Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Verify that this is the solution to an ordinary differential equation

  1. Jan 13, 2010 #1
    1. The problem statement, all variables and given/known data
    I am given dy/dt -2yt = 1
    and y(t) = (e^(t^2))[e^(-s^2)ds] + e^(t^2)
    integrate from t to 0 within the brackets.

    2. Relevant equations



    3. The attempt at a solution
    I know that the derivative of y(t) would equal e^(t^2)
    However I do not know how I am supposed to solve after I plug all the numbers in. It becomes a huge mess with 2t * the whole mess that is y(t)
    Please advise
    Thank you!
     
  2. jcsd
  3. Jan 13, 2010 #2

    Mark44

    User Avatar
    Insights Author

    Staff: Mentor

    Is this your y(t)?
    [tex]y(t)~=~\int_0^t e^{t^2}(e^{-s^2})ds~+ e^{t^2}[/tex]
     
  4. Jan 13, 2010 #3
    Yes, but the first e^(t^2) is outside and multiplying the integral.
     
  5. Jan 13, 2010 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    So
    [tex]y(t)= e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2}[/tex]?
    No, the derivative is NOT [itex]e^{t^2}[/itex].
    Differentiating,
    [tex]y'= 2te^{t^2}\int_t^0 e^{-s^2}ds- e^{t^2}e^{-t^2}+ 2te^{t^2}[/tex]
    [tex]= 2t(e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2})- 1= 2ty- 1[/tex].
    That's exactly why it satisfies dy/dt- 2ty= 1.
    But that is NOT the "general solution". The way you have set it up, when t= 0, the integral is from 0 to 0 and so is 0. [itex]y(0)= 0+ e^{0^2}= 1[/itex]. This is the solution of dy/dx- 2ty= 1 with intial condition y(0)= 1. To get the general solution, add "+ C". Then choose C to give whatever specific value it is supposed to have.

    Plug what numbers in? I have no idea what you are talking about! Do you mean differentiate it? (As I just did? I used the product rule on [itex]e^{2x}\int e^{-2s}ds[/itex] and the Fundamental theorem of Calculus to differentiate [itex]\int e^{-2s}ds[/itex] itself.)
     
  6. Jan 13, 2010 #5
    Hi, thank you very much. I guess i misread a rule from the book when I thought that differentiating it would be simple. When i said plugging numbers in I meant plugging the y and y' back into the original equation to make sure it works.
     
  7. Jan 13, 2010 #6

    Mark44

    User Avatar
    Insights Author

    Staff: Mentor

    y and y' aren't "numbers" to plug in - they are functions. It's very important to communicate clearly, even moreso in the context of a forum where you have to use text only, and there is a lag time between a question and an answer.
     
  8. Jan 13, 2010 #7
    when i do this i have the integral of e^(-t^2). which from my understanding is not a very fun integral. I dont understand the method you guys took to doing this. can someone explain??
     
  9. Jan 13, 2010 #8
    Hi, Thank you so much for helping me. I will try to be more clear in the future.
     
  10. Jan 14, 2010 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The derivative of
    [tex]\int_0^x e^{-t^2}dt[/tex]
    is just [itex]e^{-x^2}[/itex]- that's the "Fundamental Theorem of Calculus". But [itex]\int_0^x e^{x^2}e^{-t^2}dt= e^{x^2}\int_0^t e^{-t^2}dt[/itex] is a product of functions of x and you have to use the chain rule.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Verify that this is the solution to an ordinary differential equation
Loading...