = \frac{1}{2} \log{\left [ \frac{62.4}{(0.1)^5} \right ]}$$
This gives
$$t \approx 2.797$$
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