Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrating Factors Found by Inspection

  1. Jul 16, 2009 #1
    I need some explanation regarding a solution found in our textbook.

    The example starts with this DE:

    (1) y(x3 - y)dx - x(x3 + y)dy = 0.

    By regrouping, exact DEs will be found and the equation can be rewritten to:

    (2) x3 d([tex]\frac{x}{y}[/tex]) - [tex]\frac{d(xy)}{y}[/tex] = 0.

    The integrating factor is found to be: x-2y-1. Multiplying the equation by the IF results to:

    (3) [tex]\frac{x}{y}[/tex] d([tex]\frac{x}{y}[/tex]) - [tex]\frac{d(xy)}{x^{2}y^{2}}[/tex]

    Now this is the part that I don't understand. The text continues after the equation above, "of which a set of solutions is given by"

    (4) [tex]\frac{1}{2}[/tex] ([tex]\frac{x}{y}[/tex])2 + [tex]\frac{1}{xy}[/tex] = [tex]\frac{c}{2}[/tex].

    Just how did (3) turn into (4)?

    I think it's a very simple question but I'm really lost. I'm so confused with anything that has to do with differential equations. :( Thanks in advance for the help.
  2. jcsd
  3. Jul 16, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    I suppose they just integrated equation 3.
    If we call u = x / y and v = x y, then putting an integral sign in front of (3) reads
    [tex]\int u \, du - \int \frac{dv}{v^2} = 0.[/tex]
    Both integrals are elementary, giving
    [tex]\tfrac12 u^2 + \frac{1}{v} + C[/itex].
    We can now plug in the definitions of u and v again, and bring C to the other side, giving
    [tex]\frac12 \left( \frac{x}{y} \right)^2 + \frac{1}{xy} = \frac{c}{2}[/tex]
    where c = -2C is just another constant (i.e. determining C from boundary conditions is equivalent to determining c).

    If this means more to you: basically, they just did a change of variables.
  4. Jul 16, 2009 #3

    Yeah, I knew there was some integration going on, but I didn't know how to integrate the x's and y's since they were together. I get it now. The exact DE's were absorbed into the du and dv. Stupid. :( Thanks so much for taking the time to explain it step by step. :)

    I want to ask another question, though, if it's all right. (Maybe I should change the topic title to something more general--because I have so many questions regarding DE's...)

    I'm having problems with transformations of homogeneous equations as well. I get to a solution, but it's not the solution in the book. For example,

    (16x + 5y)dx + (3x + y)dy = 0

    I used the transformation y = vx, y' = x + xv'. Then I divided by [tex]\frac{1}{x^{2}}[/tex], put the fractions together and ended up with:

    [tex]\frac{dx}{x}[/tex] = [tex]\frac{(3 + v)dv}{(v + 4)^{2}}[/tex]

    Integrating using partial fractions, the result is:

    -ln|x| = ln|v + 4| - [tex]\frac{1}{v + 4}[/tex] + c

    I plugged back v = y/x, solved for c (x = 1; y = -3), and so the answer is:

    -y - 3x = (y + 4x) ln(y + 4x)

    But the book says the answer is: y + 3x = (y + 4x) ln(y + 4x).

    My answers often have incorrect signs. Was there anything wrong in the process I did? Or would they be equal in the end? I double-checked it, and still ended up with the same answer. And I have this problem for most of transformations of homogeneous equations.

    Thanks again in advance. :)
    Last edited: Jul 16, 2009
  5. Jul 17, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    I'm not sure where all those minus signs came from, but
    [tex]\int \frac{dx}{x} = \ln|x|[/tex]
    [tex]\ln|v + 4| + \frac{1}{v + 4} [/tex]
    according to Mathematica.
    Perhaps you have introduced some faulty signs in the partial fractions?

    Maybe an easier solution would be to write
    [tex]\frac{3 + v}{(v + 4)^2} = \frac{(4 + v) - 1}{(v +4)^2} = \frac{1}{v + 4} - \frac{1}{(v + 4)^2}.[/tex]
  6. Jul 23, 2009 #5
    Thanks so much, CompuChip. :)

    I worked on that equation again. BTW, sorry, I made a typo--there is supposed to be a negative sign before dx/x, so the integration -ln|x| is correct. My mistake was in the integration by partial fraction. My resulting integral was separated by a negative sign when it should have been a plus. That fixed everything! Thanks! :)

    I worked on the other homogeneous equations over the week as well. Most errors were due to integrating mistakes. :( Now I feel embarrassed. :lol:

    I do need help with other equations again, however. Sorry in advance for so much trouble. I've been practicing on a lot of DE's, so I have a rather big unanswered list. Here goes:

    [tex](1 - xy)^{-2}dx + [y^{2} + x^{2}(1 - xy)^{-2}]dy = 0[/tex]

    The above equation is not separable, not exact, not linear, and not homogeneous. Or is it? I'm out of ideas for this one. I did try to multiply it by (1 - xy)2, but it didn't seem to help.

    The next two are supposed to be answered by looking for a common exact DE by regrouping the terms, but any pattern is really difficult to find:

    [tex]y(x^{2} + y)dx + x(x^{2} - 2y)dy = 0[/tex]
    [tex]y(3x^{3} - x + y)dx + x^{2}(1 - x^{2})dy = 0[/tex]

    And the topic that stumps me most: Bernoulli equations! I haven't been able to solve even one equation. :( It'd be too much to post all equations here, so I'll just post the first one. Hopefully, after I know how to solve this one, I'll be able to get the drift of it.

    [tex]\frac{dy}{dx} = sin(x + y)[/tex]

    No matter how I look at it, it just doesn't seem to be a Bernoulli equation. It's nowhere near the correct form. :(

    All right, sorry, sorry, sorry for so many questions and thanks, thanks, thanks. :D Any help with any equation is very much appreciated!


    I have another question. *sheepish* I'm wondering what the general approach to answering DE's is. In what order should you check the form of the DE? Like, I know the first you have to check for is separability. Then? Exactness? Linearity? Homogeneity? Thanks in advance!
    Last edited: Jul 23, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook