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Need help solving Differential Equation

  1. Jan 15, 2017 #1
    Sin(3x)dx+2ycos(3x)dy=0

    So far, I have ∫2ydy=-∫sin(3x)/cos^2(3x)dx. Is that right? If so, how do I integrate sin(3x)/cos^2(3x)?
     
  2. jcsd
  3. Jan 15, 2017 #2

    fresh_42

    Staff: Mentor

    Is this homework? And do you mean ##\sin(3x) + 2\cos(3x) \cdot y(x) \cdot y'(x) = 0\;##?
     
  4. Jan 15, 2017 #3
    These are practice questions for elementary differential equations. Oh...I just realized that there was a typo.

    It's: Sin(3x)dx+2yCos^2(3x)dy=0
     
  5. Jan 15, 2017 #4
    Again,

    So far, I have ∫2ydy=-∫sin(3x)/cos^2(3x)dx. Is that right? If so, how do I integrate -∫sin(3x)/cos^2(3x)?
     
  6. Jan 15, 2017 #5

    fresh_42

    Staff: Mentor

    O.k. but please use our homework section for this kind of questions in the future. It automatically inserts a template that you should use.
    E.g. it includes a paragraph for your own efforts, which shortens the process a lot. (Normally we don't want to give away ready made solutions to students but try to teach them instead.)

    Your typo makes the entire equation more difficult, for otherwise it would have been simply the tangent.
    You can look up many integrals here: https://de.wikibooks.org/wiki/Forme...timmte_Integrale_trigonometrischer_Funktionen
    It's the wrong language but there is little beside a list of formulas. You won't need the language.
    ##\int \frac{\sin (3x)}{\cos^2 (3x)} = \frac{1}{3 \cos(3x)}## but I haven't checked whether it's correct.
     
  7. Jan 15, 2017 #6
    Thank you for the link. It translates automatically into any language so it's ok. I wanted to know the whole steps on how to solve the integral: sin(3x)/cos^2(3x) (I found the template, but don't know how to use it yet. I don't go on that much on these forums for help. It's my first time using this to ask for help)
     
  8. Jan 15, 2017 #7

    fresh_42

    Staff: Mentor

    No problem. Knowing the answer one could simply differentiate. The other way might be using one of the forms ##\frac{\sin x}{\cos^2 x}=\frac{\tan x}{\cos x} ## and integrate by parts: ##\int u'v = uv - \int uv'##.

    And thanks for the language test. I didn't know this, but surely good to know as - you might have recognized it - there are a lot more formulas for all kind of integrals.
     
  9. Jan 15, 2017 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
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    Look up Weierstrass substitution.
     
  10. Jan 15, 2017 #9
    Thanks, I know how to do it now. :smile:
     
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