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Math Help - Can't figure this one out

  1. #1
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    Can't figure this one out

    I have no clue where to start with this any advice would be really appreciated. I tried using u-substitution but that did not work.

    Solve the initial value problem

    dy/dx = 1-sin(5x)/cos(5x)+5x+6 (x>-1), y=3 when x=0
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  2. #2
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    Re: Can't figure this one out

    You need to use paretheses. Is this what you mean?

    $\dfrac{dy}{dx} = 1-\dfrac{\sin(5x)}{\cos(5x)}+5x+6$

    or did you mean:

    $\dfrac{dy}{dx} = \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}$

    Assuming it is the latter:

    Let $u = \cos(5x)+5x+6$. Then $du = \left(5-5\sin(5x)\right)dx = 5\left(1-\sin(5x)\right)dx$

    This gives:

    $\displaystyle y = \int \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}dx = \dfrac{1}{5}\int u^{-1}du = \dfrac{1}{5}\ln\left|\cos(5x)+5x+6\right| + C$

    Plugging in the initial values:

    $3 = \dfrac{1}{5}\ln(7)+C$ shows $C = 3-\dfrac{1}{5}\ln(7)$

    This gives $y = \dfrac{1}{5}\ln \left| \dfrac{\cos(5x)+5x+6}{7} \right|+3$
    Last edited by SlipEternal; April 7th 2014 at 08:17 PM.
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  3. #3
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    Re: Can't figure this one out

    Hey thanks for the reply but I can't read your text!
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  4. #4
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    Re: Can't figure this one out

    Quote Originally Posted by alexpasty2013 View Post
    Hey thanks for the reply but I can't read your text!
    This forum makes use of LaTeX. If it does not display correctly, try reloading the page.
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  5. #5
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    Re: Can't figure this one out

    Here it is reformatted. Let me know if this is better.

    Quote Originally Posted by SlipEternal View Post
    You need to use paretheses. Is this what you mean?

    \dfrac{dy}{dx} = 1-\dfrac{\sin(5x)}{\cos(5x)}+5x+6

    or did you mean:

    \dfrac{dy}{dx} = \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}

    Assuming it is the latter:

    Let u = \cos(5x)+5x+6. Then du = \left(5-5\sin(5x)\right)dx = 5\left(1-\sin(5x)\right)dx

    This gives:

    \displaystyle y = \int \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}dx = \dfrac{1}{5}\int u^{-1}du = \dfrac{1}{5}\ln\left|\cos(5x)+5x+6\right| + C

    Plugging in the initial values:

    3 = \dfrac{1}{5}\ln(7)+C$ shows $C = 3-\dfrac{1}{5}\ln(7)

    This gives y = \dfrac{1}{5}\ln \left| \dfrac{\cos(5x)+5x+6}{7} \right|+3
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