Results 1 to 8 of 8

Math Help - integration of (ax+by+c)/(ex+fy+g) = dy/dx, where a,c,c,e,f,g are constants

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    5

    integration of (ax+by+c)/(ex+fy+g) = dy/dx, where a,c,c,e,f,g are constants

    My maths book tells me that if we let X= x+k and y = Y+m, then dY/dX
    equals (aX + bY)/(eX +fY) and then proceed as for a homogenous equation. My question is what happened to c and g and how is the new expression homogenous?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2011
    From
    Islamabad
    Posts
    96
    Thanks
    1
    the values of k and m are taken such that c and g are removed
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    5
    ok. Thanks. I still don't understand how removing a constant (c) can be the same as multiplying another constant (a or e) by X over the entire range in which x operates, even if another variable (X) is used. Could you help me, please? What ensures that the new expression tracks the original one?
    Last edited by woolyhead; June 11th 2011 at 10:09 AM. Reason: omission of detail
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2011
    From
    Islamabad
    Posts
    96
    Thanks
    1
    to find the numbers k and m solve salmantaniously the equations

    ax + by + c = 0 and ex + fy + g = 0

    and substitute x = X - k and y = Y -m in given diff equation try this method with a question with numerical values
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    The substitution required depends on whether the two lines ax+by+c=0 and ex+fy+g=0 are intersecting lines, identical lines, or parallel, non-intersecting lines. If they are identical lines, then the DE simplifies greatly. If they are parallel, non-intersecting lines, then let u = ax+by+c, and rewrite the DE in terms of u and y. The resulting equation will be separable.

    Lastly, if the coefficients, when equated to zero, represent intersecting lines, then find the point of intersection (p,q). Define new coordinates u = x-p, v = y-q, and rewrite the DE in terms of these coordinates. The resulting DE will be homogeneous.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2011
    Posts
    5

    visualising your replies

    Thanks Ackbeet and Waqarhaider. I tried your suggestions and they worked, but I still cannot visualise or imagine why we let the numerator and the denominator of the original expression equal zero, even though I know the method given uses this trick. Should I just use the method and stop asking questions about the reasons why it works, then? Wouldn't that reduce the intellectual content of the approach somewhat? Suppose I came across another type of problem where a full understanding the method used in the present problem could be of help. Just knowing a method that works without knowing why it works wouldn'd be a lot of use, would it? Can you also please help me understand what's going on with the method you kindly gave me.
    As I see it, the original equation describes two 2D surfaces which can be moved up or down according to the value of the constants. These constants are in the way, so to speak, of our calling the expression homogenous, so we use your trick to get rid of them in order to be able to use the standard method for solving such equations. So what are we achieving by making the numerator and the denominator zero? Where's the logic in it?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by woolyhead View Post
    Thanks Ackbeet and Waqarhaider. I tried your suggestions and they worked, but I still cannot visualise or imagine why we let the numerator and the denominator of the original expression equal zero, even though I know the method given uses this trick. Should I just use the method and stop asking questions about the reasons why it works, then? Wouldn't that reduce the intellectual content of the approach somewhat? Suppose I came across another type of problem where a full understanding the method used in the present problem could be of help. Just knowing a method that works without knowing why it works wouldn'd be a lot of use, would it? Can you also please help me understand what's going on with the method you kindly gave me.
    As I see it, the original equation describes two 2D surfaces which can be moved up or down according to the value of the constants. These constants are in the way, so to speak, of our calling the expression homogenous, so we use your trick to get rid of them in order to be able to use the standard method for solving such equations. So what are we achieving by making the numerator and the denominator zero? Where's the logic in it?
    Instead of worrying about the general case, attempt a concrete question using the help you've been given so far. Post the question if you can't do it, showing all your work and saying where you get stuck.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Feb 2011
    Posts
    5
    ok, Mr Fantastic. Thanks. I'll do that.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Ricatti Equations and combining integration constants.
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: April 24th 2011, 05:34 AM
  2. integration -- getting constants
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: November 24th 2010, 04:39 PM
  3. Replies: 5
    Last Post: November 23rd 2009, 12:17 PM
  4. constants of D.e
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 10th 2008, 11:32 PM
  5. ode's and constants
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 7th 2007, 07:34 AM

Search Tags


/mathhelpforum @mathhelpforum