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Math Help - Some help required..

  1. #1
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    Some help required..

    Hi everyone

    I have this particular problem which I would appreciate someone would help me solve:

    Find the solution of the initial-value problem:

    dy/dx = cos(4x)
    ------------- , y=6 when x=0
    3 - sin(4x)


    Here are my workings:

    y = ∫ cos (4x) / 3 - sin (4x) dx = ∫ 1/4sin(4x)/3x+1/4cos(4x) dx

    y = sin (4x) (3x + cos (4x)) +c since 1/4 cancel each other out
    6 = sin (0) (0+cos(0)) + c
    6 = 0 + 6
    c = 6

    I am not sure whether the solution arrived at is correct. Would appreciate some help on this.

    Many thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    New York, USA
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    Quote Originally Posted by tigerdivision View Post
    Hi everyone

    I have this particular problem which I would appreciate someone would help me solve:

    Find the solution of the initial-value problem:

    dy/dx = cos(4x)
    ------------- , y=6 when x=0
    3 - sin(4x)


    Here are my workings:

    y = ∫ cos (4x) / 3 - sin (4x) dx = ∫ 1/4sin(4x)/3x+1/4cos(4x) dx

    y = sin (4x) (3x + cos (4x)) +c since 1/4 cancel each other out
    6 = sin (0) (0+cos(0)) + c
    6 = 0 + 6
    c = 6

    I am not sure whether the solution arrived at is correct. Would appreciate some help on this.

    Many thanks
    Ok, so i dont really get what you were doing, but here's the way i'd do this.

    we have dy/dx = cos(4x)/(3 - sin(4x))

    => dy = cos(4x)/(3 - sin(4x)) dx
    => y = int{cos(4x)/(3 - sin(4x))}dx
    we proceed by substitution:
    Let u = 3 - sin(4x)
    => du = -4cos(4x) dx
    => (-1/4) du = cos(4x) dx
    so our integral becomes:

    y = (-1/4)*int{1/u}du
    = (-1/4)lnu + C
    = (-1/4)ln(3 - sin(4x)) + C

    when x = 0, y = 6
    => 6 = (-1/4)ln(3) + C
    so C = 6 + (ln3)/4

    so y = (-1/4)ln(3 - sin(4x)) + 6 + (ln3)/4 is our solution
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tigerdivision View Post
    Hi everyone

    I have this particular problem which I would appreciate someone would help me solve:

    Find the solution of the initial-value problem:

    dy/dx = cos(4x)
    ------------- , y=6 when x=0
    3 - sin(4x)


    Here are my workings:

    y = ∫ cos (4x) / 3 - sin (4x) dx = ∫ 1/4sin(4x)/3x+1/4cos(4x) dx

    y = sin (4x) (3x + cos (4x)) +c since 1/4 cancel each other out
    6 = sin (0) (0+cos(0)) + c
    6 = 0 + 6
    c = 6

    I am not sure whether the solution arrived at is correct. Would appreciate some help on this.

    Many thanks
    oh, i think i see what you were doing now. You thought the integral of a rational function was just the integral of the top divided by the integral of the bottom. I'm afraid that's not so. The above method is the simplest way i see to do this problem
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