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Math Help - Separable Equation With Intial Condition.

  1. #1
    Ted
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    Separable Equation With Intial Condition.

    Hello.

    Problem: Find the solution of the differential equation that satisfies the given intial condition.
    \frac{dL}{dt}=kL^2ln(t)
    The intial condition: L(1)=-1.

    My solution:

    Well, I noticed its a separable equation,
    The problem here, is the question did not identify the K!
    so I assumed its a constant.
    then I rewrite the D.E. as:

    \frac{dL}{L^2}=k ln(t) dt

    By integrating both sides and using integration by parts, I obtained:

    \frac{-1}{L}=k ( tln(t)-t+C)

    \frac{-1}{L}=k ( tln(t)-t)+A where A=Ck.

    Solving for L,I got:

    L=\frac{-1}{k(tln(t)-t)+A}.

    since L(1)=-1:

    L(1)=\frac{-1}{-k+A}=-1.

    so:

    A-k=1, i.e. A=1+k.

    Then ????
    I stopped here!
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  2. #2
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    e^(i*pi)'s Avatar
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    From what I can gather you've gone as far as you can without knowing the value of k.

    If I were doing this question I'd leave the equation for L in terms of t and k
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  3. #3
    Ted
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    So the particular solution is:

    L=\frac{-1}{k(tln(t)-t)+k+1}

    Right?
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  4. #4
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    Quote Originally Posted by Ted View Post
    So the particular solution is:

    L=\frac{-1}{k(tln(t)-t)+k+1}

    Right?
    That's what I'd put, it's possible you might be given k but since A is pretty much the constant of integration it'd be better to have the equation in terms of k
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