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Solving for an initial condition with a DE

  1. Oct 25, 2006 #1

    I was wondering if anyone could clarify some things for me here. The problem tells me to solve the given DE subject to the indicated initial condition. In this case, the initial condition is y(1) = 3

    Here is the equation.


    So after I have solved, what do I do with the given initial condition?
  2. jcsd
  3. Oct 25, 2006 #2
    The initial condition allows you to turn that general expression (with the C), into a statement that is true for the condition given.

    What does y(1) = 3 mean?

    You could write it like this, y(t=1)=3

    So, when t = 1, then y = 3, so...

    where you plugged in the values from,

    You can then solve for C
  4. Oct 25, 2006 #3
    Ok, so you are saying that the solved DE is some function y(t). So when y(1)=3, that means that when t=1, y=3, thus we plug in 1 for t and 3 for y and solve for C. Cool. So the result, after solving for C is the initial condition? In other words, C = "the initial condition."

    So in this case, the initial condition is ln(3)-(1/2) = C or, C = 0.599.

    Thank you for your help!!
  5. Oct 25, 2006 #4
    C does not quite equal the initial condition, but if you find C with the method above, your final solution y(t) - with the proper C - will be the solution satisfying the initial condition defined in any initial value problem. Seems like you got it though.
  6. Oct 25, 2006 #5


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    It would be a good idea to go ahead and solve for y, either before or after finding C.
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