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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

    CompuChip

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    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
    Thanks!

    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

    CompuChip

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    I'm not sure where all those minus signs came from, but
    [tex]\int \frac{dx}{x} = \ln|x|[/tex]
    and
    [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!


    Edit

    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
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