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Math Help - trigonometric equation

  1. #1
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    trigonometric equation

    A)
    y'siny=cosx(2cosy-sin^{2}x)
    B)
    x^{2}y'-1=cos2y
    how to solve these
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    B)
    x^{2}y'-1=cos2y
    Does this separate?

    x^{2}y'-1=\cos 2y

    x^{2}y'=\cos 2y+1

    x^{2}\frac{dy}{dx}=\cos 2y+1

    x^{2}\frac{dy}{\cos 2y+1}=dx

    \frac{dy}{\cos 2y+1}=\frac{dx}{x^{2}}

    It does! what next?
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  3. #3
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    what is the integral of 1/(cos2y+1)
    ?

    regarding A)
    i put v=cosy

    -v'=cosx(2v-sin^2x)
    Last edited by transgalactic; November 13th 2010 at 06:57 AM.
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  4. #4
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    how to solve A
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  5. #5
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    For (A), the substitution u=\cos(y) renders the equation linear.
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  6. #6
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    from this substitution i got
    -u'=2u^2 -u(sin x)^2
    by rearanging i got
    u'/u +2u=(sin x)^2

    what to do now?
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  7. #7
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    You did the substitution incorrectly. u = sin(y). sin(x) remains as is.
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  8. #8
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    i got
    u'+2ucosx=-(sin x)^2

    (ue^{2sinx})'=(sin x)^2e^{2sinx}

    i dont know how to solve the right side integral
    Last edited by transgalactic; November 17th 2010 at 08:13 AM.
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  9. #9
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    Original DE:

    y'\sin(y)=\cos(x)(2\cos(y)-\sin^{2}(x)).

    Substitution: u=\cos(y). Then -u'=\sin(y)y'.

    Plugging into the DE yields

    -u'=\cos(x)(2u-\sin^{2}(x))\quad\to\quad -u'-2u\cos(x)=-\cos(x)\sin^{2}(x)

    \to\quad u'+2u\cos(x)=\cos(x)\sin^{2}(x).

    I think you forgot to distribute the cosine everywhere. So, propagate these changes: what do you get?
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  10. #10
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    on the left side i get cos x (sin x)^2 e^{2sinx}

    i dont know how to solve this integral
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  11. #11
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    Substitute v = sin(x). Then you'll need by parts twice.
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  12. #12
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    Quote Originally Posted by transgalactic View Post
    what is the integral of 1/(cos2y+1)?
    \frac{1}{\cos{2y}+1} = \frac{1}{2}\sec^2{y} which is the derivative of \frac{1}{2}\tan{y}.
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  13. #13
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    thanks
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  14. #14
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    So, did you finish?
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  15. #15
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    yes


    i solved just like you said it went perfect
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