Contents
MA 366: Ordinary Differential Equations
Differential equations, which state a relation between a function and its derivatives, appear constantly in physics, chemistry, engineering, biology, and whatever other sufficiently mathematical science one can think of, and the more difficult ones continue to be an active area of mathematical research. Solving differential equations, as you will see in this course, proves to be an endeavor requiring a variety of techniques, and far more frequently than you would prefer, no simple formula exists for deriving the desired solution.
The differential equations you will study in this course are broken up into different families according to which techniques lead to their solutions. It is (mostly) along these lines that this course wiki is divided.
Contents
1) What does it mean to solve a differential equation? What is an initial value problem?
2) Separable Equations
3) First Order Linear Equations
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
Separable Equations
If the differential equation in question can be written as
$ M(t)dt=N(y)dy $
then the equation is called separable. The name refers to your ability to separate the function y and the dependent variable t, with all of the t's and dt's on one side and all of the y's and dy's on the other. If, by using a little bit of algebra, you can get your equation into this form, then you can simply integrate both sides to find the solution. Integrate the left side with respect to t, and the right side with respect to y, and your differential equation will turn into something more familiar and useful. Unfortunately, very few equations are separable, and you'll soon find yourself pining for them as you're forced to use much more involved methods of solution.
Linear Equations and Integrating Factors
(Note): Firstly, in case you've forgotten, y' means dy/dt unless otherwise stated. The two expressions are interchangeable (so long as you are consistently differentiating with respect to t). Which one you use just depends on what your purposes are; you'll know which one to use in practice.
Suppose that, through some sort of algebraic manipulation, you can get your differential equation into the following form:
$ y'+p(t)y=g(t) $
This is called the "general first order linear equation". The important thing to remember here is that the functions p(t) and g(t) are NOT functions of y; they are ONLY functions of the independent variable t. There is a neat trick to solving these equations. First, make sure your equation is in the form displayed above. What you'll do now is multiply both sides of the equation by what is called the integrating factor. The integrating factor is defined as follows:
$ \mu\,=e^{\int p(t)\,dt}, $
Once you've done that, you'll find that the left side of the equation is always the derivative of a product. You can now integrate both sides of the equation, and then the most you have left is some algebra. Your text will most likely contain a thorough explanation of why the integrating factor produces the derivative of a product, and you should probably read it just to see why it works. Let's do an example to see the technique in action.
Suppose the differential equation given is
$ ty'+2y=4t^{2} $
Remember that we want our equation to be in the form mentioned above; that is, y' + p(t)y = g(t). We see that the y' term is not by itself: there's a "t" attached to it. So let's divide by t. We get:
$ y'+\frac{2}{t}\,y=4t $
Now our equation is in the form we want. We can see that
$ p(t)=\frac{2}{t}\, $
Therefore, our integrating factor is
$ \mu\,=e^{\int \frac{2}{t}\,dt} $
Evaluate the integral to get:
$ \mu\,=e^{2ln|t|}=(e^{ln|t|})^{2}=(t)^{2}=t^{2} $
So our integrating factor is t^2. Always simplify your integrating factor BEFORE you multiply both sides of your differential equation by it. Now, let's multiply both sides of our equation by t^2. We get:
$ t^{2}y'+2ty=4t^{3} $
And here's where the weird part happens. Look at the left side of the equation. I told you that it would turn out to be the derivative of a product. Remember that the derivative of a product pq is pq'+p'q. With that in mind, what function do you think, when differentiated, will give the left side of the equation? Here's the answer.
$ t^{2}y'+2ty=(t^{2}y)' $
So, we can rewrite the left side of our equation as the derivative we found above.
$ (t^{2}y)'=4t^{3} $
You know that integration, roughly speaking, "cancels" differentiation. So, once we've got our differential equation in the above form, with the left side being just one great big derivative, we can integrate both sides. Integrating both sides yields:
$ t^{2}y=t^{4}+C\qquad \qquad (*) $
I put an asterisk on this line to remind you that it is absolutely essential to remember your integration constant C! You may have, in previous calculus courses, found this detail to be a sort of notational formality of little consequence, but you will find that in differential equations it really does matter. Now we can simplify (*) and solve for y.
$ y=t^2+\frac{C}{t^2}\, $
This is your final solution. Again, I want to emphasize the integration constant. It is NOT okay to just tack on the C at the end of your calculations; you have to do it when you integrate.
Existence and Uniqueness
You may wonder, after having found a solution to a differential equation, whether you have found the only possible solution (Under construction; more to be added soon)
