# Homework Help: Differential equations - population model

1. Jun 12, 2005

### DivGradCurl

"Suppose that a certain population satisfies the initial value problem

$$\frac{dy}{dt}=r(t)y - k, \qquad y(0)=y_0$$

where the growth rate $$r(t)$$ is given by

$$r(t)=\frac{1+\sin t}{5}$$

and $$k$$ represents the rate of predation.

(a) Suppose that $$k=\frac{1}{5}$$. Plot $$y$$ versus $$t$$ for several values of $$y_0$$ between 1/2 and 1.

(b) Estimate the critical initial population $$y_c$$ below which the population will become extinct.

(c) Choose other values of $$k$$ and find the corresponding $$y_c$$ for each one.

(d) Use that data you have found in parts (a) and (b) to plot $$y_c$$ versus $$k$$."

I'm trying to apply the Method of Integrating Factors, but I'm stuck. Here's what I have:

$$\frac{dy}{dt}- r(t)y = - k$$

$$\mu = \exp \left[ -\frac{1}{5} \int \left( 1 + \sin t \right) \: dt \right] = \exp \left( \frac{\cos t}{5} - \frac{t}{5} \right)$$

$$y(t) = \exp \left( \frac{t}{5} - \frac{\cos t}{5} \right) \int -k \exp \left( \frac{\cos t}{5} - \frac{t}{5} \right) \: dt$$

$$y(t) = -k \exp \left( \frac{t}{5} - \frac{\cos t}{5} \right) \underbrace{\int \exp \left( \frac{\cos t}{5} - \frac{t}{5} \right) \: dt} _{\mbox{?}}$$

Any help is highly appreciated.

2. Jun 12, 2005

### Pyrrhus

the integral you underlined cannot be expressed in elemental functions, but you can still work the rest of the problems.

Last edited: Jun 12, 2005
3. Jun 13, 2005

### DivGradCurl

I can get the vector field of this differential equation without any problem. Still, I can't plot $$y(t)$$ for several values of $$y_0$$---part (a)---because I can't find $$y(t)$$ in the first place. I've attached a Mathematica 5 notebook that shows it.

Thanks

#### Attached Files:

• ###### DifferentialEqns_PopulationModel.zip
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4. Jun 13, 2005

### saltydog

Dude, you got Mathematica and you're sayin' you can't? What about NDSolve? Here, I know it by heart by now cus' I use it all the time:

sol1=NDSolve[{y'[t]-(1+Sin[t])/5 y[t]== -k,y(0)=ystart},y,{t,0,whatever}];

f[x_]:=Evaluate[y[x] /. sol1];

Edit: Tell you what Thiago, what about writing a complete Mathematica program (only because the integral is non-elementary) that does all the calculating, searching for the critical values, then outputs a nice plot of y0 vs. k? Whatever. Anyway, why does this happen? That would make a good e don't you think?

Last edited: Jun 13, 2005
5. Jun 13, 2005

### saltydog

Is it just me or is this a tough problem to code for other people? I don't know. I'm kinda slow. Thiago, I've attached a plot of what I came up with. Are you getting the same thing? If you got it, good. If not and want help, just post a message. Still can't say why it behaves that way. You? Don't suppose you want to work on e do you?

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• ###### thiago_ode.JPG
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6. Jun 14, 2005

### saltydog

e

Alright, I got e. Thiago, I don't want to be doing your homework alright. In fact, e is not even your homework so I think I'm ok.

First instead of integrating indefinitely, do so definitely from $t_0$ to t:

$$\int_{t=0,y=y_0}^{t=t,y=y} d[e^{1/5(Cos(t)-t)}y]=-k\int_{t_0}^t e^{1/5(Cos(t)-t)}$$

I know, the t's, the y's, y0, is confussing on the left hand side but you know what I mean.

This yields:

$$y(t)=e^{-1/5(Cos(t)-t)}[y_0e^{1/5}-k\int_0^t e^{1/5(Cos(u)-u)}du]$$

Now,

$$e^{-1/5(Cos(t)-t)}$$

is always greater than 0, so y(t) will reach zero only when the expression in brackets reaches zero. That is, when:

$$y_0=ke^{-1/5}\int_0^t e^{1/5(Cos(u)-u)}du$$

Now hear's the important point: The critical point is when $y_0$ just equals that expression and since the integral is an increasing function which reaches a limit, We can say the following about the critical point $y_c$

$$y_c=ke^{-1/5}\int_0^{\infty} e^{1/5(Cos(u)-u)}du$$

Numerically, the integral is approximately 5.089. Thus we have the following expression for the critical point as a function of k:

$$y_c=ak$$

with:
$$a\approx 5.089 e^{-1/5}$$

The three plots exhibit this. The first plot is the one I generated above using a brute force approach to search for the critical points. The second one is the expression for $y_c$. The third one is a superposition of both. That is, they are identical.

#### Attached Files:

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• ###### superposition.JPG
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