Results 1 to 10 of 10
Like Tree4Thanks
  • 1 Post By HallsofIvy
  • 1 Post By romsek
  • 1 Post By romsek
  • 1 Post By romsek

Math Help - Having problems understanding differential equations

  1. #1
    Newbie
    Joined
    Feb 2013
    From
    Norfolk
    Posts
    22

    Having problems understanding differential equations

    Hi there ,

    I'm looking for a worked example of the below question. Or for someone to walk me through what to do and what is going on please.

    Having problems understanding differential equations-untitled.jpg

    I can't seem to find anything online that helps me and I don't seem to have much on the topic in my books.

    I've got a fair few to work through but just need to see one in action as I don't know what I'm doing .

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,413
    Thanks
    1852

    Re: Having problems understanding differential equations

    What exactly do you want to know? The first thing the problem says is "in the absence of species Y (i.e. y= 0)" so the two equations become
    \frac{dx}{dt}= x\left(1- \frac{x}{12}\right)
    and 0= 0. (The right side of the second equation is 0 because it is 0 times something. The left hand is 0 because y is a constant, 0, and the derivative of a constant function is 0.)

    That equation is a "separable first order equation" which is typically the first thing you study in differential equations. You can "separate" it as
    \frac{dx}{x\left(1- \frac{x}{12}\right)}= dt

    Integrate on the left using "partial fractions".
    Thanks from jezb5
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2013
    From
    Norfolk
    Posts
    22

    Re: Having problems understanding differential equations

    Quote Originally Posted by HallsofIvy View Post
    What exactly do you want to know? The first thing the problem says is "in the absence of species Y (i.e. y= 0)" so the two equations become
    \frac{dx}{dt}= x\left(1- \frac{x}{12}\right)
    and 0= 0. (The right side of the second equation is 0 because it is 0 times something. The left hand is 0 because y is a constant, 0, and the derivative of a constant function is 0.)

    That equation is a "separable first order equation" which is typically the first thing you study in differential equations. You can "separate" it as
    \frac{dx}{x\left(1- \frac{x}{12}\right)}= dt

    Integrate on the left using "partial fractions".
    Thanks for the reply.

    Sorry for the late reply but it has taken me a while to figure it out.

    I rearrange \frac{dx}{dt}= x\left(1- \frac{x}{12}\right) to get


    \frac{dx}{x\left(1- \frac{x}{12}\right)}= dt

    From here I get the partial fractions

    \frac{1}{x(1-\frac{x}{12})} =\frac{A}{x} + \frac{B}{1- \frac{x}{12}}

    After multiplying both sides by the denominator, and collecting like terms, I get

    1 = (B - \frac{A}{12})x + A


    Equating coefficients gives

    B-\frac{A}{12} = 0

    and
    A = 1

    So
    B =\frac{1}{12}


    With knowing A and B I can integrate them now.

    \ln \left |\frac{x}{12-x}\right |=t+C

    \frac{x}{12-x}=c\cdot e^{t}


    Does that look about right? Also, how do I go about using
    x_0 =3,    and,    x_0 = 18 for the end part?

    Thanks
    Last edited by jezb5; January 21st 2014 at 10:46 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,789
    Thanks
    1147

    Re: Having problems understanding differential equations

    Quote Originally Posted by jezb5 View Post
    Thanks for the reply.

    Sorry for the late reply but it has taken me a while to figure it out.

    I rearrange \frac{dx}{dt}= x\left(1- \frac{x}{12}\right) to get


    \frac{dx}{x\left(1- \frac{x}{12}\right)}= dt

    From here I get the partial fractions

    \frac{1}{x(1-\frac{x}{12})} =\frac{A}{x} + \frac{B}{1- \frac{x}{12}}

    After multiplying both sides by the denominator, and collecting like terms, I get

    1 = (B - \frac{A}{12})x + A


    Equating coefficients gives

    B-\frac{A}{12} = 0

    and
    A = 1

    So
    B =\frac{1}{12}


    With knowing A and B I can integrate them now.

    \ln \left |\frac{x}{12-x}\right |=t+C

    \frac{x}{12-x}=c\cdot e^{t}


    Does that look about right? Also, how do I go about using
    x_0 =3,    and,    x_0 = 18 for the end part?

    Thanks
    that looks correct but you should finish by solving for x to get x = f(t,c).

    Then you can use x0 = f(0,c) to solve for c for each of your two initial conditions.
    Thanks from jezb5
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2013
    From
    Norfolk
    Posts
    22

    Re: Having problems understanding differential equations

    Quote Originally Posted by romsek View Post
    that looks correct but you should finish by solving for x to get x = f(t,c).

    Then you can use x0 = f(0,c) to solve for c for each of your two initial conditions.
    Thanks

    I'm not sure how to do this bit as I don't fully understand the notation used. Could you talk me through it please?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,789
    Thanks
    1147

    Re: Having problems understanding differential equations

    Quote Originally Posted by jezb5 View Post
    Thanks

    I'm not sure how to do this bit as I don't fully understand the notation used. Could you talk me through it please?
    were you able to solve for x?
    Thanks from jezb5
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2013
    From
    Norfolk
    Posts
    22

    Re: Having problems understanding differential equations

    Quote Originally Posted by romsek View Post
    were you able to solve for x?
    Is this correct?

     x = Ce^t(12 -x)

    Then for
     x_0 = 3 , I change x to 3 and t to zero?

    Which gives me

     3 = C(12 -3)

     C = \frac{1}{3}

    Then do the same for
     x_0 = 18
    Last edited by jezb5; January 22nd 2014 at 03:31 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,789
    Thanks
    1147

    Re: Having problems understanding differential equations

    Quote Originally Posted by jezb5 View Post
    Is this correct?

     x = Ce^t(12 -x)

    Then for
     x_0 = 3 , I change x to 3 and t to zero?

    Which gives me

     3 = C(12 -3)

     C = \frac{1}{3}

    Then do the same for
     x_0 = 18
    While what you've done is correct it's not exactly right. You want to solve for x as a function of c and t.

    You have

    \frac{x}{12-x}=c e^t \Rightarrow

    x=\frac{12 c e^t}{1+c e^t}

    In the first case

    x_0=3$ so $3=\frac{12c}{1+c}\Rightarrow c=\frac{1}{3}

    just repeat this for x_0=18
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Feb 2013
    From
    Norfolk
    Posts
    22

    Re: Having problems understanding differential equations

    Quote Originally Posted by romsek View Post
    While what you've done is correct it's not exactly right. You want to solve for x as a function of c and t.

    You have

    \frac{x}{12-x}=c e^t \Rightarrow

    x=\frac{12 c e^t}{1+c e^t}

    In the first case

    x_0=3$ so $3=\frac{12c}{1+c}\Rightarrow c=\frac{1}{3}

    just repeat this for x_0=18
    Thanks

    I've nearly got this finished as a worked example.

    When looking at the results for the below

     When, x_0 = 3, C = \frac{1}{3} Does this mean the population is growing?

    and

     When, x_0 = 18, C = -3 So the population is declining?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,789
    Thanks
    1147

    Re: Having problems understanding differential equations

    Quote Originally Posted by jezb5 View Post
    Thanks

    I've nearly got this finished as a worked example.

    When looking at the results for the below

     When, x_0 = 3, C = \frac{1}{3} Does this mean the population is growing?

    and

     When, x_0 = 18, C = -3 So the population is declining?
    that's basically correct though you really need to look at the whole function. In this case though it does turn out that negative C corresponds to a decline in population while positive C corresponds to growth in population.

    To see if you really understand what's going on see if you can determine what the population stabilizes at for any choice of C other than 0.
    Thanks from jezb5
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Partial Differential Equations Problems
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: September 22nd 2010, 05:22 PM
  2. Differential Equations Application Problems
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: September 11th 2010, 04:57 AM
  3. Separable Differential Equations Problems
    Posted in the Differential Equations Forum
    Replies: 18
    Last Post: August 31st 2010, 05:27 AM
  4. żunsolved problems in differential equations?
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: November 21st 2009, 06:12 AM
  5. Replies: 3
    Last Post: April 1st 2008, 08:17 PM

Search Tags


/mathhelpforum @mathhelpforum