0
$\begingroup$

The evolution of certain quantity Represented by differential equation $\displaystyle \frac{dz}{dt}=1.5z.$ How long it take for $z$ to be double. If time $t$ measure in hours.

What i Try: $$\frac{dz}{z}=1.5dt\Longrightarrow \int^{2z_{0}}_{z_{0}}\frac{dz}{z}=\int^{t}_{0}1.5dt$$

$$\ln(2)=1.5t\Longrightarrow t=3\ln(2)$$

I have a doubt. When time $t=0.$ Can we take population $z_{0}$.

Please tell is is my thinking is right. Also please tell me is my solution is correct. Thanks

$\endgroup$

2 Answers 2

2
$\begingroup$

I think your confusion results from a combination of imprecise/inconsistent notation (e.g. wanting to choose a general $z_0$ while fixing $t_0 = 0$) and attempting to use a definite integral to solve the equation (when a general solution is clearer and more precise). I think a more instructive way to write the same work is as follows:

We first deduce a general solution to the differential equation, \begin{align*} \frac{\mathrm{dz}}{\mathrm{dt}} = 1.5z &\Leftrightarrow \frac{\mathrm{dz}}{z} = 1.5 \mathrm{dt} \\ &\Leftrightarrow \int \frac{\mathrm{dz}}{z} = \int 1.5 \mathrm{dt} \\ &\Leftrightarrow \mathrm{ln}(z) = 1.5t + C \\ &\Leftrightarrow z = D e^{1.5t}, \end{align*} so that if we have $z_0 = De^{1.5t_0}$ and $2z_0 = De^{1.5t_1}$ are quantities occurring at distinct times $t_0$ and $t_1$, then \begin{align*} \frac{2z_0}{z_0} &= \frac{De^{1.5t_1}}{De^{1.5t_0}} \\ 2 &= e^{1.5(t_1-t_0)}. \end{align*} That is, the doubling time is $$ t_1 - t_0 = \frac{\mathrm{ln}(2)}{1.5} $$ (note, as SarGe mentions, that your answer has a small error here).

Writing things this way emphasizes the independence of the doubling time from the "free variable" $D$ (as $D$ cancels in the calculations), which is an important conceptual idea. In addition, that our final answer is a difference $t_1 - t_0$ (not a single value $t$) reinforces that for this differential equation the doubling time has nothing to do with the exact starting/stopping values, just their difference.

$\endgroup$
1
$\begingroup$

Yes, you can take initial population to be $z_0$ or anything else provided the final population is double of that. Also the general relation between population and time is given as $$\ln \left(\frac{z}{z_0} \right) =1.5t$$ Hence, there is correction in your answer, it should be $\displaystyle\frac{2}{3}\ln 2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .