Results 1 to 6 of 6

Math Help - Problem Solving First Order Differential Equation

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    95

    Problem Solving First Order Differential Equation

    I'm trying to solve the following D.E: (2yexp(y/x) + x)dy/dx = 2x + y
    Here is my current working:
    dy/dx = (2x+y) / (2yexp(y/x) + x)
    sub y = ux
    then dy/dx = (du/dx)x + u
    dy/dx - u = (du/dx) x
    (2 + u) / (2uexp(u) + 1) - u = (du/dx) x
    2(1-u^2exp(u))/(2uexp(u) + 1) = (du/dx) x

    So then I have to integrate 1/x dx (easy) but then also (2uexp(u) + 1) / 2 (1-u^2exp(u)) du . Actually this second integration is possible according to wolfram alpha but I can't see what method to use.

    Any help would be much appreciated.
    Last edited by StaryNight; March 17th 2011 at 06:20 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    [SECOND EDIT]: See next post for a correction in the OP.

    Hmm. Strange. I don't get quite the same integration you do. Using your substitution, I get

    \left(2\,\dfrac{y}{x}\,e^{y/x}+1\right)y'=2+2\,\dfrac{y}{x}\implies

    (2ue^{u}+1)(xu'+u)=2+2u\implies

    xu'+u=\dfrac{2+2u}{2ue^{u}+1}\implies

    xu'=\dfrac{2+2u}{2ue^{u}+1}-u=\dfrac{2+2u-2u^{2}e^{u}-u}{2ue^{u}+1}=\dfrac{2+u-2u^{2}e^{u}}{2ue^{u}+1}\implies

    \dfrac{(2ue^{u}+1)\,du}{2+u-2u^{2}e^{u}}=\dfrac{dx}{x}.

    The integral on the left is intractable. Have I made a mistake somewhere?

    I wonder if the opposite substitution, that of vy=x, would give you a more do-able integral. *Working it out...*

    [EDIT]: Nope, that integral is just as intractable.

    It looks to me like the best you can do is reduce the DE to quadratures.
    Last edited by Ackbeet; March 17th 2011 at 06:28 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2008
    Posts
    95
    My aplogies, the first line should be (2yexp(y/x) + x)dy/dx = 2x + y . I believe the rest of my working is correct though, i'm just left with this nasty-looking integral that I've no idea how to do! If it helps the exact question is http://www.maths.cam.ac.uk/undergrad...erNST_IA_2.pdf Q1c.

    Quote Originally Posted by Ackbeet View Post
    Hmm. Strange. I don't get quite the same integration you do. Using your substitution, I get

    \left(2\,\dfrac{y}{x}\,e^{y/x}+1\right)y'=2+2\,\dfrac{y}{x}\implies

    (2ue^{u}+1)(xu'+u)=2+2u\implies

    xu'+u=\dfrac{2+2u}{2ue^{u}+1}\implies

    xu'=\dfrac{2+2u}{2ue^{u}+1}-u=\dfrac{2+2u-2u^{2}e^{u}-u}{2ue^{u}+1}=\dfrac{2+u-2u^{2}e^{u}}{2ue^{u}+1}\implies

    \dfrac{(2ue^{u}+1)\,du}{2+u-2u^{2}e^{u}}=\dfrac{dx}{x}.

    The integral on the left is intractable. Have I made a mistake somewhere?

    I wonder if the opposite substitution, that of vy=x, would give you a more do-able integral. *Working it out...*

    [EDIT]: Nope, that integral is just as intractable.

    It looks to me like the best you can do is reduce the DE to quadratures.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Ah, I'm with you now. Right, I think your working is correct. Let me see. You have to evaluate the integral

    \displaystyle\int\frac{2ue^{u}+1}{2(1-u^{2}e^{u})}\,du.

    Here's the trick: multiply top and bottom by e^{-u}. See where that leads.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2008
    Posts
    95
    Ahh, It's now just a log as the numerator is the negative derivative of the denominator. I did think of this before but didn't consider multiplying top and bottom by exp(-u). How did you come up with this out of interest? I suppose the next time I have a similar integral I'll know this is something to try.

    Thanks for your help.

    Quote Originally Posted by Ackbeet View Post
    Ah, I'm with you now. Right, I think your working is correct. Let me see. You have to evaluate the integral

    \displaystyle\int\frac{2ue^{u}+1}{2(1-u^{2}e^{u})}\,du.

    Here's the trick: multiply top and bottom by e^{-u}. See where that leads.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Quote Originally Posted by StaryNight View Post
    Ahh, It's now just a log as the numerator is the negative derivative of the denominator. I did think of this before but didn't consider multiplying top and bottom by exp(-u). How did you come up with this out of interest?
    Well, the numerator looked like the derivative of the denominator, so I was already headed in that direction. However, when I took the derivative of the denominator, I had to use the product rule, and I got a whole bunch of terms that I didn't want. So I was wondering if I could possibly separate out the exponentials from the polynomials in order to avoid using the product rule, and then that little trick occurred to me. I've had problems before where that sort of thing comes in handy.

    Ultimately, though, the solution came out of my imagination, which is why I've long thought that by far the most important quality a mathematician can possess, is a good imagination. That's why reading great books (as in, literature), is so important. It trains the imagination.

    I suppose the next time I have a similar integral I'll know this is something to try.

    Thanks for your help.
    You're very welcome!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 8th 2011, 06:29 PM
  2. Solving a fifth order differential equation?
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: October 21st 2010, 07:24 PM
  3. Solving first order differential equation
    Posted in the Differential Equations Forum
    Replies: 9
    Last Post: August 28th 2010, 08:01 AM
  4. Solving First Order Differential Equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: August 18th 2009, 06:25 AM
  5. Solving 1st order non-linear differential Equation
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: December 6th 2008, 11:50 AM

Search Tags


/mathhelpforum @mathhelpforum