# solving differencial equations

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• Jul 17th 2007, 08:23 AM
googoogaga
solving differencial equations
These homework problems that I need help on are due tommorrow, eastern standard time. I am stuck could someone please help? Here they are:

1) Solve the initial value problem: dx/dt = 1- x^(2), x(0)= 3

2) The world population in 1990 was 5,274.320M and in 2000 was 6,073.265M. Predict the world population in 2007 and 2050 by using the Unbounded Population Model. Is your answer greater, lesser, or equal to
the current world population posted on
U.S. and World Population Clocks - POPClocks (Solve w/o Maple)

http://www.mathhelpforum.com/math-he...c3fd4448-1.gif (Unbounded population model)

3) the world population in 1980 was 4,447.068M, in 1990 was 5,274.320M
and in 2000 was 6,073.265M. Predict the world population in 2007 and
2050 by using the Bounded Population Model. Also, what is the limiting
population (carrying capacity) of the world? (Solve w/o Maple).

http://www.mathhelpforum.com/math-he...72ba832d-1.gif (logistic growth model)

where http://www.mathhelpforum.com/math-he...3c97c188-1.gif is the carrying capacity.

• Jul 17th 2007, 08:48 AM
topsquark
Quote:

Originally Posted by googoogaga

1) Solve the initial value problem: dx/dt = 1- x^(2), x(0)= 3

You've seen a number of these. By now you should be looking to see if the equation is separable:
$\displaystyle \frac{dx}{dt} = 1 - x^2$

$\displaystyle \frac{dx}{1 - x^2} = dt$

$\displaystyle \int \frac{dx}{1 - x^2} = \int dt$

$\displaystyle \frac{1}{2} ln \left | \frac{x + 1}{x - 1} \right | = t + C$

Your initial condition is $\displaystyle x(0) = 3$, so
$\displaystyle \frac{1}{2} ln \left | \frac{3 + 1}{3 - 1} \right | = 0 + C$

So
$\displaystyle C = \frac{1}{2} ln \left | \frac{4}{2} \right | = \frac{1}{2} ln(2)$

Thus the (implicit) solution to the differential equation is
$\displaystyle \frac{1}{2} ln \left | \frac{x + 1}{x - 1} \right | = t + \frac{1}{2} ln(2)$

If you wish an explicit solution:
$\displaystyle ln \left | \frac{x + 1}{x - 1} \right | = 2t + ln(2)$

$\displaystyle \left | \frac{x + 1}{x - 1} \right | = e^{2t + ln(2)} = e^{2t}\cdot e^{ln(2)} = 2e^{2t}$

Now, there is no value of t such that $\displaystyle 2e^{2t}$ is negative, so we may drop the absolute value bars:
$\displaystyle \frac{x + 1}{x - 1} = 2e^{2t}$

$\displaystyle x + 1 = (x - 1)2e^{2t}$

$\displaystyle x + 1 = x2e^{2t} - 2e^{2t}$

$\displaystyle x - x2e^{2t} = -1 - 2e^{2t}$

$\displaystyle x(1 - 2e^{2t}) = -(1 + 2e^{2t})$

$\displaystyle x = -\frac{1 + 2e^{2t}}{1 - 2e^{2t}}$

Here's the graph of the function.

-Dan
• Jul 17th 2007, 08:56 AM
topsquark
Quote:

Originally Posted by googoogaga
2) The world population in 1990 was 5,274.320M and in 2000 was 6,073.265M. Predict the world population in 2007 and 2050 by using the Unbounded Population Model. Is your answer greater, lesser, or equal to
the current world population posted on
U.S. and World Population Clocks - POPClocks (Solve w/o Maple)

http://www.mathhelpforum.com/math-he...c3fd4448-1.gif (Unbounded population model)

I'll just mention:
$\displaystyle \frac{dP}{P} = r dt$

-Dan
• Jul 17th 2007, 09:01 AM
topsquark
Quote:

Originally Posted by googoogaga
3) the world population in 1980 was 4,447.068M, in 1990 was 5,274.320M
and in 2000 was 6,073.265M. Predict the world population in 2007 and
2050 by using the Bounded Population Model. Also, what is the limiting
population (carrying capacity) of the world? (Solve w/o Maple).

http://www.mathhelpforum.com/math-he...72ba832d-1.gif (logistic growth model)

where http://www.mathhelpforum.com/math-he...3c97c188-1.gif is the carrying capacity.

This one is a bit different. We have:
$\displaystyle \frac{dP}{dt} = rP - \frac{r}{K}P^2$

$\displaystyle \frac{dP}{dt} - rP = -\frac{r}{K}P^2$

This is a Bernoulli equation. The exponent of P is 2, so substitute:
$\displaystyle w = \frac{1}{P^{2 - 1}} = \frac{1}{P}$

Then
$\displaystyle P = \frac{1}{w}$ ==> $\displaystyle \frac{dP}{dt} = -\frac{1}{w^2}\frac{dw}{dt}$

So the differential equation becomes:
$\displaystyle -\frac{1}{w^2}\frac{dw}{dt} - \frac{r}{w} = - \frac{r}{K}\frac{1}{w^2}$

$\displaystyle \frac{dw}{dt} + rw = \frac{r}{K}$
which is a linear differential equation in w(t), which you should be able to solve.

-Dan
• Jul 17th 2007, 10:15 AM
CaptainBlack
Quote:

Originally Posted by googoogaga

This is also seperable:

$\displaystyle \int \frac{1}{P(1-P/K)} dP =\int r dt$

RonL
• Jul 17th 2007, 01:38 PM
Krizalid
Quote:

Originally Posted by topsquark
$\displaystyle \frac{dP}{dt} - rP = -\frac{r}{K}P^2$

$\displaystyle \frac{dP}{dt} - rP = -\frac{r}{K}P^2\iff\frac{P'}{P^2}-\frac{r}{P}=-\frac{r}{K}$

It's faster if you set $\displaystyle w=\frac1{P}$