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Solve differential equation

  1. Nov 29, 2011 #1

    sharks

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    The problem statement, all variables and given/known data
    Solve differential equation
    4xdy - ydx = (x^2)dy

    The attempt at a solution

    (4x - x^2)dy = ydx

    (1/y)dy = (1/(4x - x^2))dx

    Integrating both sides:

    For integrating 1/(4x - x^2) i completed the square for the denominator part and got 4-(x-2)^2 then used substitution; let (x-2) = 2sint

    lny = sin^-1((x-2)/2) + lnA

    But the answer wrong for some reason unknown to me.
     
    Last edited: Nov 29, 2011
  2. jcsd
  3. Nov 29, 2011 #2

    I like Serena

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    Hi sharks! :smile:

    I'm not sure how you integrated, but you should have integrated dt/2cost.
    But this is not so easy, and it is not what you wrote.

    Instead I recommend integrating by using partial fraction decomposition.
    See for instance: http://en.wikipedia.org/wiki/Partial_fraction#Illustration
     
  4. Nov 29, 2011 #3

    sharks

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    Hi "I like Serena"! :approve:

    If you would allow me to expand a little bit from my first trial above:

    Using substitution; let (x-2) = 2sint
    Indeed, i do get 1/2cost but then i also need to change the derivative from dx to dt
    I do that by differentiating x-2 = 2sint
    which gives me dx = 2cost.dt
    and i replace in the integral to give me:

    (1/4cos^2t).2cost.dt and this gives me (you are correct!) an integral of (1/2cost).dt = (1/2).sect.dt which i then integrate to (1/2).ln(pi/4 + t/2)
    and then i replace from the substitution.

    2sint = x-2
    sint = (x-2)/2
    t = arcsin ((x-2)/2)

    Wow.... OK, i see why you said it's not so easy. But is it even possible at all? As i don't see how the arcsin would go away.


    Using your proposed method, since i cannot factorize x^2 - 4x
    I've done this: x(x - 4)

    So, the partial fraction is...

    1/(4x) + 1/(4(4-x))

    ... and behold, i got the answer! y = A(x/(4-x))^(1/4)
     
    Last edited: Nov 29, 2011
  5. Nov 29, 2011 #4

    ehild

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    The first factor should be 1/(2cost)2.

    ehild
     
  6. Nov 29, 2011 #5

    hunt_mat

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    Note that [itex]4x-x^{2}=2^{2}-(x-2)^{2}[/itex] I think that this will help you a great deal.
     
  7. Nov 29, 2011 #6

    sharks

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    OK, i went over it again, found the mistake as you pointed out and i edited my previous post, as i got stuck with my alternate method. It really appears to be quite complicated if i use a trig substitution, but i wonder how/if it'll work out.
     
  8. Nov 29, 2011 #7

    I like Serena

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    Oh well, it says here what the integral of sec t is:
    http://en.wikipedia.org/wiki/Lists_of_integrals#Trigonometric_functions


    You can get rid of the arcsin, by using the definition of sine, cosine and tanget.

    arcsin(x) means you have a triangle with hypotenuse 1, and opposing side x.
    So for instance cos(arcsin(x)) is the adjacent/hypotenuse = √(1-x2)/1.


    Btw, you could also have integrated your fraction with the completed square, since it's the derivative of (1/4)artanh((x-2)/2).
     
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