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Math Help - Initial value problem

  1. #1
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    Initial value problem

    hi

    got the problem solve the initial value problem

    dy/dx=(sin(5x)+6x)/(4-cos(5x)+15x^2) when y=7 and x=0

    thought i would try and use formula integration f'(x)/f(x) dx= ln(f(x))+c

    i got (-1/5cos(5x)+3x^2)/(4-cos(5x)+15x^2)

    = -5ln(4-cos(5x)+15x^2)

    i am in real mess please help
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  2. #2
    MHF Contributor alexmahone's Avatar
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    \frac{dy}{dx}=\frac{(sin(5x)+6x)}{(4-cos(5x)+15x^2)}

    \frac{dy}{dx}=\frac{1}{5} \frac{5(sin(5x)+6x)}{(4-cos(5x)+15x^2)}

    y=\frac{1}{5}\int \frac{5(sin(5x)+6x)}{(4-cos(5x)+15x^2)}

    Using the formula for integration: \int \frac{f'(x)}{f(x)} dx= ln(f(x))+C, we get

    y=\frac{1}{5} ln (4-cos(5x)+15x^2)+C

    At x=0,\ y=7

    7=\frac{1}{5} ln (4-1)+C

    7=\frac{1}{5} ln 3+C

    C=7-\frac{1}{5}ln 3

    y=\frac{1}{5} ln (4-cos(5x)+15x^2)+7-\frac{1}{5}ln 3

    y=\frac{1}{5} ln (\frac{4-cos (5x)+15x^2}{3})+7
    Last edited by alexmahone; July 5th 2009 at 04:30 AM.
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  3. #3
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    many thanks for the clear presentation of your working.

    thanks
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  4. #4
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    But... where does the rest of the equation "go"?

    Sorry, this is probably a silly question, but where does the 1/5 & the 5 come from (second image - 1/5 ((5 sin (5x)+6x)/(4-cos(5x)+15x^2))?

    And why does (5 sin(5x)+6x) then disappear? Is something cancelling it out, or am I just not understanding the formula completely?

    It's so confusing!!

    Many thanks for your help,

    S
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  5. #5
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    You need the 1/5 because the derivative of 4-cos(5x)+15x^2 is 5sin(5x)+30x which is 5 times more than the what you are given so you multiply what you are given by 5 and multiply the whole equation by 1/5. Then using the constant multiple rule you can take the 1/5 outside when you integrate.
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