1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

ODE, Integrating Factor method

  1. Feb 9, 2009 #1

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    I'm trying to follow this example from class notes.

    [tex]x^2 \frac{{dy}}{{dx}} + xy = 1,\,\,y\left( 1 \right) = 2[/tex]

    Divide through by x^2 to set it up in a form for using the integrating factor method.

    [tex]
    \frac{{dy}}{{dx}} + \frac{y}{x} = - \frac{1}{{x^2 }}

    [/tex]
    I'm not sure where the minus sign came from on the right side. I just copied the problem from the board.

    Seperate the functions of x to make it in I.F. form
    [tex]
    \frac{{dy}}{{dx}} + y\left( x \right)\frac{1}{x} = - \frac{1}{{x^2 }}
    [/tex]

    Compute the Integrating Factor with from the term that doesn't contain a y. Is this correct? Is this why 1/x was chosen to integrate and raise e to the power of it?

    [tex]
    I.F.\,\,e^{\int_{}^{} {\frac{1}{x}dx} } = e^{\ln x} = x
    [/tex]

    The next step in the notes is
    [tex]
    x\frac{{dy}}{{dx}} + y = \frac{1}{x}
    [/tex]
    Is this simply from multiplying the Integrating Factor, in this case, x, by all 3 terms in the 2nd equation? If so, what happened to my minus sign (or did it never exist and I inadvertantly put it in by mistake)?

    The problem is longer than this, but I want to understand it up to this point for the moment. Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 9, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes. You have the right idea. I don't know why the minus sign is appearing and disappearing on the right side. Somebody is just being careless.
     
  4. Feb 9, 2009 #3

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    Thanks. That careless person may be me. I don't always copy flawless notes :)

    So I left off at [tex]x\frac{{dy}}{{dx}} + y = \frac{1}{x}[/tex]

    The next step (assuming I copied it correctly) is
    [tex]
    \left( {xy} \right)^\prime = \frac{1}{x}
    [/tex]

    The right side remained the same. So how is it that [tex]
    x\frac{{dy}}{{dx}} + y = \left( {xy} \right)^\prime
    [/tex]
    ?
    This is the same as saying [tex]
    xy' + y = \left( {xy} \right)^\prime

    [/tex], isn't it?
     
  5. Feb 9, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure. Just use the product rule to figure out d/dx(x*y(x)).
     
  6. Feb 9, 2009 #5

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    still a bit confused. Product rule makes sense as a check. I can go from (xy)' = x'y+xy'=y+xy'. When we learned the product rule, we learned to take a product and turn it into two terms. But what's the method for doing the product rule in reverse? How do I start with two terms and work them into a product?
     
  7. Feb 9, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's the magic of the integrating factor. If the integrating factor is M(x) and the left side of the ode is y'+a(x)*y(x), you know that the derivative (M(x)*y(x))' will come out to be M(x)y'(x)+M(x)*a(x)*y(x). That's why it's called an 'integrating factor'.
     
  8. Feb 9, 2009 #7

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    Thanks. I'll take me a little while to grasp that. There's one final question in this problem. After integrating both sides and ending up with
    [tex]
    y = \frac{{\ln x + C}}{x}

    [/tex]

    I plug in 1 for x and set the equation to 2, as per the instructions.
    [tex]
    y\left( 1 \right) = \frac{{\ln \left( 1 \right) + C}}{1} = 2
    [/tex]

    The next step (which is the answer) from the class notes is simply
    [tex]
    y = \frac{{\ln x + 2}}{x}
    [/tex]

    How did this last step come about? I hate to say that the 1's magically turned back into x's, and the answer replaced C.

    gotta run out the door now, so if I don't says 'thanks' for a few hours, I still appreciate your explanations.
     
  9. Feb 9, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    There's nothing magic about it. You solved y(1)=(ln(1)+C)/1=2 for C. Then you plugged that value of C back into y(x)=(ln(x)+C)/x.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: ODE, Integrating Factor method
  1. Integrating factor ode (Replies: 5)

Loading...