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Differential Equations Help

  1. Feb 19, 2014 #1
    Hello

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

    Find the general solution of the following diff erential equations. In each case if
    y = 2 when x = 1, fi nd y when x = 3.

    [itex]x \frac{dy}{dx} = \frac{1}{y} + y [/itex]

    2. Relevant equations



    3. The attempt at a solution

    [itex]x \frac{dy}{dx} = \frac{1}{y} + y [/itex]

    [itex]x . \frac{1}{dx} =( \frac{1}{y} + y) \frac{1}{dy}[/itex]

    [itex] \frac{1}{x} dx = y + \frac{1}{y} dy[/itex]

    [itex]\int \frac{1}{x} dx = \int y + \frac{1}{y} dy[/itex]

    [itex] ln|x| + c_{1} = \frac{y^{2}}{2} + ln|y| + c_{2}[/itex]

    [itex] ln|x| - \frac{y^{2}}{2} - ln|y| = c_{2} - c_{1}[/itex]

    [itex] ln|x| - \frac{y^{2}}{2} - ln|y| = c[/itex]

    If y = 2 and x = 1, c = -2

    (assuming I have found the general solution correctly...)

    But I don't see how I can find y, when x = 3 because there are two terms involving y.

    I think the most likely explanation is that i've done something wrong, so i'd really appreciate some help!

    thanks,

    BOAS :)
     
  2. jcsd
  3. Feb 19, 2014 #2

    vela

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    $$\left(y+\frac1y\right)^{-1} \ne \frac{1}{y} + y$$
     
  4. Feb 19, 2014 #3

    Ray Vickson

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    Always substitute your alleged 'solution' into the original DE, to see if it works. My guess is that you have not done that yet.
     
  5. Feb 19, 2014 #4
    I've only just started learning DE's and this may sound silly, but how do I substitute it in?
     
  6. Feb 19, 2014 #5
    [itex]( \frac{1}{y} + y)^{-1} = \frac{1}{\frac{1}{y} + y} [/itex]

    I'll see if I can solve it now
     
  7. Feb 19, 2014 #6

    vela

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    No, it wouldn't. You can't distribute the exponent over the sum.

    Just try it with y=1. On the lefthand side, you end up with 1/2; on the righthand side you end up with 2.
     
  8. Feb 19, 2014 #7

    Ray Vickson

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    In the obvious way: try it and see. Your solution says that x and y are related by
    [tex] ln|x| - \frac{y^{2}}{2} - ln|y| = c, [/tex]
    while the DE says they are related by
    [tex] x dy = \left( y + \frac{1}{y} \right) dx .[/tex]
    Do these two statements agree? Can you take the first form and show that it satisfies the second?
     
  9. Feb 19, 2014 #8
    I think I have identified and fixed the problem, or the second problem that was stopping me from making sense of your help :)

    [itex]x \frac{dy}{dx} = \frac{1}{y} + y [/itex]

    [itex]x . \frac{1}{dx} =( \frac{1}{y} + y) \frac{1}{dy}[/itex]

    [itex] \frac{1}{x} dx = ( \frac{1}{y} + y)^{-1} dy[/itex]

    [itex]\int \frac{1}{x} dx = \int ( \frac{1}{y} + y)^{-1} dy[/itex]

    [itex] ln|x| + c_{1} = \frac{ln|y^{2} + 1|}{2} + c_{2}[/itex]

    [itex] ln|x| - \frac{ln|y^{2} + 1|}{2} = c_{2} - c_{1}[/itex]

    [itex] ln|x| - \frac{ln|y^{2} + 1|}{2}| = c[/itex]

    (whilst trying to implement your help, I was trying to differentiate that section when I needed to be integrating it)
     
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