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Math Help - initial-value problem and integrating..... my brain just isn't grasping help

  1. #1
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    initial-value problem and integrating..... my brain just isn't grasping help

    Hi, i have this initial-value problem....i need to use the reverse quotient rule, but i also need to use an integration formula too:

    dy/dx= (cos(5x))/(7+sin(5x)) when y=6 when x=0

    f'(x)/f(x)dx=in(f(x))+c (f(x)>0)


    i don't want the answer i would just like to know how??? if that makes sense lol. I cannot find an example that is similar to this so i can work through it as i find this is the best way of learning....can anyone help pretty please??
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  2. #2
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    Re: initial-value problem and integrating..... my brain just isn't grasping help

    Use a direct substitution by let u(x) = 7 + sin(5x). Then du = 5cos(5x) dx. Then (1/5) du = cos(5x) dx. Then make the replacements.

    dy/dx = (cos (5x)) / (7 + sin (5x))

    y = INT {cos (5x) / (7 + sin (5x))} dx * I don't know how to make the integral sign on here *
    y = INT {(1/5) 1/u} du
    y = (1/5) ln |u| + C
    y = (1/5) ln |7 + sin(5x)| + C

    To find C, let y = 6 and x = 0.

    6 = (1/5) ln |7 + sin(5*0)| + C
    6 = (1/5) ln |7 + 0| + C
    6 = (1/5) ln 7 + C
    6 - 1/5 ln 7 = C

    Then y = (1/5) ln |7 + sin(5x)| + 6 - 1/5 ln 7

    y = (1/5) ln |(7 + sin(5x))/7| + 6
    y = (1/5) ln |1 + (1/7)sin(5x)| + 6

    As for the second question, simply let y = f(x) and so then dy/dx = f ' (x). Then you see that INT {f'(x) / f(x)} dx = INT {1/y} dy = ln |y| + C = ln |f(x)| + C. Since f(x) > 0 we see that |f(x)| = f(x). Thus, we see the answer is ln (f(x)) + C as needed.
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