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Math Help - Separation of Variables Problem

  1. #1
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    Separation of Variables Problem

    Hi. I am trying to solve dy/dx= -6y - 8.

    I get:

    -1/6Ln (-6y-8)=x +c.
    Dividing by -1/6, you get Ln (-6y-8)= -6x-6c.

    I next take the e of both sides to get -6y-8= e^(-6x-6c).

    How do I proceed from here, as the correct answer is y= -4/3+ce^-6x?
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  2. #2
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    Quote Originally Posted by jaijay32 View Post
    Hi. I am trying to solve dy/dx= -6y - 8.

    I get:

    -1/6Ln (-6y-8)=x +c.
    Dividing by -1/6, you get Ln (-6y-8)= -6x-6c.

    I next take the e of both sides to get -6y-8= e^(-6x-6c).

    How do I proceed from here, as the correct answer is y= -4/3+ce^-6x?
    Try dividing both sides by 3y + 4.


    So you get

    \left(\frac{1}{3y + 4}\right)\frac{dy}{dx} = -2

    \int{\left(\frac{1}{3y + 4}\right)\frac{dy}{dx}\,dx} = \int{-2\,dx}

    \int{\frac{1}{3y + 4}\,dy} = -2x + C_1

    \frac{1}{3}\ln{|3y + 4|} + C_2 = -2x + C_1

    \frac{1}{3}\ln{|3y + 4|} = -2x + C, where C = C_1 - C_2

    \ln{|3y + 4|} = -6x + 3C

    |3y + 4| = e^{-6x + 3C}

    |3y + 4| = e^{3C}e^{-6x}

    3y + 4 = \pm e^{3C}e^{-6x}

    3y + 4 = Ae^{-6x} where A = \pm e^{3C}

    3y = Ae^{-6x} - 4

    y = \frac{A}{3}e^{-6x} - \frac{4}{3}.


    Now by letting \frac{A}{3} equal some other constant, the problem is solved.
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  3. #3
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    How are you seperating this equation? Did your instructor say you HAVE to use seperation of variables? The differential is linear:

    \frac{dy}{dx}+6y=8
    \mu = e^{\int 6dx}\Rightarrow \mu = e^{6x}
    (y*e^{6x})'=\int 8e^{6x}dx
    y*e^{6x}=\frac{4}{3} e^{6x}+C
    y=\frac{4}{3}+ Ce^{-6x}

    The thing about these differentials, is that more often than not there are several ways of solving it. Some are way more easier to solve than not.
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  4. #4
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    Actually, I would consider "separating variables" to be easier than finding an integrating factor!

    jaijay32, from -6y-8= e^(-6x-6c), you can separate the exponential as -6y- 8= e^(-6x)e^(-6c). Since c is an "unknown constant", e^(-6c) is also some unknown constant so just call it "C": -6y- 8= Ce^(-6x).

    (In fact, that is better: the integral of 1/x dx is ln|x|, not ln(x). What you should have got was ln|-6y-8|= ln|6y+8|= -6x- 6c so |6y+ 8|= e^(-6x-6c)= e^(-6x)e^(-6c). The point is that, since an exponential is always positive, "-6y- 8= e^(-6x)e^(-6c)" implies that -6y- 8 is positive. But what you really have is |6y+8|= e^(-6x)e^(-6c) so that 6y+8 can be either positive or negative. Writing "-6y- 8= Ce^(-6x)" or "6y- 8= Ce^(-6x)" includes C being either positive or negative.)

    Now, with -6y- 8= Ce^(-6x), -6y= 8+ Ce^(-6x) and then y= -4/3+ (C/-6)e^(-6x)= -4/3+ C'e^(-6x) with C'= C/-6.

    If you had written it 6y+ 8= Ce^(-6x), you would have 6y= -8+ Ce^(-6x) and then y= -4/3+ (C/6)e^(-6x) or y= -4/3+ C"e^(-6x), exactly the same answer with C"= C/6 now.
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