Linear growth

Unlimited growth: Linear and quadratic growth

Linear growth

For linear growth of a quantity $y(t)$ at time $t$ :

the quantity $y(t)$ changes with the same value per unit of time;
$y(t)$ can be described by a linear function $y(t)=a\cdot t+b$ , in which $a$ is the increase or decrease per unit of time and $b$ the initial value of $y$ at time $t=0$ ;
the corresponding graph of $(y,t)$ is a straight line.

The first two characteristics of linear growth mean that the derivative of $y(t)$ , $y'(t)$ , is constant and equal to $a$ . This constant $a$ is the same as the slope of the straight line which represents $y$ .

For linear growth of a quantity $y(t)$ with slope $a$ holds: $y'(t)=a$

Such an equation for the function $y(t)$ , in which the derivative $y'(t)$ plays a role, is called a differential equation. The unknown in a differential equation is, in contrast to a traditional equation, not a number, but a function. We say that $y(t)=a\cdot t+b$ , for certain constant $b$ , is a solution of the differential equation $y'(t)=a$ . This is a solution, because this function has all the properties described in the differential equation, namely, that the derivative is equal to the constant $a$ .

In mathematical models of growth and other change, the differential equation is often just the beginning of the modelling process. Often, first a dynamic model is designed on the basis of measurement data, i.e., one makes a formula in which the rate at which a parameter changes is expressed in the magnitude and / or the time. In linear growth, you suppose for example, that the property 'constant change per unit of time' is true. Such a dynamic model is an example of a differential equation of the first order: this is a relationship between the derivative of a variable and the variable and/or the time. Next, you still will have to find all the solutions of the differential equation. We call this solving a differential equation. This is similar to solving a simple equation such as $t^2=4$ , except that now you are not looking for numbers as a solution of the equation (in the example $t=2$ and $t=-2$ ), but for functions with the desired properties of the derivative as defined by the differential equation. In other words, you must find the general solution of the differential equation. Finally, you specialise the general solution to the given problem situation. This usually means that you are trying to find suitable values for the parameters in the general solution, such that the solution describes the data well.

Suppose that the differential equation $y'(t)=0$ is given and you want to find all the solutions. So you are looking for a function $y(t)$ whose derivative is zero. You can obviously make an educated guess: the function $y(t)=0$ has the desired derivative. But this is not the only solution: $y(t)=1$ and $y(t)=1\!\tfrac{1}{2}$ are two other examples of solutions of the differential equation. It can be shown that function of the form $y(t)=b$ with constant $b$ is a solution. The unknown function $y(t)$ is thus generally constant.

Suppose that you want to solve the differential equation $y'(t)=2$ So you are looking for a function $y(t)$ of which the derivative is constant and equal to 2. You can obviously make an educated guess: the function $y(t)=2t$ has the desired property. But this is not the only solution: $y(t)=2t+1$ and $y(t)=2t-3$ are two other examples of solutions of this differential equation. It can be shown that any function of the form $y(t)=2t+b$ with constant $b$ is a solution. Only when an initial value $y(0)$ or the value of $y$ at another time $t$ is known, the value of $b$ can be calculated. For example, if it is known that $y(2)=5$ , then we know $5=2\times 2+b$ , or $b=1$ . The solution is now completely specified as $y(t)=2t+1$ .

The general solution of the differential equation $y'(t)=a$ is equal to $y(t)=a\cdot t + b$ .

For math enthusiasts, we give the proof of the assertion that $y(t)=a\cdot t+b\;$ is the general solution of the differential equation $y'(t)=a$ .
In this proof we use the following lemma: if $f'(t)=0$ for a certain function $f(t)$ , then $f(t)$ is a constant function.
Suppose $y(t)$ is a solution of the differential equation $y'(t)=a$ . We can now define a new function $f(t)$ by $f(t)=y(t)-a\cdot t$ . The difference rule for differentiating yields: $f'(t) = y'(t)-(a\cdot t)'= a - a = 0$ . It follows from the auxiliary theorem that function $f(t)$ is constant, say with function value $b$ . So $y(t)-a\cdot t = b$ , or $y(t)=a\cdot t + b$ .
This is what had to be proved!