Results 1 to 9 of 9

Math Help - One last Diff Eq

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    19

    Post One last Diff Eq

    y' = -2(3y +4)

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

    y = -4/3 + Ce^-6x

    I tried and made it to 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Write as

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

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

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

    Integrate:

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

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

    \boxed{y=\frac{-4}{3}+Ce^{-6x}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2008
    Posts
    13
    \frac{dy}{-6y-8}=dx

    Integrating both sides yields

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

    Exponentiating:

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

    And finally:

    y=Ce^{-6x}-\frac 4 3 where C=-\frac{C_2}{6}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2008
    Posts
    19
    Quote Originally Posted by cyph1e View Post
    \frac{dy}{-6y-8}=dx

    Integrating both sides yields

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

    Exponentiating:

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

    And finally:

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

    Ahh that makes an little more sense Good Work!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,933
    Thanks
    336
    Awards
    1
    Quote Originally Posted by vodka View Post
    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:
    y' = -2(3y + 4)

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

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

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

    Now to integrate the left hand side.

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

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

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

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

    ln | 3y + 4 | = -6x + 3A

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

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

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

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

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

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

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2008
    Posts
    13
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    A bogaboo of mine is posters who thank one while shunning the other. It may be trifling, but that is how I see it.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,933
    Thanks
    336
    Awards
    1
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Diff Eq. x' = t * cos(t)
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: April 2nd 2010, 05:16 PM
  2. Diff eq IVP
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: December 10th 2009, 01:05 PM
  3. diff eqn, sin(2x).
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: March 12th 2009, 09:30 AM
  4. diff eq help
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 25th 2009, 01:16 PM
  5. diff eq help
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 31st 2008, 03:41 PM

Search Tags


/mathhelpforum @mathhelpforum