1. Not finding help here? Sign up for a free 30min 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!

Exact ODE

  1. Dec 14, 2005 #1
    I missed a course and now I have to solve an ODE, I don't freaking know how :)

    [tex](ax+by)dx+(kx+ly)dy=0[/tex]

    So, I have to know on what condition this equation is exact (1), and I have to find the exact equation (2). Well if a, b, k and l = 0 then it's certainly over but i guess it's not really the point.

    Let's say M(x,y) = ax+by and N(x,y) = kx+ly, IF i'm right, the ODE is exact when;

    [tex]\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}[/tex]

    Then;

    [tex]\frac{\partial (ax+by)}{\partial y}=\frac{\partial (kx+ly)}{\partial x}[/tex]

    [tex]b=k[/tex]

    So my answer to the first question is; is it exact when b = k. And to show how sure I am (not...) I'll say c = b = k for the next part of this question. However, intuitively, i'd say the winning condition is when a and l = 0, and b = -k, (b=c,k=-c) then by integrating you'll got bxy+kxy = 0, so cxy-cxy=0, 0=0.

    And then, I don't know what method to use to find the exact ODE.
     
    Last edited: Dec 14, 2005
  2. jcsd
  3. Dec 14, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, it is true that the differential M(x,y)dx+ N(x,y)dy is "exact" when there exist some function f(x,y) such that
    [tex]\frac{\partial f}{\partial x}= M(x,y)[/tex]
    and
    [tex]\fract{\partial f}{\partial y}= N(x,y)[/tex]
    and that requires, in order that
    [tex]\frac{\partial^2f}{\partial x\partial y}= \frac{\partial^2f}{\partial y\partial x}[/tex]
    that
    [tex]\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}[/tex]

    So your differential equation (ax+by)dx+(kx+ly)dy=0
    will be an exact equation if and only if b= k. But I don't know what you mean by "find the exact equation"! Obviously, the exact equation is
    (ax+ by)dx+ (bx+ ly)dy when b= k. Is it possible that you mean "find the general solution to the exact equation"??

    An equation is exact if and only if the left hand side is an exact differential: For some f(x,y), df= M(x,y)dx+ N(x,y)dy which means that we must have [tex]\frac{\partial f}{\partial x}= M(x,y)[/tex] and that [tex]\frac{\partial f}{\partial y}= N(x,y)[/tex].

    In particular, if df= (ax+ by)dx+ (bx+ ly)dy we must have
    fx= ax+ by and fy= bx+ ly. Since the partial derivative wrt x is taken by treating y as a constant, we can "back out" by taking the anti-derivative treating y as a constant: f(x,y)= (1/2)ax2+ bxy+ g(y). "g(y)" is the constant of integration". Since we are treating y as a constant, it could be any function of y. Now differentiate that with respect to y: fy= bx+ g'(y)= by+ ly. Precisely because we set k= b, the x terms cancel and we have g'(y)= ly. Integrating that, g(y)= (1/2)ly+ C (since g is a function of y only, that C really is a constant). Putting that into the equation of f(x,y),
    f(x,y)= (1/2)ax2+ bxy+ (1/2)ly. Since df= 0 according to the equation, f(x,y) must equal a constant:
    (1/2)ax2+ bxy+ (1/2)ly= C is the general solution to the equation.
     
  4. Dec 15, 2005 #3
    wow. Thanks, you really took time to answer and it's appreciated.

    So...

    [tex](ax+by)dx + (kx+ly)dy = 0[/tex]

    To be exact, b must be equal to k

    And the general answer is;

    [tex]f(x,y) = \frac{ax^2}{2}+byx+\frac{ly^2}{2}[/tex]

    f(x,y) is equal to a constant because df = 0 (but it could still be written, f(x,y) = ... + C no ?)

    [tex]C = \frac{ax^2}{2}+byx+\frac{ly^2}{2}[/tex]

    That's clear. However you wrote at the general answer "ly/2" and I got "ly^2/2". It doesn't make more sence ? If "ly/2", then how can you get "ly" when you derive f(x,y) ? And another thing, is there a way to test that answer numerically with Maple (don't know MatLab or Mathematica yet...) ?

    Thanks a lot for your time. If you ever need a letter of recommendation from an unknown "non-math major" non-american undegraduate student with a lot of trouble writting in english, I'm your man.
     
    Last edited: Dec 15, 2005
  5. Dec 15, 2005 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    "f(x,y) is equal to a constant because df = 0 (but it could still be written, f(x,y) = ... + C no ?"

    No. f(x,y)= (1/2)ax2+ bxy+ (1/2)cy2+C is simply a function in two variables and does not define y, even implicitly, as a function of x. Either (1/2)ax2+ bxy+ (1/2)cy2+C= 0 or, same thing, (1/2)ax2+ bxy+ (1/2)cy2= C does.

    "However you wrote at the general answer "ly/2" and I got "ly^2/2". It doesn't make more sence ? "

    Yes! I messed up the superscripts in "(1/2)ax2+ bxy+ (1/2)ly= C"!

    By the way, your English is far better than my (put practically any language here!).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Exact ODE
  1. Exact ODE (Replies: 11)

  2. ODE exact equations (Replies: 4)

  3. Exact ODE (Replies: 2)

Loading...