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Implicit Solution To Differential Equation

  1. Feb 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Verify that the indicated expression is an implicit solution of the given first order differential equation. Find at least one explicit solution in each case. Give an interval I of definition of each solution.

    The differential equation is: [itex]\displaystyle \frac{dX}{dt} = (X -1)(1-2X)[/itex]

    and the solution is [itex]\displaystyle \ln \left( \frac{2X-1}{X-1} \right) = t[/itex]


    2. Relevant equations



    3. The attempt at a solution

    Implicitly differentiating gives

    [itex]\displaystyle \ln(2X -1) - \ln(X-1) = t [/itex]

    [itex]\displaystyle \frac{\dot{X}}{2X-1} - \frac{\dot{X}}{X-1} = 1[/itex]

    [itex]\displaystyle \frac{2 \dot{X}(X-1)}{(2X-1)(X-1)} - \frac{\dot{X}(2X-1)}{(X-1)(2X-1)} = 1[/itex]

    [itex]\displaystyle \frac{\dot{X}(2X-1-2X +1)}{(2X-1)(X-1)}[/itex]

    [itex]\displaystyle \frac{\dot{X} \cdot 0}{(2X-1)(X-1)} = 1[/itex]

    What happened? What did I do wrong? According to this link http://rmower.com/s_diff_eq/Examples/0101p2.pdf I am incorrect.
     
  2. jcsd
  3. Feb 1, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    You error is in this step. You didn't distribute the factor of 2 correctly.


    In that link, I don't understand how they can go from
    $$
    \frac{2 dX}{2X-1} - \frac{dX}{X-1} = dt
    $$
    to
    $$
    2dX - dX = (2X-1)(X-1) dt
    $$
     
  4. Feb 1, 2014 #3
    Try leaving $$ ln(\frac{1-2X}{X-1}) $$ as a a single logarithm and implicitly differentiating by the quotient rule. You should end up with
    $$ 1 = (\frac{(-2X+2)-(1-2X)}{(1-2X)(X-1)})\frac{dX}{dt} $$
    where you can simplify to prove that $$ t = ln(\frac{1-2X}{X-1})$$ is a particular solution where C = 0
     
    Last edited: Feb 1, 2014
  5. Feb 1, 2014 #4
    $$\frac{-2dX}{1-2X} - \frac{dX}{X-1} \equiv \frac{dX}{(1-2X)(X-1)}$$
    by partial fractions.

    Therefore, by multiplying by $$(1-2X)(X-1)$$ we're left with
    $$-2dX(X-1)-dX(1-2X) = (1-2X)(X-1)dt $$
    $$ -2XdX + 2dX - dX + 2XdX = (1-2X)(X-1)dt $$ and our 2XdX's cancel out.
     
    Last edited: Feb 1, 2014
  6. Feb 1, 2014 #5

    DrClaude

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    Staff: Mentor

    Geez. I read ##(2X-1)## where it was written ##(1-2X)##. :redface:
     
  7. Feb 1, 2014 #6
    The initial differential equation, independent of the assumption that X < 0 was $$ \frac{dX}{dt} = (1-2X)(X-1) $$
     
    Last edited: Feb 1, 2014
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