X^2 y^''-x(1+x) y^'+y=0

Name: Problem 14, Chapter 8: Fundamentals of Differential Equations
Uploaded: 2019-09-09T21:09:50-08:00
Duration: 17 min 23 s
Channel: Brian Ketelobeter
Description: x^2 y^''-x(1+x) y^'+y=0

Key Concepts

Ordinary Differential Equations

An ordinary differential equation (ODE) is an equation involving a function of one independent variable and its derivatives. Understanding ODEs involves knowing how to classify them by order, linearity, and whether they are homogeneous or non-homogeneous, which is crucial for selecting an appropriate method of solution.

Linear Differential Equations

A linear differential equation is one in which the unknown function and all its derivatives appear to the power 1 and are not multiplied together. This property allows for the principle of superposition to apply, meaning that combinations of solutions are also solutions, and provides avenues for systematic solution methods.

Homogeneous Differential Equations

A homogeneous differential equation is one in which every term involves the unknown function or its derivatives, and typically no independent forcing term is present. In the context of linear ODEs, this implies that the zero function is always a solution and that the general solution can be expressed as a linear combination of independent solutions.

Variable Coefficient Equations

Differential equations with variable coefficients have coefficients that are functions of the independent variable rather than constants. These equations are generally more complex to solve than their constant-coefficient counterparts, and they often require specialized methods such as power series or transformation techniques to find solutions.

Transcript

00:03 Good day, everyone.

00:05 Today, we're looking at problem number 14 here.

00:10 We have to find two with two linearly independent solutions using the provenius method.

00:20 Okay.

00:22 First off, i have a few notes here.

00:24 So this problem was done in 8 .6, number 32.

00:32 There, there, we found that r1 was a 1 equals 1 was a double root.

00:43 And so, and we found that y1 equals to this guy here was our solution.

00:53 And because r1 is a double root, theorem 7 says that r1, our solution is of this form.

01:09 Okay, so now we know the form of solution, so we just have to go about plugging y2 here into 14 and then solving it.

01:30 One other comment i want to make, and i reiterate this from previous videos but it's important to note the b ends here start at 1 so this this index starts at 1 i mentioned this explicitly because i made the mistake not realizing that bn started one caused me a great deal of problems in other in the well, anyhow, you should look at theorem number seven here carefully just to make sure that you know what the form of the solution should look like in the various r cases.

02:19 Okay, so here we go.

02:23 We substitute it in and we get this, you know, this ridiculous looking sum here.

02:34 And it is ridiculous, i have to admit, it's very...

02:38 But what i'm doing here is all i'm doing, so if you look at what i've done here, all this is each of these terms represent the subsequent substitution in each of these.

02:59 So in particular what i mean here, maybe to try and be a little bit clear is this here this represents to plugging in why because y2 is a solution i can just plug it in and so that this corresponds to the first term here this corresponds to the second term here so on and so forth so i mean really all i'm doing is just plugging it in and i broke it up in this form because it's a little easier to do it like this and then when i get is i come to this guy here so again this is very you know it's very large but this is just an element of simplifying here and crucially you'll remember that why is a solution so the why or more particularly maybe why one is a solution so this part here is zero because it it already solves you'll notice that this and the inside here is actually the 14 so this already this already is zero and now what we have to do is carry on sort of our analysis then on the remainder of this guy.

04:41 Okay.

04:42 And the next step then is to just really plug in the y function.

04:50 So just plug in.

04:51 We know why is this form.

04:54 Okay.

04:55 And so when we plug that in, we get this.

04:59 So you get this nasty little equation here again.

05:04 You know this is just plugging in for why here and i haven't done anything yet to the bees okay and then now what i've done is just a little simplifying so this next line here i'm just simplifying it a little bit so there's nothing really new here just just simplified the you know, cancel the two here, cancel this term, and i think it's this term, or not this term, sorry.

05:46 This term, cancel this term here.

05:53 Okay.

05:56 And i'm just, okay.

06:00 Well, there should be another term here that seems to have disappeared that i forgot.

06:05 But anyhow, so this is what i get here.

06:11 Okay, this is what i get in the end.

06:15 Okay, next.

06:19 Now we want to do index some, what i, okay, yeah.

06:25 So now what we want to do is some index, what i call index wrangling, because when we get the solutions, we want the x to be, you know, to the same power because we want to set the coefficients to zero.

06:42 And now this in particular here is where you need to note that the bn start at one.

06:50 So this, this here starts at one.

06:54 This starts at one.

06:55 This starts at zero.

06:58 And this starts at zero.

07:02 But on, but you'll notice that we are, we have this two.

07:07 And a -n.

07:10 So when we go to the next line, since this starts at zero, this term is really just repeated.

07:20 It now starts at one, but when you multiply zero times a zero, you're just going to get zero.

07:29 So it doesn't change when we go to this one here.

07:35 But now this one does, this next one does.

07:39 Why? because when we have to factor out the a -0 term.