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ODE Homogeneous Eqn - What did I do wrong this time?

  1. May 26, 2009 #1
    Hey guys, I've been stuck on this problem for a good hour... I have no idea how to finish it up.

    1. The problem statement, all variables and given/known data

    Solve: dy/dx = 4y-3x / (2x-y)
    Use homogeneous equations method.

    2. Relevant equations
    Answer : |y - x| = c|y + 3x|5 (also y = -3x)


    3. The attempt at a solution
    dy/dx = 4y-3x / (2x-y)
    dy/dx = 4y-3x / (2x-y) * [(1/x) / (1/x)]
    dy/dx = (4y/x)-3 / (2-(y/x))

    let v = y/x
    y = v(x) x
    dy/dx = x dv/dx + v

    x dv/dx + v = 4v-3 / (2-v)
    x dv/dx = [4v-3 / (2-v)] - v
    x dv/dx = 4v-3 -[2v-v2] / (2-v)
    x dv/dx = 2v-3+v2] / (2-v)
    dx/x = [(-v + 2) / (v^2 + 2v -3)] dv
    dx/x = [(-5/4)(1/(v+3)) + (1/4)(1/(v-1))] dv (partial fractions)

    Integrating both sides:

    ln |x| = (1/4) ln |v-1| - (5/4) ln |v+3| + c
    ln |x| = (1/4) [ln |v-1| - 5 ln |v+3| + c]
    |x| = (1/4) exp {[ln |v-1| - 5 ln |v+3| + c]}
    |x| = c |v-1| |v+3|-5
    c|v+3|5 = |v-1|/|x|
    |(y/x)-1| / |x| = c|(y/x)+3|5

    Answer from book : |y - x| = c|y + 3x|5 (also y = -3x)

    Much appreciated.
     
    Last edited: May 26, 2009
  2. jcsd
  3. May 26, 2009 #2

    djeitnstine

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    Gold Member

    c|v+3|^5 = |v-1|/|x|

    What happened to your 1/4!! it disappeared

    rework it, you should have
    c|v+3|^-5 = |v-1|/|x|^4

    rearrange and substitute

    |x|^4 = |y/x-1|/c|y/x+3|^-5

    factor out a 1/x^-5 from the right

    |x|^4 = |y/x-1| / (c|y+3x|^-5 * 1/x^-5)

    Multiply both sides

    c 1/x = |y/x-1| / (|y+3x|^-5 )

    c 1/x*(|y+3x|^-5 ) = |y/x-1|

    facor out 1/x from the right and everything should be clear

    - srry it was just quick and dirty but bear with me =]

    Edit! O man sorry OP, that ^5 was ^-5, recheck your partial fractions, now it all works out
     
    Last edited: May 27, 2009
  4. May 26, 2009 #3
    Thanks! But just making sure I'm getting this.

    Code (Text):
    Multiply both sides

    cx = |y/x-1| / (|y+3x|^5 )

    cx*(|y+3x|^5 ) = |y/x-1|
    Should be

    Code (Text):
    Multiply both sides

    c(1/|x|) = |y/x-1| / (|y+3x|^5 )

    c*(|y+3x|^5 ) = |y/x-1|*|x|

    c*(|y+3x|^5 ) = |y-x|
     
    Right?

    Also, is there any easy way to remember that (y/x + 3)^5 = (y + 3x)^5 * (1/x)^5?
     
  5. May 26, 2009 #4

    djeitnstine

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    Gold Member

    Yes you're right (check my edit - my net is giving me probs) umm I don't know what to say, you just have to train yourself to spot it I guess.
     
  6. May 27, 2009 #5

    HallsofIvy

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    I don't think it is a matter of "remembering". Just add the fractions:
    y/x+ 3= y/x+ 3x/x= (y+ 3x)/x.
     
  7. May 30, 2009 #6
    I understand that, however, what I can't wrap my head around is doing the same thing if the term is being raised to the power of something. Namely, ((y/x)+3)N where N is integer > 1

    Let's say I had N=2, ((y/x)+3)2

    Expanding gives y2/x2 + 6(y/x) + 9

    y2/x2 + 6(y/x)(x/x) + 9(x2/x2)

    [y2 + 6(yx) + 9x2] (1/x2)

    (y+3x)2 (1/x)2

    .'. ((y/x)+3)2 = (y+3x)2 (1/x)2

    Now, since N=1 is true (intuitive), N=2 is true (proven), and you guys claim N=5 is also true, does that mean ((y/x)+3)N = (y+3x)N (1/x)N for 1..2..3..N by induction?

    If the above is true, am I overthinking this all? Is it mathematically correct to do this:

    ((y/x)+3)N
    ((y/x)+3(x/x))N
    ((y+3x)/x))N
    (y+3x)N (1/x)N
     
  8. May 30, 2009 #7

    Cyosis

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    Homework Helper

    You are definitely over thinking this. What you've done is correct. Remember [itex]a^n/b^n=(a/b)^n, a^nb^n=(ab)^n[/itex]. It is just basic arithmetic.
     
  9. May 30, 2009 #8

    djeitnstine

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    Gold Member

    I concur, definitely over thought. Simple algebra.
     
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