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Math Help - First order differential equations

  1. #1
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    First order differential equations

    Obtain the solution that satisfies the given condition of the differential equation.

    (1 + cos^2 x) dy/dx = y(y + 1) sin 2x, given that y = 2 at x = 0


    Please help me to solve that.
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    Quote Originally Posted by geton View Post
    Obtain the solution that satisfies the given condition of the differential equation.

    (1 + cos^2 x) dy/dx = y(y + 1) sin 2x, given that y = 2 at x = 0


    Please help me to solve that.
    Separate the variables.
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  3. #3
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    But how can I solve this part?

    ∫ 1/(y)(y+1) dy
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  4. #4
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    Quote Originally Posted by geton View Post
    But how can I solve this part?

    ∫ 1/(y)(y+1) dy

    Partial fractions.
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  5. #5
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    Quote Originally Posted by geton View Post
    (1 + cos^2 x) dy/dx = y(y + 1) sin 2x, given that y = 2 at x = 0
    \left( {1 + \cos ^2 x} \right)\,\frac{{dy}}<br />
{{dx}} = y(y + 1)\sin 2x \iff \frac{1}<br />
{{y(y + 1)}}\,dy = \frac{{\sin 2x}}<br />
{{1 + \cos ^2 x}}\,dx.

    Now integrate both sides. (Don't forget integration constant.)

    RHS integral is solved here, second one can be easily solved by noting that (y+1)-y=1, form there you just split the original ratio into two fractions.
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  6. #6
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    I tried on that way, please check it right or wrong.

    After integrating,

    1/y – 1/(y+1) = - ln |1 + cos^2 x| + C

    Since y =2 at x = 0 this gives
    1/2 - 1/3 = - ln 2 + C
    C = 1/6 + ln 2

    So the solution is,
    1/y – 1/(y+1) = - ln |1 + cos^2 x| + 1/6 + ln 2
    1/y – 1/(y+1) = ln |2/(1 + cos^2 x)| + 1/6


    I did another solution, please check it too.

    Can we consider this?
    ∫ (sin 2x)/(1+cos^2 x)
    = ∫2(sin 2x)/(cos 2x + 3)


    But still I can’t find my book’s answer, which is (y+1)/y = (1+cos^2 x)
    Last edited by geton; January 25th 2008 at 07:15 AM.
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  7. #7
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    Quote Originally Posted by geton View Post
    I tried on that way, please check it right or wrong.

    After integrating,

    1/y – 1/(y+1) = - ln |1 + cos^2 x| + C
    You made a big mistake here

    You only integrated the RHS, what about the LHS? It's not integrated

    Try again.

    (See my signature.)
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  8. #8
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    Quote Originally Posted by Krizalid View Post
    You made a big mistake here

    You only integrated the RHS, what about the LHS? It's not integrated

    Try again.

    (See my signature.)
    Yes that was great mistake.

    But I also told about
    ∫ (sin 2x)/(1+cos^2 x)
    = ∫ 2(sin 2x)/(cos 2x + 3)

    Is it wrong way?
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