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Help with Integration

  1. Sep 6, 2008 #1
    Help with Integration :(

    Ok guys i've been trying to solve this problem for a couple of days but no luck :( Please help me out!! I think I have part of it done.

    I've added the problem as an attachment wasn't sure how to type it up properly. But when i solve for y' i get the answer 2t+e^(t^2) I am however having trouble solving the integral in order to get a value for y to plug into the equation.

    Thanks for the help!!

    Attached Files:

  2. jcsd
  3. Sep 6, 2008 #2
    Re: Help with Integration :(

    i am confused here, what are you really trying to do?

    I mean, solve for y, or t, at the end?

    It would be a good idea to post the entire problem, with the text, so it will clarify things.
    Last edited: Sep 6, 2008
  4. Sep 6, 2008 #3
    Re: Help with Integration :(

    Oh my bad here is the problem:

    Verify that each given function is a solution of the differential equation.

    So I was assuming to solve for y' and plug it into the initial equation and check to see if the condition equals 1.
  5. Sep 6, 2008 #4


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    Homework Helper

    Re: Help with Integration :(

    You are asked only to verify that the expression for y(t) satisfies the DE y'-2ty=1. If you work out what is y' (using the fundamental theorem of calculus) and substitute it into the DE, you'll find that it all cancels out nicely to give you 1 as expected.
  6. Sep 6, 2008 #5
    Re: Help with Integration :(

    Thats where i am running into some problems defender. I solved for y' and i got 2t+e^(t^2) was wondering if someone can check that. But also do I have to solve the integral in y before I can plug it back into the original DE? Those are the step that I am taking so far but I am having trouble solving the integral in y in order to plug it back into the DE. :(
  7. Sep 7, 2008 #6


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    Re: Help with Integration :(

    There isn't any way to evalute the integral in elementary functions and you do not have to. Your expression for y' should have the integral inside. I don't know how you got that but you have to use the product rule for differentiation.
  8. Sep 7, 2008 #7


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    Re: Help with Integration :(

    You are given, I suppose,
    [tex]y= e^{t^2}\int_0^t e^{-s^2}ds+ e^{t^2}[/tex]
    I think I would be inclined first to write it as
    [tex]y= e^{t^2}\left(\int_0^t e^{-s^2}ds+ 1\right)[/tex]

    Now, to see if that function satisfies the differential equation differentiate that to find y' and substitute those formulas for y' and y into the equation. As Defennder said, you cant actually "find" y in terms of elementary functions and you don't need to. Since you have a product of functions, use the product rule. Of course, the derivative of [itex]e^{t^2}[/itex] is, by the chain rule, [itex]2te^{t^2}[/itex] and the derivative of the integral is, by the fundamental theorem of calculus, [itex]e^{-t^2}[/itex].
  9. Sep 7, 2008 #8
    Re: Help with Integration :(

    Awesome that works out well, thanks a lot guys!
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