step 1 Well, good afternoon, folks. We are going to solve another problem today. And it's number four h

X y^''+y^'-4 y=0 | Numerade

Name: Problem 4, Chapter 8: Fundamentals of Differential Equations
Uploaded: 2019-08-18T04:47:47-08:00
Duration: 14 min 21 s
Channel: Brian Ketelobeter
Description: x y^''+y^'-4 y=0

Key Concepts

Variable Coefficient Differential Equations

In a variable coefficient differential equation, the coefficients of the function and its derivatives are not constant but can depend on the independent variable. The presence of the term ?xy' in the equation indicates that the coefficient of y' is a function of x. Such dependencies can complicate the search for analytic solutions and often require specialized techniques or series expansions.

Linear Homogeneous Differential Equations

A linear homogeneous differential equation has the property that the function and its derivatives appear linearly and there is no forcing function (i.e., the equation is set equal to zero). This implies that the set of solutions forms a vector space. The equation y'' - xy' + 4y = 0 is linear and homogeneous because all terms are either a multiple of y or its derivatives, and they sum to zero.

Series Solution Methods

When dealing with differential equations with variable coefficients, series solution methods, such as the power series method or the method of Frobenius, are often employed to find solutions in the form of an infinite series. These methods are particularly useful when the equation cannot be solved by standard techniques for constant coefficient equations.

Ordinary Differential Equations

Ordinary differential equations (ODEs) involve functions of a single variable and their derivatives. They are used to model phenomena where the rate of change of a quantity is related to the quantity itself and possibly to independent variables. The given equation is an example of an ODE because it relates y, its first derivative, and its second derivative with respect to x.

Second-Order Differential Equations

Second-order differential equations contain the second derivative of the unknown function and are common in modeling physical systems, such as in mechanics and electrical circuits. The order of a differential equation is determined by the highest derivative present, and in this case, the presence of y'' makes the equation second-order.

Transcript

00:05 Well, good afternoon, folks.

00:09 We are going to solve another problem today.

00:14 And it's number four here.

00:18 You should read the book if you want to know the whole statement of the problem.

00:24 Because personally, it's too long for me to state.

00:28 Let me be forthright.

00:33 Before we begin, that my final answer, that i get does not exactly match what the book gets.

00:47 And i have tried to figure out where the mistake is.

00:53 I cannot.

00:55 And i know sometimes that the answer in the book is not quite in line with what the answer should be.

01:02 So i don't know what's wrong.

01:05 I've tried to figure it out.

01:06 It just isn't quite the right number.

01:09 So maybe we can figure it out here somewhere.

01:12 I don't know.

01:13 But i'm going to work you through the method.

01:17 The method is, you know, what it is.

01:20 It's a lot of work, a lot of algebra, and, you know, mistakes somewhere can happen.

01:29 So but the idea is right.

01:31 Okay.

01:32 So just trust me on that.

01:34 The idea is right.

01:35 Somehow or other, i'm not sure there might be a mistake.

01:38 I don't know.

01:39 But okay, so the first thing is to check what the r values are and it turns out that the r values are zero.

01:51 So you apply theorem 7 .b.

01:58 And so the solutions are of this form then.

02:06 So our solutions are this form where we have this sum.

02:10 Here and we have this sum here and so our task we have really two tasks in this problem and the first one is to find the a ends here even though if you'd done this problem before you'd already have the a ends which you could skip that part and the second part which i think is the almost a more tedious and more annoying part is the b ends right now i think that's, boy, that is really annoying.

02:46 Okay, so i'm going to take you through the first part, which you really should know already, and it's find the coefficients for y1.

03:00 Okay, so sorry, there is a bit of a, okay, so just finding the coefficients for y1 here, and this is doing just, you know, the method, if you will.

03:20 Funny thing is i seem to forget his name all the time.

03:26 And so the first step, i'm just rewriting all the terms.

03:31 And the second one, i'm simplifying that down, putting all the n minus 1 plus r's over here, and then this lonely little n2.

03:44 Plus r over here again as you know i'm sure you're pretty used to by now if you've done these a few times okay so then index wrangling again as usual you want to get in the end you want the x to the k plus r term and then the you want the x to the k plus r and then you want the x to the k plus r and then multiplied by this term here okay.

04:18 And then now we have a recursive formula for the coefficients.

04:24 We're just solving the a .j.

04:33 Equals zero equation where, of course, the a .j came from, oops.

04:41 Sorry about that.

04:42 The a .j is this guy here.

04:44 So we're just solving.

04:48 And so we end up with well we end up with the first four terms being this right here they really said to just find the first three terms i sort of found a closed form for the entire thing but it's not really necessary and i guess in this and this subset it looks like you know it's safe to just take a not to be one so this would be your a not and b not to b1 because you're actually looking at linear combination.

05:22 So, you know, it's just a0, just take it to be one.

05:28 Okay, so that's, this is our first one.

05:31 And let me just say, this answer is correct.

05:34 Okay, so you have the, you have the first, you know, the first three terms are correct there.

05:45 Okay, so it's really in the second part here.

05:49 That and i think you'll see why pretty quickly.

05:56 It's actually fairly frustrating to do.

05:59 Okay, so what you have to do is you assume you have, because were you, because of the, what's gonna call, because we're, this is seven, this theorem seven b that we use, we assume that the solution has this form, or we know that a solution has this form, form.

06:22 And now what we have to do is we really have to input it's in.

06:29 And i just want to mention that in this algebra here, i've sort of deleted the twos.

06:38 So just assume that, yeah, maybe i should have left it there.

06:44 But this is y2 up here.

06:48 And then this actually is y1.

06:50 Down here.

06:52 Okay, so the y, this is down here, this is a really a y1, and this, of course, is a y2 here.

07:04 The main reason i stopped using that index form is because it's just too confusing, too many indices and everything.

07:15 But, i mean, past this point, everything is y1.

07:20 So it's past this point.

07:21 Point here, everything is y1...