Series Solutions to a 2nd Order Diff. Eq.

In summary, the conversation discusses a problem involving a differential equation, y'' + y' + xy = 0. The person asking for help is unsure if their solution is correct and is seeking clarification. The conversation also mentions the importance of verifying solutions and correcting a mistake in the solution's third term. The person asking for help is advised to revise their solution and post it for further feedback.
  • #1
Destroxia
204
7

Homework Statement



y'' + y' + xy = 0

Just want to make sure I understand this completely, I had a bit of trouble towards the end, and thought the -29/600 was a little weird of a fraction to be right. I wasn't given a correct answer to base mine off, so I'm not sure if I'm doing this all right.

Homework Equations


The Attempt at a Solution



upload_2015-7-15_19-19-23.png
 
Physics news on Phys.org
  • #2
First off, you should always be able to verify that the solution you've derived is actually a solution by plugging it back into the DE and comparing terms order by order. Second, and more directly relevant, in your 5th line, you have lost the factor of ##x## in the 3rd, ##xy## term. Correcting this will significantly change the recursion relations.
 
  • #3
I think you forgot to increase the exponent of x in the last summation of line 5 when you multiplied through by x.
Already answered above. How come I can't delete this?
 
  • #4
fzero said:
First off, you should always be able to verify that the solution you've derived is actually a solution by plugging it back into the DE and comparing terms order by order. Second, and more directly relevant, in your 5th line, you have lost the factor of ##x## in the 3rd, ##xy## term. Correcting this will significantly change the recursion relations.

FactChecker said:
I think you forgot to increase the exponent of x in the last summation of line 5 when you multiplied through by x.
Already answered above. How come I can't delete this?

Thank you both, I will rework it and post my answer here.
 

Related to Series Solutions to a 2nd Order Diff. Eq.

1. What is a series solution to a 2nd order differential equation?

A series solution to a 2nd order differential equation is a method used to solve a differential equation by representing the solution as a power series. This allows for an infinite number of terms to be included in the solution, providing a more accurate and precise answer.

2. When is a series solution method used?

A series solution method is typically used when it is difficult or impossible to find an explicit solution to a differential equation using other methods, such as separation of variables or variation of parameters.

3. How does a series solution to a 2nd order differential equation work?

A series solution to a 2nd order differential equation involves expanding the unknown function into a power series and substituting it into the differential equation. This results in a recursive relationship between the coefficients of the power series, which can be solved to find the values of the coefficients.

4. What are the advantages of using a series solution method?

One advantage of using a series solution method is that it can provide a more precise and accurate solution compared to other methods. Additionally, it can be used to solve a wide range of differential equations, including those that cannot be solved using other methods.

5. Are there any limitations to using a series solution method?

Yes, there are some limitations to using a series solution method. It may not work for all types of differential equations, and the convergence of the series may be slow for certain equations. It also requires a good understanding of power series and series operations to use effectively.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
384
  • Calculus and Beyond Homework Help
Replies
25
Views
594
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
17
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
Back
Top