Finding general solution to Euler equation via variation of parameters

In summary, the author discusses how to solve a differential equation in terms of a constant coefficient equation, using Euler's equation as an example. He explains how to do this by referencing equations (3) and (4). However, he is still having trouble understanding the notation and is requesting help in understanding it.
  • #1
s3a
818
8

Homework Statement


The problem is attached as TheProblemAndSolution.png, and everything is typewritten, so it should be easily legible (but you will likely need to zoom into read the text since the image's height is significantly larger than its width).

Homework Equations


Differential operator method.
Wronskian.
Variation of parameters.
Integration.

The Attempt at a Solution


What I understand:
1. I understand what an Euler equation is.

2. The Cramer's rule part.

3. The Wronskian part.

4. The variation of parameters part.

5. The integrating part.

6. The part about forming the general solution given the previous data.

What I don't understand:
1. Why is x^4 ln x in the question? Using WolframAlpha.com to confirm, the general solution presented at the end of the solution of this problem is that of x^2 y'' – 4x y' + 6y = x^4 sin x and not x^2 y'' – 4x y' + 6y = x^4 ln x.

2. I don't get the phi(t) = y(e^t) part.

3. I don't get the [D(D – 1) – 4D + 6] phi(t) = [(D – 3)(D – 2)] phi(t) = 0 part.

4. I don't get the y_1(x) = |x|^3 and y_2(x) = |x|^2 part.


Any input helping me fully understand this problem would be GREATLY appreciated!
 

Attachments

  • TheProblemAndSolution.png
    TheProblemAndSolution.png
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  • #3
Thanks, we can now "cross out" #2. and #4. off my list of things I don't understand.

I'm still confused about the rest, though.

About the image, you can press Ctrl and scroll up, or Ctrl and the + sign, and your browser should zoom in, but alternatively, here (in the attachment of this post) is the image I uploaded in my opening post, converted to a pdf. Is this better for you?
 

Attachments

  • TheProblemAndSolution.pdf
    47.4 KB · Views: 285
  • #4
It is really hard to discuss problems when they are posted in separate attachments because we can't edit and refer to particular equations. That plus the fact it is hard to know what is really confusing you. As far as the ##x^4\ln x## versus ##x^4\sin x## is concerned, the problem has a right side of ##b(x) = x^4\sin x## or ##x^4
\ln x##. Apparently the author wants you to work it with two different NH terms (two different problems), and he has only worked one of them.
 
Last edited:
  • #5
I had a feeling that that's what was going on (for #1 of the list of what I don't get), but I wasn't sure if it was just something fancy I didn't understand, and I didn't want to overlook it.

So, now, I can cross out #1 from my list of what I don't get, which leaves only #3!

I'd say we're doing pretty good (with knocking out things that are confusing me). ;)

Here's the last thing remaining that confuses me.:
3. I don't get the [D(D – 1) – 4D + 6] phi(t) = [(D – 3)(D – 2)] phi(t) = 0 part.

Put differently, how do I got from substituting x with e^t to the stuff with the differential operator?
 
  • #6
s3a said:
I had a feeling that that's what was going on (for #1 of the list of what I don't get), but I wasn't sure if it was just something fancy I didn't understand, and I didn't want to overlook it.

So, now, I can cross out #1 from my list of what I don't get, which leaves only #3!

I'd say we're doing pretty good (with knocking out things that are confusing me). ;)

Here's the last thing remaining that confuses me.:


Put differently, how do I got from substituting x with e^t to the stuff with the differential operator?

Look again at the reference I gave in post #2. In particular, look at equations (3) and (4). If we let ##\mathcal D = \frac d {dt}## and ##D = \frac d {dx}##, equation (3) becomes ##t\mathcal D = D## and equation (4) becomes ##t^2\mathcal D^2 = D^2-D = D(D-1)##. Once you understand how the substitution works, you can quickly change the Euler equation expressed with ##\mathcal D's## to a constant coefficient expressed with ##D's##.

Here's how you would do it with his example$$
t^2y''-3ty'+7y=0$$ $$
( t^2 \mathcal D^2 - 3t\mathcal D + 7)y(t) = 0$$ $$
(D(D-1) -3D + 7)y(x) =0$$ $$
(D^2 - 4D + 7)y(x) = 0$$ $$
y'' - 4y' + 7y = 0$$where this is for ##y## in terms of ##x##. Now you solve this constant coefficient equation in terms of ##x## and get the solution to the Euler equation by substituting ##x=\ln t##.
 
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  • #7
I'm still having some trouble with this.

Before I get back to the notation that I find challenging, I was hoping that I could do this problem in a more familiar way, so could you please point out what I did wrong in my DifferentNotation.pdf file (which is my latest attachment, which is attached in this post)?
 

Attachments

  • DifferentNotation.pdf
    47.3 KB · Views: 330

Related to Finding general solution to Euler equation via variation of parameters

What is the Euler equation?

The Euler equation is a mathematical equation that describes the relationship between a function and its derivatives. It is written as y'' + p(x)y' + q(x)y = g(x), where p(x), q(x), and g(x) are functions of the independent variable x.

What is the general solution to the Euler equation?

The general solution to the Euler equation is a set of functions that satisfy the equation for all values of x. It is written as y(x) = c1y1(x) + c2y2(x), where c1 and c2 are constants and y1(x) and y2(x) are linearly independent solutions to the equation.

What is variation of parameters?

Variation of parameters is a method for finding the general solution to a non-homogeneous linear differential equation, such as the Euler equation. It involves replacing the constants in the general solution with functions of x and solving for these functions using the given non-homogeneous term.

How does variation of parameters work for the Euler equation?

To use variation of parameters for the Euler equation, we first find the general solution to the associated homogeneous equation. Then, we use this solution to find the particular solutions that make up the general solution using a set of specific formulas. Finally, we combine these particular solutions to find the complete general solution.

What are the advantages of using variation of parameters for the Euler equation?

Variation of parameters allows us to find a general solution to the Euler equation even when the non-homogeneous term is not a simple function. It also gives a more general solution compared to other methods, which only give particular solutions. Additionally, it can be used for other types of non-homogeneous linear differential equations.

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