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Integrating factor

  1. Feb 16, 2010 #1
    Integrating factor!!

    As promised I'm back with integrating factor differential equation.


    (x^2 + 1)dy/dx -2xy = 2x(x^2+1) y(0)=1

    First put into standard from by dividing thru by (x^2 +1 )


    dy/dx -2xy/(x^2 + 1) = 2x

    Integrating factor is given by exp( integral of -2x(x^2 + 1))

    After some working out I get the IF to be 1/(x^2 + 1)

    Now the solution is given by

    y(x)=1/IF(integral of 2x(x^2 + 1)

    Hopefully I'm on the right track so far...

    After doing the integration by parts and some tidying up I have

    y(x)= (x^2 + 1){(2(x^2 + 1)x^3)/3 - 4x^5/15} + C

    After plugging in the values I have 1=C

    What do you think??
     
  2. jcsd
  3. Feb 16, 2010 #2
    Re: Integrating factor!!

    On checking my work I think I've mad a mistake

    the equation should be

    y(x) = 1/IF{integral of 2x/(x^2 + 1)}

    Which makes it different...

    My final revised answer is

    y(x) = (x^2 + 1)(ln{x^2 + 1}) + C

    After plugging in values y(0)=1

    I have y(x)=(x^2 + 1)(ln{x^2 + 1}) + 1


    Cheers!!!
     
  4. Feb 16, 2010 #3

    vela

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    Re: Integrating factor!!

    That's almost right. Did you try plugging it back into the original equation to see if it worked?
     
  5. Feb 16, 2010 #4
    Re: Integrating factor!!

    I haven't and I'm not really sure how to do it or what you mean! Forgive my ignorance but can you show me....

    James
     
  6. Feb 16, 2010 #5

    Mark44

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    Re: Integrating factor!!

    You have your differential equation back in your original post in this thread. In post 2 you have a solution, y(x)=(x^2 + 1)(ln{x^2 + 1}) + 1.

    Does this function satisfy the initial condition? I.e., is y(0) = 1?
    Does this function satisfy the differential equation? I.e., if you replace y and dy/dx in the differential equation with the function above and its derivative, do you get a true statement in this equation: (x^2 + 1)dy/dx -2xy = 2x(x^2+1)?

    You should always check your solutions to differential equations.
     
  7. Feb 16, 2010 #6
    Re: Integrating factor!!

    Yes my solution satisfies condition y(0)=1

    So I now need to differntiate my solution and substitute back into the equation? I'm not sure I get what you mean sorry? Bear in mind I'm trying to learn this...! lol

    James
     
  8. Feb 16, 2010 #7

    Mark44

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    Re: Integrating factor!!

    Yes, that's what I mean. Take the derivative of your solution. Multiply it (the derivative) by (x^2 + 1). Subtract 2x times your solution. If you get 2x(x^2 + 1), your solution satisifies the DE.
     
  9. Feb 16, 2010 #8
    Re: Integrating factor!!

    I don't! After doing what you said I arrive at (x^2 + 1) - 1

    Which is x^2.

    So my solution is wrong?
     
  10. Feb 16, 2010 #9

    Mark44

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    Re: Integrating factor!!

    Yes, it's wrong. That's what vela was suggesting that you do back in post #3. Now that you know you have a mistake, go back and take another look at your work and see if you can spot an error.
     
  11. Feb 16, 2010 #10
    Re: Integrating factor!!

    Thanks for your help Mark, I've been over my solution several times but always get the same - and its worng? I just can't see where Ive gone wrong...
    Is it my integrating factor?
    James
     
  12. Feb 16, 2010 #11

    vela

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    Re: Integrating factor!!

    Hint: Your mistake has to do with when you introduced the constant of integration.
     
  13. Feb 16, 2010 #12
    Re: Integrating factor!!

    Hmmm I'm not sure, but at a guess should it be ln(C) not just C ??
     
  14. Feb 16, 2010 #13
    Re: Integrating factor!!

    Hold on! I think I may have it...?

    I should have multiplied at all by 1/IF making my constant thus

    C(x^2 + 1)

    Or am I way off again...

    James
     
  15. Feb 16, 2010 #14

    vela

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    Re: Integrating factor!!

    Yup, that's it. You can, of course, check your answer by plugging it back into the original differential equation.
     
  16. Feb 16, 2010 #15
    Re: Integrating factor!!

    Thanks for your help everyone, my revised solution works! I've learned a lot, your gentle pointers eventually made the penny drop.

    At least now I know how to check my solutions!

    Thanks again.

    James

    P.S

    I'll be moving onto 2nd order differential equations next and looking forward to your help again...
     
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