 New Zealand
Level 8 - NCEA Level 3

Introduction to Differential Equations

Lesson

Differential Equations - what are they?

An ordinary differential equation is an equation involving an unknown function, say $y$y, and at least one of its derivatives. So for example, if $y=f(x)$y=f(x), so that $\frac{dy}{dx}=f'(x)=y'$dydx=f(x)=y, then $y-y'=0$yy=0 is a differential equation.

The task of a mathematician is to discover the general function $y$y that makes the equation true. In the simple case of the function $y-y'=0$yy=0, we notice one solution is given by $y=e^x$y=ex, since $y'=e^x$y=ex and therefore $y-y'=0$yy=0. But is this the only solution or are there more?

Laws of nature usually express themselves in the form of a differential equation. They are the language of nature. In any process of nature, the variables involved are usually related to their rates of change by the basic scientific principles that govern the process.

The power to reveal nature

You may be familiar with models of population growth. As the population increases, the rate of change of population quickens so that large populations get larger more quickly than small populations. The same applies to money in the bank - the more you have the more quickly the balance grows with interest payments based on the most recent balances. In both of these cases we might write that the rate of change of the amount $A$A is proportional to value of $A$A at any given time. Thus we naturally form the differential equation $\frac{dA}{dt}=kA$dAdt=kA for some amount $A$A and time $t$t.  We could write this more succinctly as $A'=kA$A=kA and the task before us is to find the general function $A$A that makes this true.

The details concerning how we go about solving a differential equation like this need not concern us here. Suffice to say that the general solution to $A'=kA$A=kA is given by $A=A_0e^{kt}$A=A0ekt where $A_0$A0 is some initial population value.  In other words, if we assume that the rate can be described as $A'=kA$A=kA, then solving the differential equation reveals the exponential law that lies behind it. It's like pulling the curtains up and revealing natures secret.

Of course we can take this law and differentiate it. Thus if $A=A_0e^{kt}$A=A0ekt then:

$\frac{dA}{dt}=\frac{d}{dt}A_0e^{kt}=kA_0e^{kt}=k[A_0e^{kt}]=kA$dAdt=ddtA0ekt=kA0ekt=k[A0ekt]=kA

So by differentiation we reveal the correct rate of change $A'=kA$A=kA

Nature most often reveals itself by rates of change, and so the study of differential equations becomes incredibly important, and like most areas of mathematics, the search for the solution techniques of differential equations has a colourful history.

Newton and his second Law

Newton's second law of motion states that $F=ma$F=ma so that the acceleration a of an object of mass m is proportional to the total force acting on it. Suppose we call the constant $g$g the acceleration of a body near the surface of the Earth (taken as negative). If we drop an object of mass m from a height $h$h and let the function y be the height of the object from the ground at any time $t$t, then we know that:

$mg=m\frac{d^2y}{dt^2}$mg=md2ydt2

This means that $\frac{d^2y}{dt^2}=g$d2ydt2=g, and since $g$g is a negative constant it's not too difficult to integrate twice to reveal the law of falling after being dropped (ignoring air resistance). Integrating the first time we reveal that $\frac{dy}{dt}=gt+C_1$dydt=gt+C1 and since at $t=0$t=0, $v=0$v=0 we must have that $C_1=0$C1=0 and $\frac{dy}{dt}=gt$dydt=gt. Integrating again we reveal that $y=\frac{1}{2}gt^2+C_3$y=12gt2+C3 and because at $t=0$t=0, $y=h$y=h, we have that $y=\frac{1}{2}gt^2+h$y=12gt2+h

The law that this simple exercise reveals is that an object falling to the Earth from a height $h$h will fall a distance proportional to the square of the time taken. Taking $g=-9.8$g=9.8 metres per second and $h$h as $100$100 metres, after $t=1$t=1 second the height $y$y will be $y=100-4.9(1^2)=95.1$y=1004.9(12)=95.1 metres of the ground. After $t=2$t=2 seconds the height will be $y=100-4.9(2^2)=80.4$y=1004.9(22)=80.4 metres  of the ground. A little algebra shows that it will hit the ground after about $4.5$4.5 seconds. You might like to verify that yourself.

Suppose we now get on top of a $100$100 metre building, drop a rock (safely) and have someone measure the time it takes to fall to the ground. If this number is $4.5$4.5 seconds as predicted, we have to some extent demonstrated the reasonableness of Newton's second law of motion.

Imagine it for a moment. A young Isaac Newton using little else but the power of thought to deduce a law of nature that he could express as a differential equation - specifically identifying the rate of change of velocity of an object as a constant. He solves the equation and reveals, inescapably, that an object must fall a distance proportional to the square of the time of falling. By, say, climbing up a $100$100 metre tower or standing near a cliff edge, he could confirm that the actual falling time as measured by a stop watch agrees with the fall predicted by his function $y$y. This confirmation could be interpreted as a revelation of a natural law and indeed Isaac Newton believed it to be so.

Other examples

Through the years great strides have been made by mathematicians to solve various types of differential equations and this has lead to general techniques of solution many of which go well beyond your present requirements. Here are two special differential equations:

1. $(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+p(p+1)y=0$(1x2)d2ydx22xdydx+p(p+1)y=0
2. $\frac{d^2y}{dx^2}+xy=0$d2ydx2+xy=0

Equation 1 is Legendre's equation, after the French mathematician Adrien Marie Legendre (1752-1833). Equation 2 is Airy's equation, after the British astronomer George Biddell Airy (1801-1892). Their solutions are difficult and beyond the scope of most students.

However many of the basic techniques are accessible, and these will be discussed in later chapters. But before we do that, we need to look at a powerful tool that can be applied to certain differential equations.  It is referred to as a differential equation's direction field, and even without solving the differential equation we can learn a great deal about the unknown function by the shape and form of its direction field. That will be our next foray into this fascinating topic.

Outcomes

M8-12

Form differential equations and interpret the solutions

91579

Apply integration methods in solving problems