# Ordinary Differential Equations/Separable 2

First-Order Differential Equations

# Separable Variables: Real-World Examples

This page gives some examples of where simple separable variable DEs are found in the world around us.

## Acceleration, velocity, and position

The classic real world example of differential equations is the relationship between acceleration, velocity, and position.

$a(t)=v'(t)=x''(t) \,$

So if you're given an equation for acceleration, you can figure out both velocity and position.

### Example 1 - Constant Acceleration

Lets say that accelearation is a constant, g (the acceleration due to gravity, or about 10 m s-2. The initial velocity at t=0 is v0. The initial position is x0. Solve for v and x.

First, you need to solve for v. I.w.r.t.x.

$v'=a=g \,$
$\int v' dv=\int g dt \,$
$v=gt+C \,$

Now plug in to find C.

$v_0=g \times 0+C \,$
$C=v_0 \,$
$v(t)=gt+v_0 \,$

Now we solve for x.

$x'=v=gt+v_0 \,$
$\int x' dt=\int (gt+v_0)dt$
$x=\frac{1}{2}gt^2+v_0t+C$

Again, now we solve for C.

$x_0=\frac{1}{2}g \times 0^2+v_0 \times 0+C$
$x_0=C \,$
$x(t)=-\frac{1}{2}gt^2+v_0t+x_0$

Anyone who has studied physics will recognize this as the basic equation for position for an object undergoing a constant force in one dimension.

### Example 2 - Resistive Medium

Lets say we're traveling through a medium that resists movement. In this medium, $a=-v^2$. Solve for v and x, given that the initial velocity was 10 m/s and the initial displacement was 0 m.

$a=v'=-v^2 \,$
$\frac{dv}{dt}=-v^2$
$\int \frac{1}{v^2}dv=\int -1 dt$
$-\frac{1}{v}=-t+C$
$v=\frac{1}{t+C}$

Notice that as t increases, velocity decreases. This is what you'd expect if the medium was resisting your movement and slowing you down over time. Now substitute in the initial velocity at t=0

$10=\frac{1}{0+C}$
$C=\frac{1}{10}$
$v=\frac{1}{t+0.1}$

Now to solve for x.

$v=x'=\frac{1}{t+0.1}$
$\frac{dx}{dt}=\frac{1}{t+0.1}$
$\frac{dx}{1}=\frac{dt}{t+0.1}$
$\int 1 dx=\int \frac{1}{t+0.1} dt$
$x=ln(t+0.1)+D \,$

The position increases throughout, but it increases ever more slowly as time goes on. This is again what you'd expect for a medium resisting motion. Since velocity is never less than 0, we never stop going forward, but we go exponentially less far over time. Putting in our boundary conditions,

$0=ln(0+0.1)+D \,$
$D=-ln(0.1) \,$
$x=ln(t+0.1)-ln(0.1)=ln(10t+1) \,$

## Exponential Growth and Decay

One of the most common differential equations in science is

$y'=ky \,$.

The solution to this is

$y=Ce^{kt} \,$.

If k is positive, this is called exponential growth. If k is negative, its exponential decay. Both are used in science, for very different reasons.

### Population Growth

Lets say we have a group of animals in the wild. We want to know how many animals there will be in t years. We know how many there are now. We also know the birth rate and death rate. Can we solve this problem?

Of course we can. First, we need to figure out the rate of growth. If the birth rate is B, and the death rate is D, the total rate of change is (B-D). Since this is the rate, we need to multiply it by the current population to get the population growth. The final equation looks like

$\Delta P=(B-D)P \,$

where P is the population. That looks like the equation for exponential growth, doesn't it? As a matter of fact, change the 'delta' to a differential, and it is. The growth factor is (B-D).

#### Example 3

In a certain population of rabbits, the birth rate is 10%. The death rate is 15%. The initial population is 100. How many rabbits are there after 10 years? Will we always have rabbits?

From our solution to the exponential equation:

$P=Ce^{(B-D)t} \,$
$100=Ce^{(B-D)\times 0}=Ce^0 \,$
$C=100 \,$
$P=100e^{-0.05t} \,$
$P(10)=100e^{-0.5}\approx 61$

Unfortunately, we will not always have rabbits. Since the growth rate is negative, they will eventually go extinct. (Note: we will never actually hit 0, but in real life you can't have less than 1 rabbit. If we were measuring a continuous property instead of a discrete one, we would always have something, it would just get very small).

$\frac{dA}{dt}=\lambda A$