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Thread: One last Diff Eq

  1. #1
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    Post One last Diff Eq

    $\displaystyle y' = -2(3y +4)$

    It is separable im lost in how to get the answer,

    $\displaystyle y = -4/3 + Ce^-6x$

    I tried and made it to $\displaystyle dy = (-6y - 8)dx$ but im unsure as how to get my dy/y for 6lny sigh now im really getting confused. Anyone feel free to shed some light on the case in point.
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  2. #2
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    Write as

    $\displaystyle \frac{dy}{dx}+6y=-8$

    The integrating factor is $\displaystyle e^{\int 6dx}=e^{6x}$

    $\displaystyle \frac{d}{dx}[e^{6x}y]=-8e^{6x}$

    Integrate:

    $\displaystyle \int\frac{d}{dx}[e^{6x}y]=-8\int e^{6x}dx$

    $\displaystyle e^{6x}y=\frac{-4}{3}e^{6x}+C$

    $\displaystyle \boxed{y=\frac{-4}{3}+Ce^{-6x}}$
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  3. #3
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    $\displaystyle \frac{dy}{-6y-8}=dx$

    Integrating both sides yields

    $\displaystyle -\frac 1 6 \ln(-6y-8)=x+C_1$

    Exponentiating:

    $\displaystyle -6y-8=C_2e^{-6x}$ where $\displaystyle C_2=e^{-6C_1}$

    And finally:

    $\displaystyle y=Ce^{-6x}-\frac 4 3$ where $\displaystyle C=-\frac{C_2}{6}$
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  4. #4
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    Quote Originally Posted by cyph1e View Post
    $\displaystyle \frac{dy}{-6y-8}=dx$

    Integrating both sides yields

    $\displaystyle -\frac 1 6 \ln(-6y-8)=x+C_1$

    Exponentiating:

    $\displaystyle -6y-8=C_2e^{-6x}$ where $\displaystyle C_2=e^{-6C_1}$

    And finally:

    $\displaystyle y=Ce^{-6x}-\frac 4 3$ where $\displaystyle C=-\frac{C_2}{6}$

    Ahh that makes an little more sense Good Work!
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by vodka View Post
    $\displaystyle y' = -2(3y +4)$
    galactus' (and cyph1e's) solution is much simpler and much shorter, but I think separation of variables in this case is quite instructive:
    $\displaystyle y' = -2(3y + 4)$

    $\displaystyle dy = -2(3y + 4)~dx$

    $\displaystyle \frac{dy}{3y + 4} = -2~dx$
    (or you could keep the -2 with the y's. It makes no difference.)

    $\displaystyle \int \frac{dy}{3y + 4} = -2 \int dx$

    Now to integrate the left hand side.

    Let $\displaystyle u = 3y + 4 \implies du = 3~dy$

    So
    $\displaystyle \int \frac{dy}{3y + 4} = \int \frac{du}{3u} = \frac{1}{3}ln|u| = \frac{1}{3}ln | 3y + 4 |$

    So
    $\displaystyle \int \frac{dy}{3y + 4} = -2 \int dx$

    $\displaystyle \frac{1}{3}ln | 3y + 4 | = -2x + A$
    (A is my undetermined constant.)

    $\displaystyle ln | 3y + 4 | = -6x + 3A$

    $\displaystyle | 3y + 4 | = e^{-6x + 3A}$

    $\displaystyle | 3y + 4 | = e^{3A}e^{-6x}$

    Now $\displaystyle e^{3A}$ is just a (positive) undetermined constant. Let's call this B.

    $\displaystyle | 3y + 4 | = Be^{-6x}$

    The way to deal with the absolute value bars is to split this into two equations:
    $\displaystyle 3y + 4 = Be^{-6x}$
    and
    $\displaystyle 3y + 4 = -Be^{-6x}$

    Since B is an undetermined (positive) constant anyway, let's say that we can have B be any constant (positive or negative) and just write this as one equation:
    $\displaystyle 3y + 4 = Ce^{-6x}$

    $\displaystyle y = -\frac{4}{3} + Ce^{-6x}$

    -Dan
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  6. #6
    Eater of Worlds
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    but I think separation of variables in this case is quite instructive
    That is what I thought. That is why I used the integrating factor instead of SOV. Apparently vodka didn't appreciate it, so I deleted it. Last time I answer a question by that poster.
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  7. #7
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    Quote Originally Posted by galactus View Post
    That is what I thought. That is why I used the integrating factor instead of SOV. Apparently vodka didn't appreciate it, so I deleted it. Last time I answer a question by that poster.
    Even if he didn't appreciate it, please don't delete your post. It was a nice solution and other people viewing this thread might learn from it.
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  8. #8
    Eater of Worlds
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    A bogaboo of mine is posters who thank one while shunning the other. It may be trifling, but that is how I see it.
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  9. #9
    Forum Admin topsquark's Avatar
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    I have used my phenomenal cosmic powers to restore galactus' post. My reason is that other users can benefit from seeing the solution if they look up this thread. As I think galactus' solution was not only perfectly acceptable, but efficient and clearly explained I feel it should be included in this thread. (No offense to cyph1e's solution which I think is just as good.)

    -Dan
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