Initial Value Problems for Growth and Decay
UBC Calculus Online Course Notes

Introducing a Differential Equation

Growth and Decay Phenomena

Applications of the Exponential Functions and Logarithms


Remember the Exponential function?
In a previous discussion we made a remarkable discovery about the exponential function with the special base $ e=2.71828 $ . We found out that the function


\[ 
y= f(x)=e^x 
\]

has a special relationship to its own derivative, namely


\[ 
\frac{dy}{dx}= = e^x = y 
\]

Since we want to talk about applications in which the independent variable is time, we will be thinking of these same ideas but with the notation $ y=y(t)=e^t $ where $ t $ stands for time. The relationship of the function to its derivative is then:


\[ 
\frac{dy}{dt} = e^t = y 
\]

or, more simply,


\[ 
\frac{dy}{dt} = y. 
\]


What is a differential equation ?

By studying the exponential functions, and picking a convenient base, we have inadvertantly stumbled on a relationship satisfied by the function and its derivative. That relationship is called a differential equation . The topic of differential equations is an extremely important one in mathematics and science, as well as many other branches of studies (economics, commerce) in which changes occur and in which predictions are desirable. In most such circumstances, the systems studied come with some kind of "Natural Laws", or observations that, when translated into the language of mathematics, become differential equations. It is then the job of the mathematician to try to figure out what are the predictions, i.e. to find the functions that satisfy those equations.

In our work we have been lucky enough to start with a function $  y(t) = e^t $ and show that there is a differential equation that it satisfies, $  dy/dt = y $ .

DEFINITION
We say that a function is a solution to a differential equation if, when we plug it (and its various derivatives) into the equation, we find that the equation is satisfied.

Comment:
It is important to notice right off, that a solution to a differential equation is a function , unlike the solution to an algebraic equation which is (usually) a number, or a set of numbers. This makes differential equations much more interesting, and often more challenging to understand, than algebraic equations.

Example
Another differential equation, closely related to the previous one we studied is


\[ 
\frac{dy}{dt} = 2y 
\]

The function $ y=e^t $ will no longer satisfy this equation. (Why not ? Try plugging it in and find out !) However, maybe if we experiment a little bit with similar functions, we will find one that works. Suppose we modify the simplest exponential and try out the possible solution


\[ 
y(t) = C e^{kt} 
\]

where $ k, C $ are constants. To find whether this function satisfies the above differential equation, we will need to compute its derivative, which we do using the Chain Rule. (We need to use the Chain Rule because the function above has an expression $ u=kt $ in the exponent, not just $ t $ alone.)


\[ 
\frac{dy}{dt} = \frac{d}{dt} \left(C e^{kt}\right) \\ 
 = C \frac{d e^{kt}}{dt} = C \frac{de^{u}}{du} \frac{du}{dt}\\ 
  = C e^u k = k (C e^{kt}) = k y \\ 
\]

Notice that we have succeeded to show that the function $ y(t) = C e^{kt} $ will satisfy the equation


\[ 
\frac{dy}{dt} = k y \\ 
\]

Thus, if we pick $ k=2 $ , i.e select $ y(t) = C e^{2t} $ as the candidate for the right function, it should indeed satisfy the differential equation in our example.


The solutions to a differential equation

Let us summarize the important result that we have just discovered and state it in a more general form:

$ y(t) = C e^{kt} $ is a solution to the differential equation $ \frac{dy}{dt} = k y  $ .

This is true for any value of the constant $ C $ .

Comment 1:
The constant $ k $ is determined by the equation (for example, in the case we just looked at, we had to pick $ k=2 $ for the function to satisfy the differential equation.) $ k $ is called a rate constant. For consistency, it has to have units of 1/time (Why ? Check out the units of the term on the left hand side of the equation and remember that in order for the equation to make sense, the two sides of the equation should have the same units.)

Comment 2:
Since $ C $ is not determined by the differential equation, it means that we actually have a whole family of possible solutions, one for every possible value of this constant. A sketch of some of the members of this family is shown below.

Comment 3:
Each of the curves shown in this picture respresents one possible solution of the differential equation. Notice that each solution describes the way that $ y(t) $ behaves as $ t $ varies smoothly over a whole range of values (in this picture, only the values of $ t $ in the range $ 0<t<2 $ are shown. As we said before, the solution is a function, not a number !


An initial value picks out one solution

Why do we get so many possible solutions that all satisfy the same differential equation? Well, this stems from the fact that the differential equation is only telling us something about the way that our function $ y(t) $ is changing, and not where it "starts". If we had more information, for example, if we knew one particular point through which the "right" curve goes, we could select one of the members of the family as the single solution of interest to us. For example, if we specify the point on the y axis through which the curve passes (i.e. $ y(0) = y_o  $ where $ y_o $ is some given number) this will do the job ! Such additional information is called the initial value of the solution, since it is specifying the value of y at time $  t=0 $ .

Example Given that $ y(0)=1 $ , find the appropriate solution of the differential equation.

Solution: Setting $ y(0)=1 $ , and noting that $ y(0)=c e^{k 0} = C e^0 = C 1 = C $ we find that $ C=1 $ . Thus the "right" function, in this case, is $ y(t)=e^{kt} $ .

The problem consisting of a differential equation together with an initial value is called an initial value problem. To summarize, we have just made a connection as follows:

INITIAL VALUE PROBLEM SOLUTIONGRAPH

\[ 
\frac{dy}{dt}=ky 
\] 
\[ 
y(0)=y_o 
\] 
 
\[ 
y(t) = y_o e^{kt} 
\]


For your consideration:

  • (1) Find the solution to the differential equation $  dy/dt = -5 y $ with the initial value $  y(0)=10 $ . What would the graph of this function look like, and how is it related tothe examples we explored ?

  • (2) Find the solution to the same differential equation, but wih the initial value $  y(0)=0 $ . Explain what is special about this initial value. What does the solution look like? Does it increase or decrease with time?