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Initial value problem

  1. May 19, 2004 #1

    phy

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    hi everyone. i need help solving this question

    ydx - (ytan(x/y) +x)dy = 0, where y(1) = pi/4

    i know how to do the question but my problem is just i dont know how to get all the x's on one side and the y's on the other. any help would be appreciated. thanks a lot.
     
  2. jcsd
  3. May 19, 2004 #2

    AKG

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    [tex]ydx\ -\ (y\tan\left(\frac{x}{y}\right)\ +\ x)dy\ =\ 0[/tex]
    [tex]ydx\ =\ (y\tan\left(\frac{x}{y}\right)\ +\ x)dy[/tex]
    [tex]\frac{dx}{dy}\ =\ \tan\left(\frac{x}{y}\right)\ +\ \frac{x}{y}[/tex]
    [tex]let\ a\ =\ \frac{x}{y}[/tex]
    [tex]\frac{d(ya)}{dy}\ =\ \tan(a)\ +\ a[/tex]
    [tex]a\ +\ y\frac{da}{dy}\ =\ \tan(a)\ +\ a[/tex]
    [tex]y\frac{da}{dy}\ =\ \tan(a)[/tex]

    From here you should know what to do. I probably gave away the real trick to the problem, which was the proper rearrangment of the equation. :frown:
     
  4. May 19, 2004 #3

    phy

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    ok thanks a lot. i didn't know i was supposed to make the substitution a=x/y. but how do you know when to use it?
     
  5. May 19, 2004 #4

    AKG

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    :wink: It's not a general rule or anything, it was just a helpful substitution. I think 40% of the problem was the getting the 2nd and 3rd lines, 50% of the problem was lines 4-7, and 10% was the rest. As for knowing when to make such substitutions, I didn't "know" that I was supposed to make it either. Sometimes you just see it. However, if you practice enough, the chances that you'll "just see it" tend to increase, for some reason. :wink:

    Oh, and I suppose I should mention that this only holds for [tex]y\ \neq\ 0[/tex] because you divide by "y" at some point during the first three steps. However, I think it's simple enough to see from the original equation that y can never be zero anyways.
     
  6. May 19, 2004 #5
    Mostly when it is a homogeneous in x and y
     
  7. May 20, 2004 #6

    phy

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    ok thanks guys :)
     
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