Differential Equation help

**craig** · Dec 5th 2009, 01:16 PM

I have the DE, $\frac{dP}{dt} = B_0e^{-at}P$ , which I've solved to give $P = P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)}$ .

I'm also given that $B(t) = B_0e^{-at}$ and that $P(0) = P_0$ .

Q) Suppose that at time $t = 0$ there are $P_0 = 10^6$ cells and that P(t) is increasing at the rate of $3 \times 10^5$ cells per month. After 6 months the tumor has doubled (size and number of cells). Find a.

A)
We can say that $P = 10^6 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)}$ .

If the tumor has doubled in 6 months, then does this mean that $P(6) = 2P(0)$

$10^6 e^{\frac{B_0}{a}\left(1 - e^{-a6}\right)} = 2(10^6 e^{\frac{B_0}{a}\left(1 - e^{0}\right)})$ , therefore $e^{\frac{B_0}{a}\left(1 - e^{-6a}\right)} = 10^6(1)$

Also if $P(t)$ is increasing at a rate of $3 \times 10^5$ , I presume that $\frac{dP}{dt}=3 \times 10^6$ , or is it $B(t)= 3 \times 10^6$ ?

If someone could give me a gentle push in the right direction it would be much appreciated

Thanks a lot

Craig

**shawsend** · Dec 5th 2009, 02:01 PM

From your information, I can say:

$\ln(2x10^6)=-\frac{B_0}{a}e^{-6a}+c$

and you can easily find $c$ from the initial population at t=0. But the initial derivative means $3x10^5=-\frac{B_0}{a}+c$ . Now, taking this expression and substituting it into the first expression, can you not find a?

**craig** · Dec 7th 2009, 05:29 AM

Originally Posted by shawsend

From your information, I can say:

$\ln(2x10^6)=-\frac{B_0}{a}e^{-6a}+c$

and you can easily find $c$ from the initial population at t=0. But the initial derivative means $3x10^5=-\frac{B_0}{a}+c$ . Now, taking this expression and substituting it into the first expression, can you not find a?

Hi Shawsend thanks for the reply. If you don't mind could you explain where you got your values from, having a little difficultly in following along.

Also, do you mean $\ln(2x10^6)=-\frac{B_0}{a}e^{-6a}+c$ or $\ln(2 \times 10^6)=-\frac{B_0}{a}e^{-6a}+c$ ?

Thanks again

Craig

**craig** · Dec 7th 2009, 07:08 AM

Here's what I've come up with so far.

Using the fact that $P(6) = 2P(0)$ ,

$10^6 e^{\frac{B_0}{a}\left(1 - e^{-6a}\right)} = 2 \cdot10^6 e^{\frac{B_0}{a}\left(1 - e^{0}\right)}$

$e^{\frac{B_0}{a}\left(1 - e^{-6a}\right)} = 10^6$

$ln{(10^6)} = \frac{B_0}{a}\left(1 - e^{-6a}\right)$

Having trouble finding a second equation in terms of just $a$ and $B_0$ .

We also know that when $t=0$ , P(t) is increasing at $3 \times 10^5$ cells per month.

Using the expression for $\frac{dP}{dt}$ , and using $t=0$ , we get the following:

$3 \times 10^5 = B_0 \cdot P$ .

Not sure what to do from here, how to eliminate the $P$ ?

Thanks again for anyone reading through this

**craig** · Dec 7th 2009, 07:21 AM

Sorry I'm just carrying on with myself here aren't I, every time I post I seem to get a new idea how to tackle this problem.

From what I've said earlier, I've got the two following equations:

$ln{(10^6)} = \frac{B_0}{a}\left(1 - e^{-6a}\right)$

$B_0 \cdot P = 3 \times 10^5$

The second equation is true when $t=0$ , so what I thought is to use the value of $P$ when $t=0$ , ie $P_0$ .

This results in $B_0 10^6 = 3 \times 10^5$ , $B_0 = 0.3$

Now if I put this value into my first equation, this gives me:

$ln{(10^6)} = \frac{0.3}{a}\left(1 - e^{-6a}\right)$

Using Maple I've solved this to give me a value of $a=-0.5430780576$ .

However I know that both $a$ and $B_0$ are positive and real constants, so obviously something has gone wrong??

**shawsend** · Dec 7th 2009, 09:53 AM

Hi. Craig. Sorry. I made a mistake up there. I get:

$\ln(2)=\frac{3}{10a}\left(1-e^{-6a}\right)$

That's a very interesting equation to solve for a explicitly in terms of the Lambert W function. We can show:

$a=\frac{9+5 \text{Log}[2] \text{ProductLog}\left[-\frac{9 e^{-\frac{9}{5 \text{Log}[2]}}}{5 \text{Log}[2]}\right]}{30 \text{Log}[2]}\approx 0.3915$

Here's how I did it:

We have: $k=\frac{b}{a}(1-e^{-6a})$

$ake^{6a}-be^{6a}=-b$

$(6ka-6b)e^{6a}=-6b$

$(6a-\frac{6b}{k})e^{6a}=-\frac{6b}{k}$

$(6a-\frac{6b}{k})e^{6a-\frac{6b}{k}}=-\frac{6b}{k}e^{-6b/k}$

$6a-\frac{6b}{k}=\text{W}\left(-6b/k e^{-6b/k}\right)$

$a=\frac{b}{k}+\frac{1}{6}\text{W}\left(-\frac{6b}{k}e^{-6b/k}\right)$

**craig** · Dec 7th 2009, 10:03 AM

Hi thanks for the quick reply. When I tried to solve through Maple there was something about the Lambert W function.

Originally Posted by shawsend

$\ln(2)=\frac{3}{10a}\left(1-e^{-6a}\right)$

I understand the RHS of the equation, using the value of $B_0 = 0.3$ . But how did you get $\ln(2)$ and I got $\ln(10^6)$ ?

Thanks again

**shawsend** · Dec 7th 2009, 10:14 AM

We have the two equations:

$\ln(10^6)=-\frac{3}{10a}+c\rightarrow c=\ln(10^6)+\frac{3}{10 a}$

and:

$\ln(2\cdot 10^6)=-\frac{3}{10 a}e^{-6a}+c$

then:

$\ln(2)=\frac{3}{10 a}(1-e^{-6a})$

**craig** · Dec 7th 2009, 11:41 AM

Sorry about all these questions, you have the patience of a saint

I understand how you got from the two equations to give the answer for $a$ . I don't get how you formed these two equations though?

You have a $2 \times 10^6$ , but didn't these cancel out in this equation, $10^6 e^{\frac{B_0}{a}\left(1 - e^{-6a}\right)} = 2 \cdot10^6 e^{\frac{B_0}{a}\left(1 - e^{0}\right)}$ ?

Also I'm not sure where the $+ c$ has come from either? Obviously we need a constant of integration but didn't all this get dealt with when I first solved the differential equation in the first place to get an equation for $P$ ?

Sorry for the confusion, clearly something completely obvious I've failed too see

All I can say is thank you so much for the help, you've been an absolute life saver.

**shawsend** · Dec 7th 2009, 12:18 PM

We start with:

$P'=B_0 e^{-at}P$

the solution I'll leave in log form is:

$\ln(P)=-\frac{B_0}{a} e^{-at}+c$

Now, using the three given conditions:

At t=0, the derivative information tells us: $3\cdot 10^5=B_0 10^6\Rightarrow B_0=\frac{3}{10}$

At t=0, the initial conditions tells us: $\ln(10^6)=-\frac{3}{10 a}+c$

And at t=6, we have:

$\ln(2\cdot 10^6)=-\frac{3}{10 a} e^{-6t}+c$

**craig** · Dec 7th 2009, 12:31 PM

Originally Posted by shawsend

We start with:

$P'=B_0 e^{-at}P$

the solution I'll leave in log form is:

$\ln(P)=-\frac{B_0}{a} e^{-at}+c$

Now, using the three given conditions:

At t=0, the derivative information tells us: $3\cdot 10^5=B_0 10^6\Rightarrow B_0=\frac{3}{10}$

At t=0, the initial conditions tells us: $\ln(10^6)=-\frac{3}{10 a}+c$

And at t=6, we have:

$\ln(2\cdot 10^6)=-\frac{3}{10 a} e^{-6t}+c$

Thankyou, that's really helped explain it.

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