Results 1 to 2 of 2

Math Help - Find solutions

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    6

    Find solutions

    Additional processes due to biotic and abiotic degradation contributing to souce decay can be accounted for by adding a decay term in (3) that is proportional to m(t),
    , (4)
    where λ is the associated decay rate constant. Assume that α(0) and λ(0) are greater than or equal to zero, then choose one pairs of the vaules of α and λ by yourself and find solutions of Equation (4) using the initial condition m(0)=m0 for the following cases: (i) γ=1, (ii) γ=2, (iii) γ= 0.5. Then , graph the solution for m(t) for each γ

    (3)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,232
    Thanks
    1795
    Quote Originally Posted by cowracer3 View Post
    Additional processes due to biotic and abiotic degradation contributing to souce decay can be accounted for by adding a decay term in (3) that is proportional to m(t),
    , (4)
    where λ is the associated decay rate constant. Assume that α(0) and λ(0) are greater than or equal to zero, then choose one pairs of the vaules of α and λ by yourself and find solutions of Equation (4) using the initial condition m(0)=m0 for the following cases: (i) γ=1, (ii) γ=2, (iii) γ= 0.5. Then , graph the solution for m(t) for each γ

    (3)
    I assume that "M" in equation 4 is supposed to be "m" as in equation 3 so that you have just added the term -\lambda m.

    The problem does not require you to solve that equation in general and I suspect it can't be. In (case i) \gamma = 1 you just have \frac{dm}{dt}= (\alpha + \lambda )m which can be separated as \frac{dm}{m}= (\alpha+ \lambda)dt and can be integrated easily.

    In (case ii) \gamma = 2 you have \frac{dm}{dt}= \alpha m^2+ \gamma m which, again can be separated: \frac{dm}{\alpha m^2+ \gamma m}= dt. The denominator on the left can be factored as m(\alpha m+ \gamma) and it can be integrated using "partial fractions".

    In (case iii) \gamma = 2 you have \frac{dm}{dt} = \alpha m^{1/2}+ \gamma m which can be separated as \frac{dm}{\alpha m^{1/2}+ \gamma m}= dt. If you let u= m^{1/2} then du= (1/2)m^{-1/2}dm so dm= 2m^{1/2}du= 2udu and the left side becomes \frac{2udu}{\alpha u+\gamma u^2}= \frac{2du}{\alpha+ \gamma u} and the substitution v= \alpha+ \gamma u reduces it even further. (which means I should have suggested u= \alpha+ \gamma x^{1/2} to begin with!)
    Last edited by mr fantastic; February 11th 2009 at 07:03 PM. Reason: Fixed the latex
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find all solutions
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: May 15th 2010, 03:34 PM
  2. Find all solutions of...
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 2nd 2010, 08:47 PM
  3. Find all solutions
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: February 28th 2010, 10:14 PM
  4. find all solutions
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: May 1st 2009, 07:43 AM
  5. find all solutions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 23rd 2009, 05:26 PM

Search Tags


/mathhelpforum @mathhelpforum