Second Order Linear ODE - Power Series Solution to IVP

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Homework Help Overview

The discussion revolves around finding a power series solution for a second-order linear ordinary differential equation (ODE) given by (x² - 1)y'' + x³y' + y = 2x, with initial conditions y(0) = 1 and y'(0) = 0. The original poster notes that x = 0 is an ordinary point and expresses uncertainty about incorporating the term "2x" into the power series setup.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss how to incorporate the "2x" term into the power series equation. The original poster is unsure about treating "2x" as a series and considers different ways to express it. Others suggest equating powers of x to establish relationships between the coefficients, while questioning the role of the constant term in the series.

Discussion Status

Participants are actively exploring how to set up the series equation correctly. Some guidance has been provided regarding the treatment of the "2x" term, with suggestions to consider it as a constant coefficient term. There is an ongoing dialogue about how to combine this with other terms in the series.

Contextual Notes

The original poster indicates that this scenario was not covered in class and that they could not find relevant examples in their textbook, suggesting a potential gap in their understanding of the topic.

ChemistryNat
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Homework Statement


Let y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞) be a power series solution of

(x2-1)y''+x3y'+y=2x, y(0)=1, y'(0)=0

Note that x=0 is an ordinary point.

Homework Equations


y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞)
y'(x)=[itex]\sum[/itex](kckxk-1) (k=1 to ∞)
y''(x)=[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞)

The Attempt at a Solution



(x2-1)[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞) +[itex]\sum[/itex](kckxk) (k=1 to ∞)+[itex]\sum[/itex]ckxk (k=0 to ∞) -2x=0 ??

I'm not having an issue with the power series themselves, I'm just not sure how to incorporate in the "2x" term when I'm setting up the series equation. We didn't cover this scenario in class and I couldn't find anything like it in my textbook.

I've been trying to incorporate it as a series itself
ie. 2[itex]\sum[/itex]ckxk (k=0 to ∞) where C0=0, or 2[itex]\sum[/itex]ckxk (k=1 to ∞)
but I'm not sure I can even do that mathematically?Thank you!
 
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ChemistryNat said:

Homework Statement


Let y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞) be a power series solution of

(x2-1)y''+x3y'+y=2x, y(0)=1, y'(0)=0

Note that x=0 is an ordinary point.

Homework Equations


y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞)
y'(x)=[itex]\sum[/itex](kckxk-1) (k=1 to ∞)
y''(x)=[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞)

The Attempt at a Solution



(x2-1)[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞) +[itex]\sum[/itex](kckxk) (k=1 to ∞)+[itex]\sum[/itex]ckxk (k=0 to ∞) -2x=0 ??

I'm not having an issue with the power series themselves, I'm just not sure how to incorporate in the "2x" term when I'm setting up the series equation. We didn't cover this scenario in class and I couldn't find anything like it in my textbook.

I've been trying to incorporate it as a series itself
ie. 2[itex]\sum[/itex]ckxk (k=0 to ∞) where C0=0, or 2[itex]\sum[/itex]ckxk (k=1 to ∞)
but I'm not sure I can even do that mathematically?Thank you!

Your are going to equate powers of x to get a relation between the c_k values, right? The -2x will only contribute the x^1 term, yes?
 
Dick said:
Your are going to equate powers of x to get a relation between the c_k values, right? The -2x will only contribute the x^1 term.

So I can leave it in as a constant coefficient term, say 2C1x? and then use that in combination with the other constants I've pulled out?
 
ChemistryNat said:
So I can leave it in as a constant coefficient term, say 2C1x? and then use that in combination with the other constants I've pulled out?

Mmm. Sort of, but the coefficient your x^1 term is only going to have a -2 in it. Without any C_k term in front of it. It's just a constant.
 
Last edited:
Dick said:
Mmm. Sort of, but the coefficient your x^1 term is only going to have a -2 in it. Without any C_k term in front of it. It's just a constant.

Oh! so I can just leave it be and put it with the other x1 coefficient terms?
 
ChemistryNat said:
Oh! so I can just leave it be and put it with the other x1 coefficient terms?

Sure.
 
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