Verify that this is the solution to an ordinary differential equation

  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. Mark44

    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. Yes, but the first e^(t^2) is outside and multiplying the integral.
     
  5. HallsofIvy

    HallsofIvy 40,201
    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. 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. Mark44

    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. 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. Hi, Thank you so much for helping me. I will try to be more clear in the future.
     
  10. HallsofIvy

    HallsofIvy 40,201
    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.
     
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