Lecture 1
1.1 Introduction
This material is covered in Stewart, Section 9.1.
In almost all fields of scientific endeavour, quantifying time-dependent occurrences almost inevitably comes down to a differential equation (DE).
1.1 Introduction
A DE is an equation which contains one or more derivatives of an unknown function. There are two types of DEs:
1.1.1 Examples of ODEs
Source: Enciclopedia Britanica 2012
1.1.1 Examples of ODEs
Source: Falling |
Source: Svjo |
1.1.1 Examples of ODEs
In ODEs, one often takes $t$ instead of $x$ for the independent variable, where $t$ denotes time.
Also, derivatives, such as \[x'(t) \;\;\text{ and }\;\; x''(t)\] with respect to time are often written as \[\dot{x} \;\;\text{ and }\;\;\ddot{x},\] respectively.
1.1.2 How do ODEs arise?
Population modelling. Assume a population grows at a rate proportional to the size of the population. If $P=P(t)$ stands for the population at time $t$, then the model states that
$ \ds \dif{P}{t} \propto P, $
i.e.,
$ \ds \dif{P}{t}=k P, $
where $k$ is the growth constant. If $k>0$ the population is growing (think humans) and if $k\lt 0$ the population is decreasing (think tigers).
1.1.2 How do ODEs arise?
Population modelling. Assume a population grows at a constant rate proportional to the size of the population. If $P=P(t)$ stands for the population at time $t$, then the model states that
$ \ds \dif{P}{t} \propto P, $
i.e.,
$ \ds \dif{P}{t}=k P, $
where $k$ is the growth constant. If $k>0$ the population is growing (think humans) and if $k\lt 0$ the population is decreasing (think tigers).
Question: What happens for $k=0$?
$P$ is constant. (equilibrium solution)
1.1.2 How do ODEs arise?
Newton's eureka moment. Newton's second law of motion states that $$ \text{mass }\times \text{ acceleration } = \text{ force.} $$
1.1.2 How do ODEs arise?
Newton's eureka moment. Newton's second law of motion states that $$ \text{mass }\times \text{ acceleration } = \text{ force.} $$
Let $y=y(t)$ be the vertical displacement from ground at time $t$ of an apple of mass $m,$ soon to land on Newton's head. Then the acceleration is $\displaystyle{\frac{\dup^2y}{\dup t^2}}$ so that $$ m\, \frac{\dup^2y}{\dup t^2} = -mg, $$ where $g$ is the constant of acceleration at the surface of the earth due to gravity.
1.1.3 Solution to an ODE
Suppose that we are given an ODE for $y$ which is an unknown function of $x$. Then $y=f(x)$ is said to be a solution to the ODE if the ODE is satisfied when $y=f(x)$ and its derivatives are substituted into the equation.
Example: Show that $y=y(x)=A \exp(x^2/2)$ is a solution to the ODE $y'=xy$.
1.1.3 Solution to an ODE
Example: Show that $y=y(x)=A \exp\left(x^2/2\right)$ is a solution to the ODE $y'=xy$.
We need to show that $y=Ae^{x^2/2}$ satisfies the ODE. Calculate
$ y'$ $= \ds\frac{d}{dx}\left(Ae^{x^2/2}\right)$ $= Ax e^{x^2/2}$ $= x \left(A e^{x^2/2}\right)$ $= x y,$
which is true for all $A\in \R.$
1.1.3 Solution to an ODE
When asked to solve an ODE, you are expected to find all possible solutions. This means that you need to find the general solution to the ODE. For an ODE that involves only the unknown function $y$ and its first derivative, the general solution will involve one arbitrary constant.
You should already know how to solve ODEs of the form $$ \dif{y}{x} = f(x) \quad\text{or}\quad \frac{\dup^2y}{\dup x^2} = g(x). $$
1.1.3 Solution to an ODE
Example: Find the general solution to the differential equation $y'=x^2.$
$\ds \frac{dy}{dx} = x^2$ $\ds\;\Ra\;y = \int x^2 dx$ $\ds = \frac{x^3}{3}+C.$
This is the general solution.