Results 1 to 5 of 5

Math Help - separation of variables

  1. #1
    Junior Member
    Joined
    Aug 2008
    Posts
    38

    separation of variables

    Hello!

    In this problem, find the equation of the graph which passes through the given point and has the indicated slope:

    m = y' = dy/dx = (2y)/(3x)

    (1,1)

    So, after cross multiplying to separate, I have:

    (1/2y) (dy) = (1/3x) (dx)

    After integrating both sides:

    Ln(y)/2 = Ln(x)/3

    Simplifying:

    Ln(y) = (2Ln(x))/3

    Is this the right track? I can't remember how to get rid of the Ln on the left side to solve for y. How do I use the point (1,1) ?

    Thank you so much!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by 2clients View Post
    Hello!

    In this problem, find the equation of the graph which passes through the given point and has the indicated slope:

    m = y' = dy/dx = (2y)/(3x)

    (1,1)

    So, after cross multiplying to separate, I have:

    (1/2y) (dy) = (1/3x) (dx)

    After integrating both sides:

    Ln(y)/2 = Ln(x)/3

    Simplifying:

    Ln(y) = (2Ln(x))/3

    Is this the right track? I can't remember how to get rid of the Ln on the left side to solve for y. How do I use the point (1,1) ?

    Thank you so much!!
    Yes, you're on the right track, but when you integrate both sides, don't forget to throw the constant C on the side with the x's. So you should end up with

    \ln y=\tfrac{2}{3}\ln x+C\implies y=e^{\frac{2}{3}\ln x+C}\implies y=e^{\ln\left(x^{\frac{2}{3}}\right)}e^C\implies y=Cx^{\frac{2}{3}}

    Now use the point as the initial condition: (1,1)\implies y(1)=1

    Thus, 1=C(1)^{\frac{2}{3}}\implies C=1

    So the final answer would be \color{red}\boxed{y=x^{\frac{2}{3}}}

    Does this make sense?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2008
    Posts
    38
    Thank you, Chris.

    It does make sense, mostly - I just can't reason why the C only goes on the one side?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by 2clients View Post
    Thank you, Chris.

    It does make sense, mostly - I just can't reason why the C only goes on the one side?
    It really appears on both sides after integration:

    \tfrac{1}{2}\ln y+C_1=\tfrac{1}{3}\ln x +C_2

    Solving for y, you end up with y=e^{\frac{2}{3}\ln x+{\color{blue}C_2-C_1}}

    But C_2-C_1 is just another constant; let's call it C

    So, y=e^{\frac{2}{3}\ln x+{\color{blue}C}}=\dots

    Does this clarify things?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2008
    Posts
    38
    yes perfectly, thanks again!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Separation of Variables
    Posted in the Differential Equations Forum
    Replies: 7
    Last Post: January 5th 2011, 04:12 PM
  2. Separation of Variables
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: June 6th 2010, 02:51 PM
  3. DE by Separation of Variables
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 9th 2008, 11:00 PM
  4. Separation of variables...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 20th 2008, 05:46 PM
  5. Separation of variables
    Posted in the Calculus Forum
    Replies: 6
    Last Post: January 12th 2008, 07:09 PM

Search Tags


/mathhelpforum @mathhelpforum