$x(t)$
grows in time according to $\frac{dx}{dt}=(x1)(2x1).$
Knowing that $x(0)=0,$
after how much time does it reach reach $50\%$
of its ultimate value as time passes?

 Integrate
$\int (6x4)^2 dx.$
 Solve
$\frac{dy}{dx}=\cos x$
to find$y(x),$
given that$y(0)=1.$
 Evaluate
$\lim\limits_{x \to \infty} \big(\frac{3x^2+7x+2}{x^2+1}\big).$
 Integrate

How could you manipulate this differential equation into quantities you can integrate?

How can you rewrite the product of the resulting two fractions as a sum/difference?

What does “Ultimate value” mean in terms of time?

… how about letting time go to infinity?

Which limit to infinity do you need to evaluate to find the ultimate value of the population?

Separating the variables (
$x$
on one side,$t$
on the other), we can rewrite the differential equation as$dt = \frac{dx}{(x1)(2x1)}.$
Use partial fractions to split the product:$dt=\big(\frac{1}{x1}  \frac{2}{2x1}\big)dx.$
We can now integrate (directly or otherwise):$$\begin{aligned} \int{dt} &= \int{\frac{dx}{x1}}  2\int{\frac{dx}{2x1}} \\ t &= \lnx1  \ln2x1 + C \\ &= \ln \left \frac{x1}{2x1} \right + C \end{aligned}$$
Using the initial condition
$x(0)=0,$
we find that that$C=0.$
Therefore$t=\ln \big\frac{x1}{2x1}\big,$
which we can rearrange to get$x(t)=\frac{1}{2}  \frac{1}{4e^t  2}.$
Evaluating
$\lim\limits_{t\to\infty} x(t)$
gives us the ultimate population. As$t\to$
approaches$\infty,$
$\frac{1}{4e^t  2}$
approaches$0,$
so the ultimate population is$\frac{1}{2},$
and$50\%$
of this is$\frac{1}{4}.$
Substituting, we find that$t_{50\%}=\ln\frac{3}{2}.$