Integrating factor for a 2nd order homogeneous linear ODE

In summary, the author attempted to solve a general linear homogeneous second order equation using the product rule of differentiation and equating coefficients of y', but was unable to get the answer correct.
  • #1
danny_91
10
0

Homework Statement


Consider the general linear homogeneous second order equation:
P(x)y'' + Q(x)y' + R(x)y = 0 (1)

We seek an integrating factor μ(x) such that, upon multiplying Eq. (1) by μ(x), we can write the resulting equation in the form
[μ(x)P(x)y']' + μ(x)R(x)y = 0 (2)

(a) By equating coefficients of y' in Eqs. (1) and (2), show that μ must be a solution of:
Pμ' = (Q - P')μ (3)

Homework Equations


I guess the only relevant equation is the product rule of differentiation:
[a(x)b(x)]' = a'(x)b(x) + a(x)b'(x)

The Attempt at a Solution


Here is what I have done so far: By setting (1) = (2)

P(x)y'' + Q(x)y' + R(x)y = [μ(x)P(x)y']' + μ(x)R(x)y
Py'' + Qy' + Ry = μPy'' + μP'y' + μ'Py' + μRy
Py'' + Qy' + Ry = μPy'' + (μP' + μ'P)y' + μRy

Equating coefficients of y':

Q = μP' + μ'P
So: Pμ' = Q - P'μ, which is not the form that the book stated, Pμ' = (Q - P')μ

So am I missing something in the derivation, or is it that the book's statement is wrong?
By the way, this is Problem 11 in Section 11.1 of the book called Elementary Differential Equations and Boundary Value Problems by Boyce & DiPrima (9th Edition).

Part (b) of the problem is just solving the differential equation to find μ(x), and I'm pretty sure that I can do that part. Anyway, I would really appreciate any input.
 
Physics news on Phys.org
  • #2
I don't think you can equate the coefficients of the differentials of y, because of equation (1) which gives an explicit relationship between them. So you can't just treat them as separate things as you normally would.

Your method is good, apart from that. You should try again. There are 3 things y,y',y'' that you don't want in your final formula, so try to get rid of one of them by substituting it from one equation into the other.
 
  • #3
Thank you so much, BruceW! :)

So I went back to this step: Py'' + Qy' + Ry = μPy'' + (μP' + μ'P)y' + μRy, and then grouping like terms of y'', y', and y. And then I use the relationship from Eq. (1) then I'm left with just y'. So, yes, I got the answer as the book stated.

Thanks so much for pointing out that simply equating the coefficients of y' is not the right approach!
 
  • #4
no problem :) in a way, this question is pretty tricky, since it is really pretty lucky that the y'' and y happen to both fall out. (It is not obvious from the beginning what is the right way to go). But you've got it, so all's good!
 

Related to Integrating factor for a 2nd order homogeneous linear ODE

1. What is the purpose of an integrating factor in a 2nd order homogeneous linear ODE?

An integrating factor is used to transform a non-exact second order homogeneous linear differential equation into an exact one, making it easier to solve. It helps to eliminate the need for a particular solution and instead allows for a direct solution to the equation.

2. How do you determine the integrating factor for a 2nd order homogeneous linear ODE?

The integrating factor is determined by multiplying the entire ODE by an appropriate function, typically denoted as μ(x), such that the resulting equation is exact. This function is found by comparing the coefficient of the second derivative term with the coefficient of the first derivative term.

3. Can an integrating factor be used for non-homogeneous linear ODEs?

No, an integrating factor is only applicable for homogeneous linear ODEs. For non-homogeneous linear ODEs, other methods such as variation of parameters or the method of undetermined coefficients must be used.

4. Is there a specific method for finding the integrating factor for a 2nd order homogeneous linear ODE?

Yes, there is a specific formula for finding the integrating factor for a 2nd order homogeneous linear ODE. It involves taking the exponential of the integral of the ratio of the coefficient of the first derivative term and the coefficient of the second derivative term.

5. Are there any limitations to using an integrating factor for solving 2nd order homogeneous linear ODEs?

There are some limitations to using an integrating factor, such as the equation must be linear, homogeneous, and have constant coefficients. Additionally, the method may not always work for all equations, and alternative methods may need to be used.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
333
  • Calculus and Beyond Homework Help
Replies
7
Views
759
  • Calculus and Beyond Homework Help
Replies
3
Views
613
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
973
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
369
  • Calculus and Beyond Homework Help
Replies
21
Views
913
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Back
Top