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    Solve the initial value problem:

    dx/dt = (1-t)x with initial value being x(2)=0

    Answer:

    dx/x = (1-t)dt

    Then integrate both sides:

    1/x = t - t^2/2 + c

    Applying x(2)=0:

    0 = 1/(t - t^2/2 + c)

    0 = 1/c

    c = ?
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    When you integrate both sides is when the mistake pops up. \displaystyle \int \frac{1}{x}~dx=ln|x|+C yet I see no logarithms in your answer.
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    when you integrate both sides you have a log:

    LHS: ln|x|

    on the RHS, introduce the constant of integration in log form: ln(k)

    \displaystyle ln(x)=t- \frac{t^{2}}{2}+ln(k)

    you can exponentiate and then use the IC, or move ln(k) to the LHS, use the IC, solve for k, then exponentiate
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    you could also use the limits of integration thus:

    1) move the expression involving t to the LHS and integrate between respective limits:

    \displaystyle \int_{0}^{x} \frac{1}{x}dx -  \int_{2}^{t}(1-t)dt=k
 
 
 
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