# Math Help - solving differencial equations

1. ## 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)

(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).

(logistic growth model)

where is the carrying capacity.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Here's the graph of the function.

-Dan

3. 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)

(Unbounded population model)
I'll just mention:
$\frac{dP}{P} = r dt$

-Dan

4. 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).

(logistic growth model)

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

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

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

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

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

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

-Dan

5. Originally Posted by googoogaga

(logistic growth model)

where is the carrying capacity.

This is also seperable:

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

RonL

6. Originally Posted by topsquark
$\frac{dP}{dt} - rP = -\frac{r}{K}P^2$
$\frac{dP}{dt} - rP = -\frac{r}{K}P^2\iff\frac{P'}{P^2}-\frac{r}{P}=-\frac{r}{K}$

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