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Differential Equations - Verifying a solution of a given DE

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Verify that the indicated funciton is a solution of the given Differential Equation. c1 and c2 denote constants where appropriate.

    [tex]\frac { dX }{ dt } =(2-x)(1-x);\quad \quad \ln { \frac { 2-x }{ 1-x } } =t[/tex]


    3. The attempt at a solution
    I'm not quite sure how to really start this problem. If I'm reading the question right, the differential equation is

    [tex]\frac { dX }{ dt } =(2-x)(1-x)[/tex]

    and that the solution I'm checking is
    [tex]\ln { \frac { 2-x }{ 1-x } } =t[/tex]

    I was thinking of perhaps integrating
    [tex]\frac { dX }{ dt } =(2-x)(1-x)[/tex]
    then maybe plug in the the given value of t?

    I've read through the section and looked at my notes from class, but I can't seem to fully understand what I should be doing in this problem. The solution in the back of the book says

    [tex]\frac { d }{ dt } \ln { \frac { 2-X }{ 1-X } } =1,\quad \left[ \frac { -1 }{ 2-X } +\frac { 1 }{ 1-X } \right] \frac { dX }{ dt } =1[/tex]
    Simplifies to
    [tex]\frac { dX }{ dt } =(2-X)(1-X)[/tex]

    ...Not entirely sure what it necessarily means by that though. Any help would be GREATLY appreciated.
     
  2. jcsd
  3. Feb 19, 2012 #2
    I think they just want you to take the derivative of [tex] \ln { \frac { 2-x }{ 1-x } } [/tex] and find that it does indeed equal [tex] (2-x)(1-x) [/tex]

    EDIT: although when I take the derivative I get the answer upside down. I get [tex] \frac{1}{(1-x)(2-x)} [/tex]
     
    Last edited: Feb 19, 2012
  4. Feb 20, 2012 #3
    You just need to take the derivative of the natural log function. Some things you might want to keep in mind when doing this are that you can split this function into two natural logs, and also the chain rule.
     
  5. Feb 20, 2012 #4
    Alright, I'll give that a go and try to make sense of the question, thanks for the help you two
     
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