Author Topic: differential equation question (Read 923 times) Tweet Share

Bhootnike · « **on:** May 19, 2012, 06:06:22 pm »

A population of plants is decreasing because of the effects of a virus. If the population is
decreasing according to the differential equation $\frac{dP}{dt} = -\frac{P}{50}$ , where t is the time in days, find
the time taken for the population to decrease to 25% of its original number

how do you start this question ?

all i could do was, flip over both sides, and find t = ....log... , but then yeah, stumped.

b^3 · « **Reply #1 on:** May 19, 2012, 07:29:40 pm »

$t=\int-\frac{50}{P}dP$ (flipping it over)
$\begin{alignedat}{1}t & =-50\log_{e}(P)+C_{1} \\ -\frac{1}{50}(t-C_{1}) & =\log_{e}(P) \\ P & =e^{-\frac{1}{50}(t-C_{1})} \end{alignedat}$
Let $e^{C_{1}}=A$
$P=Ae^{-\frac{1}{50}t}$
Now at $t=0$ , the population is $P_{0}$
$\begin{alignedat}{1}P_{0} & =Ae^{0} \\ A & =P_{0} \end{alignedat}$
Now the other piece of information we have is that when $t=t_{1}$ , then $P=\frac{1}{4}P_{0}$
$\begin{alignedat}{1}\frac{1}{4}P_{0} & =P_{0}e^{-\frac{1}{50}t_{1}} \\ \frac{1}{4} & =e^{-\frac{1}{50}t_{1}} \\ -\frac{1}{50}t_{1} & =\log_{e}(\frac{1}{4}) \\ t_{1} & =-50\log_{e}(\frac{1}{4}) \\ t_{1} & =50\log_{e}(4) \end{alignedat}$

Alternatively you could have worked out $C_{1}$ and subbed it into the formula you found for t originally, that way you wouldn't have had to re-rearange, but sometimes its nice to find the general solution to the differential equation.

EDIT: Fixed up slight error.
EDIT2: Fixed the major mistake, thanks Mao.

b^3 · « **Reply #2 on:** May 19, 2012, 10:05:38 pm »

Quote from: Mao on May 19, 2012, 09:47:54 pm

@b^3. your integrand in line 1 is upside down.

Oh right, it is too....... (ah the wonders of stupid mistakes, and yet I still seemed to integrate the P right, just with the wrong coefficient... -_-)
Thanks for picking that up Mao

, I'll re-edit and fix it up now.

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