Results 1 to 4 of 4

Math Help - exact DE (with integrating factor)

  1. #1
    Newbie
    Joined
    Mar 2007
    Posts
    10

    exact DE (with integrating factor)

    Okay here's the problem, I can't seem to find an integrating factor for this differential equation. As the original equation doesn't come out to be exact, we need an integrating factor to make it exact then we can solve it as an exact DE.

    The equation is;
    [sin(y)cos(y) + xcos^2(y)]dx + x.dy = 0 ---(1)

    which is the same as;
    [1/2sin(2y) + xcos^2(y)]dx + x.dy = 0

    now if we let;
    A = 1/2sin(2y) + xcos^2(y)
    and
    B = x

    and take the partial derivatives of both;

    dA/dy = cos(2y) - xsin(2y)
    dB/dx = 1

    and since dA/dy does not equal dB/dx we use either

    i. (1/A)[dB/dx - dA/dy]
    or
    ii. (1/B)[dA/dy - dB/dx]

    to get a function dependant only on y or only on x


    ...and I can't get that. I've tried various trig identities with no success, maybe I've messed up the basic differentiation there, or maybe I've missed an identity, I'm not sure. If someone can help me with the integrating factor, I'm fairly certain I can find the general solution to the equation easily.


    ----------------

    let me add 2 more;

    1. using separation of variables, solve;
    (1+2x)cosy .dx + dy/cosy = 0

    for this one, I end up doing t-substitution partial fractions to integrate it :/ is that right?? Because I get;

    INT [1+2x] dx = INT [-1/cos^2(y)] dy
    = INT [-2/(cos(2y) +1)] dy


    2. using the change of variable u=y^3, solve;
    y' + yx^2 = [sinh(x) . e^(-x^3)]/3y^2

    and this one I don't get very far on...I can't get rid of the dy/dx when i substitute in for u and du/dy. I get;

    du/dy . dy/dx +3ux^2 = sinh(x). e^(-x^3)



    and FYI: these are from a tutorial sheet, but they need to be done, and I'd much prefer to learn how to do them. As you can see, i've given all of them a crack.
    Last edited by DistantCube; March 17th 2007 at 03:37 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by DistantCube View Post
    Okay here's the problem, I can't seem to find an integrating factor for this differential equation. As the original equation doesn't come out to be exact, we need an integrating factor to make it exact then we can solve it as an exact DE.

    The equation is;
    [sin(y)cos(y) + xcos^2(y)]dx + x.dy = 0 ---(1)

    which is the same as;
    [1/2sin(2y) + xcos^2(y)]dx + x.dy = 0

    now if we let;
    A = 1/2sin(2y) + xcos^2(y)
    and
    B = x

    and take the partial derivatives of both;

    dA/dy = cos(2y) - xsin(2y)
    dB/dx = 1

    and since dA/dy does not equal dB/dx we use either

    i. (1/A)[dB/dx - dA/dy]
    or
    ii. (1/B)[dA/dy - dB/dx]

    to get a function dependant only on y or only on x


    ...and I can't get that. I've tried various trig identities with no success, maybe I've messed up the basic differentiation there, or maybe I've missed an identity, I'm not sure. If someone can help me with the integrating factor, I'm fairly certain I can find the general solution to the equation easily.

    .
    I think you did everything correctly. It seems no integrating factor to turn the ODE into an exact one does not exist (at least not dependent on one variable).

    1. using separation of variables, solve;
    (1+2x)cosy .dx + dy/cosy = 0
    You have, for some interval.

    (1+2x)cos y + (1/cos y)y' = 0

    Divide by cos y to get,

    (1+2x) + sec^2 y * y' = 0

    Integrate,

    INT (1+2x) dx + INT sec^2 y *y' dx = C

    x+x^2 + tan y =C
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Heir.
    Attached Thumbnails Attached Thumbnails exact DE (with integrating factor)-picture3.gif  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2007
    Posts
    10
    Quote Originally Posted by ThePerfectHacker View Post
    I think you did everything correctly. It seems no integrating factor to turn the ODE into an exact one does not exist (at least not dependent on one variable).
    I found the integrating factor just now anyway. You can change the identities on the top to becomes;

    [1-1+2sin^2y +2xsinycosy]/[sinycosy + xcos^2y]

    then take out a factor of 2siny on the top, cosy on the bottom, the rest cancels and you're left with 2tany.

    -----------

    edit: Also, for the question where you attached the image, I can't believe I didn't realise before - that's a Bernoulli equation! Could have simply used that form :/

    -----------

    Thank you very much for your help, I shall work through these this afternoon. I shouldn't come across any trouble, but if I do, I'll know who to ask
    Last edited by DistantCube; March 17th 2007 at 11:19 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integrating factor on Exact, Homogeneous DE
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: September 12th 2011, 08:23 PM
  2. [SOLVED] Integrating Factor
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 14th 2011, 05:38 PM
  3. Non-Exact Equation - Integrating Factor
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: October 10th 2010, 07:03 PM
  4. Exact Equation, integrating factor
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: September 29th 2010, 01:56 PM
  5. Integrating Factor making DE Exact
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: August 29th 2009, 07:23 AM

Search Tags


/mathhelpforum @mathhelpforum