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Homework Help: A diff. eq. or trigonometry related problem?

  1. Mar 28, 2006 #1
    y' = 2y/x + x cos (y/x^2), with y=z*x^2
    => y' = 2zx + x cos z

    and y=z*x^2 => y' = 2zx + x^2 * dz/dx

    So that leaves x^2 * dz/dx = x cos z => dz/cos z = dx/x

    I integrate both sides so that:

    sec z + tan z = x + C

    But I don't have a clue on how to get past that point. Should I start from the beginning with another technique to avoid getting sec z + tan z?
  2. jcsd
  3. Mar 28, 2006 #2


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    What exactly is the problem? What are you trying to do?
  4. Mar 29, 2006 #3
    I'm trying to solve y, so I have to solve z in terms of x first and then substitute it in the equation y=z*x^2.
  5. Mar 29, 2006 #4


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    You skipped over ln parts and have a minor error: the right side should be, after taking exponentials of both sides, Cx, not x+ C.

    Also, rather than using the "Table" formula
    [tex]\int sec z dz= ln|sec z+ tan z|+ C[/itex]
    I did it as
    [tex]\int \frac{dz}{cos z}= \int \frac{cos z dz}{cos^2 z}= \int \frac{cos z dz}{1- sin^2 z}[/tex]
    Now the substitution u= sin z gives a rational integral that can be integrated by partial fractions. The result is, of course, similar to the above but in terms of sin z only and so easier to solve for z.
  6. Mar 29, 2006 #5
    Ok, I think I can carry on from that.

    Thank you!
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