$x(t)$ grows in time according to $\frac{dx}{dt}=(x-1)(2x-1).$ Knowing that $x(0)=0,$ after how much time does it reach reach $50\%$ of its ultimate value as time passes?
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- Integrate
$\int (6x-4)^2 dx.$ - 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
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How could you manipulate this differential equation into quantities you can integrate?
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How can you rewrite the product of the resulting two fractions as a sum/difference?
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What does “Ultimate value” mean in terms of time?
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… how about letting time go to infinity?
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Which limit to infinity do you need to evaluate to find the ultimate value of the population?
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Separating the variables (
$x$on one side,$t$on the other), we can rewrite the differential equation as$dt = \frac{dx}{(x-1)(2x-1)}.$Use partial fractions to split the product:$dt=\big(\frac{1}{x-1} - \frac{2}{2x-1}\big)dx.$We can now integrate (directly or otherwise):$$\begin{aligned} \int{dt} &= \int{\frac{dx}{x-1}} - 2\int{\frac{dx}{2x-1}} \\ t &= \ln|x-1| - \ln|2x-1| + C \\ &= \ln \left| \frac{x-1}{2x-1} \right| + C \end{aligned}$$Using the initial condition
$x(0)=0,$we find that that$C=0.$Therefore$t=\ln \big|\frac{x-1}{2x-1}\big|,$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$t_{50\%}=\ln\frac{3}{2}.$