Difference between revisions of "MA366-1" - Rhea

Latest revision as of 10:15, 26 October 2009

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What does it mean to solve a differential equation?

To solve a differential equation is to find a function which, when "plugged in" to the given differential equation, produces a true statement. Suppose I tell you that I'm thinking of a function that, when differentiated with respect to x, is equal to one. What is the function I'm thinking of? Well, the word problem I've given you can be written this way:

$ y'=1 $

(where differentiation is with respect to x). We can simply integrate both sides of the equation to find the answer. We get:

$ y=\int 1dx $

$ y=x+C $

(where C is an arbitrary constant.) Clearly, the function y=x satisfies the differential equation y'=1. But the function y=x+2 also satisfies the differential equation, because the two gets eliminated by differentiation. Similarly, y=x+37 is yet another solution. To represent all of these functions, we simply call the solution y=x+C, where C is any constant. Rather than having one particular solution to the differential equation, then, we have a family of functions. This is something you should get used to.

Initial Value Problems

An initial value problem is just a differential equation with an additional statement about the mystery function y. Take, once again, the differential equation

$ y'=1 $

We know that solving for y leads us to a family of functions. But what if I told you that I want the function y to satisfy the following requirement: y(2)=5? Recall our solution to the differential equation:

$ y=x+C $

The additional statement I have provided tells you that when x equals two, y must equal five. To utilize this information, simply plug the values of x and y into the solution to the differential equation:

$ 5=2+C $

As you can see, we now have the ability to solve for C. In this case, C=3. We can now go back to the solution to our differential equation and substitute the number 3 for C:

$ y=x+3 $

This is no longer a family of functions, but a single, particular function. This is a trend that you will see throughout the course. The additional statement I gave you, y(2)=5, is called an initial value. Without an initial value of some kind, you will always wind up with a family of functions. But with an initial value, your solutions will be narrowed down. In some cases, only one function will be left; in other cases, you'll have another family of functions that is a subset of the original family.

A slightly more annoying initial value problem

Suppose I told you that I was looking for a function y whose second derivative is equal to one. In other words, we have the differential equation

$ y''=1 $

This is another differential equation that can be solved simply by integrating both sides. Let's do that. We obtain:

$ y'=x+C\qquad \qquad (*) $

As you can see, we still have a derivative in the equation. We do not yet know what the function y is. We can integrate once more to get:

$ y=\frac{x^2}{2}+Cx+D \qquad \qquad (**) $

C was a constant, and so integration changed it to Cx. D is a SECOND arbitrary constant produced by integrating a SECOND time. So we now have two distinct arbitrary constants. We now have the family of functions that satisfy the given differential equation. But what if we wanted to find out what C and D are? It turns out, if you have TWO arbitrary constants, you need TWO additional pieces of information. Suppose I tell you that y'(4)=5 and that y(2)=3. Start off by using y'(4)=5: plug it into (*) to get:

$ 5=4+C $

$ C=1 $

Now, use the second piece of new information I gave you: y(2)=3. Plug it into (**):

$ 3=\frac{2^2}{2}+C(2)+D $

$ 3=2+2C+D $

Finally, use the information we gained a moment ago: that C=1. Plug that information in to get:

$ 3=2+2(1)+D $

$ D=-1 $

Now that we have both C and D, plug both of these values back into (**) to obtain:

$ y=\frac{x^2}{2}+x-1 $

This is our final answer, and the solution to the initial value problem.

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