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About this Lesson
 Type: Video Tutorial
 Length: 7:38
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 81 MB
 Posted: 06/26/2009
This lesson is part of the following series:
Calculus (279 lessons, $198.00)
Calculus: Differential Equations (10 lessons, $13.86)
Calculus: 1st Order Linear Differential Equations (2 lessons, $2.97)
Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hopital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, "Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas" and of the textbook "The Heart of Mathematics: An Invitation to Effective Thinking". He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The "Journal of Number Theory" and "American Mathematical Monthly". His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
About this Author
 Thinkwell
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11/13/2008
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Differential Equations
Solving First Order Linear Differential Equations
Using Integrating Factors Page [1 of 2]
Okay, so let's take a look at the following differential equation: . Notice this is of the form , plus something that has the x's in it, times the y, equals stuff with just x's in it. Whenever I see that form, I immediately think to myself, "This is a firstorder linear differential equation." So that's a good thought because now we just make a substitution here and say, "Okay, this whole blob right here, that must be the P(x), and that old blob right here, that must be Q(x)." So all I have to do is apply how we derived this method. The answer is just literally this once I figure out the integrating factor. So first I figure out the integrating factor, and the integrating factor is just what? Well, I have to take e and raise it to the integral of P(x). So step one is to integrate P(x). So integrate that function  not a problem at all!
So first I integrate that function  not really a huge deal. So to find the integrating factor, I first integrate P(x) dx, and what I see is that's the integral of 2(x) dx, and even I can do this, this is just x^2. So therefore the integrating factor is going to be e raised to that power. What I'm going to do now is I'm going to take that thing and multiply it through. So this is the technique. You find the integrating factor, which is that not hard to do in this example, and now just multiply it through. Let's see what happens when we multiply it through.
If we multiply that through, what I see now is e^x2^^^^+2x  and then I'm going to stick in this term right here  e^x2  because you can put it anywhere you want  times the y. So I just took that factor and stuck it in right there, and that equals (x)e^x2. Why is this a good idea? Well, this actually is a great idea. It's a great idea. Let's see why. Look at this lefthand side. This left hand side might look pretty, pretty complicated, but actually that's just the derivative of some product, and this is by design. This integrating factor allows us to write the lefthand side as the derivative of a product. It's just a matter of looking at it a little bit and realizing that it's just the derivative of something times y, and in particular the integrating factor times y, and it's always that. This is great. Check it for yourself. Let's take the derivative of this. That's the first times the derivative of the second  that's  plus the second  there's y  times the derivative of this. And what's the derivative of this? That's e^(blah^)  the derivative is e^(blah^)  times derivative of the (blah)2x. It's amazing. This integrating factor allows us to write this as really it should be, namely, a derivative of a product.
So with that simplification, we should be in great, great shape because now if I multiply through by the differential dx, what I see now is the following: The derivative of e^x2y = x e^x2dx, and now this is crying out to be integrated on both sides. The integral of the derivative, they annihilate themselves, and so I'm just left with e^x2y. And what do I get on the right? On the right, I have to integrate that. Well, that's actually a little u to u substitution. I'll actually show this to you. It's not that big of a deal. Maybe you can do it in your head. In fact, you can just let u = x^2. So du = 2x(dx). Unfortunately, I just want x(dx), so I'll divide both sides by two. So I see du = x(dx). So in place of x(dx), that integral now becomes...well, x(dx) is now just du, and then I have e^u. So that's just e^u  it's a derivative of itself. So that's e^x2, and that's the integral of this side. So that integral turns out just to be e^x2.
Now I can solve this. If I solve this for y  by the way, maybe I didn't draw this too well. This y is just multiplying this...maybe you think it should be exponent...sorry about that if you did  it's e^x2 times y. Now, if you divide both sides by e^x2...oh! Plus the constant! I made a mistake because I forgot to write +C. Now you might say, "What's a constant between friends?" Let's pretend that we didn't write that constant, +C. Then look what you'd say  oh, this is great, I'm so happy I did this! What would you say? Well, you'd say, "Oh, look! e^x2 on this side and e^x2 on this side, they cancel, so the answer is y = ." Let's see what happens if we actually put in the +C. So we're actually going to put in the +C, and I'm going to put it in a very vibrant...brown C. Now what happens when you divide both sides by ex2? When you divide both sides by e^x2, what you see is y =... look, if you divide by e^x2, it does cancel here, so I see a , but I then have +C e^x2 (if I'm dividing by e^x2, that's a negative exponent). Look what happened: Without that C I lost sort of all of the subtlety here. So it's a great, great lesson that when you integrate here at the end, don't forget the constant. And notice that the constant, just as we've seen in other examples, actually sort of entangles itself in the actual function. If you would had forgotten the constant, you would have sort of lost this piece of the function and you wouldn't have produced the most general solution to the differential equation that we were given.
So the most general solution for this differential equation turns out to be y = plus some constant times e^x2. Neat!
So whenever you see something of this from, think firstorder linear differential equation; and even if you don't think that, the important thing to do is to realize that all you've got to do is create the appropriate integrating factor by just integrating that term right there and consider the integrating factor e to that integral. Once you do that, multiply both sides, and the lefthand side will be the derivative of a product, and then you're home free.
Congratulations, and try these on your own. Don't forget the constant. See you soon.
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