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Differential Equation - Disappearing Term

  1. Oct 16, 2015 #1
    Hi,

    I am struggling to find the solution to the following equation. I can't account for the exponential term, so clearly something is going wrong...

    1. The problem statement, all variables and given/known data

    Find the general solution to ##x' = tx + 6te^{-t^2}## where ##x(t)##.

    2. Relevant equations


    3. The attempt at a solution

    Consider ##x' - tx = 0##

    ##\frac{dx}{dt} = tx##

    ##\frac{dx}{t} = t dt##
    Mod note: the above should be ##\frac{dx}{x} = t dt##
    integrating and exponentiating gives ##x = Ae^{\int t dt}##. Let ##I = \int t dt##

    ##x = A e^{I}## and rearranging gives ##A = x e^{-I}##

    Differentiating yields ##A' = x'e^{-I} - t x e^{-I}##

    Factoring out the exponential term, and recognising that this is the same as the initial equation.

    ##A' = e^{-I}(x' - tx) = e^{-I}(6te^{-t^{2}})##

    ##A = \int e^{-I}(x' - tx) dt = \int e^{-I}(6te^{-t^{2}}) dt## which simplifies to

    ##A = \int e^{-\frac{3}{2} t^{2} + c} 6t dt##

    Using the substitution ## u = - \frac{3}{2} t^{2} ## I find that

    ##A = - 2 \int e^{u} du = -2e^c e^{- \frac{3}{2} t^{2}} + c##

    ##x = e^I A = e^{- \frac{1}{2} t^2}(-2 e^{- \frac{3}{2} t^{2}} + c)##

    This is not a solution to my differential equation. The ##6t e^{-t^2}## term is nowhere to be seen.

    What am I doing wrong?
     
    Last edited by a moderator: Oct 16, 2015
  2. jcsd
  3. Oct 16, 2015 #2

    epenguin

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    I found easier than going through your arguments, just noticing that differentiating e-t2 gives you something very suggestive for a solution which I got.
     
  4. Oct 16, 2015 #3
    The steps to solve a first order differential equation of the type ##y' + a y = b ##, where ##a,b## are continuous, real valued functions, are
    1 - Find the general solution ##y_0## of the equation when ##b=0##, these are ##\{ \lambda e^{-A}, \lambda\in\mathbb{R}, A \text{ primitive of } a \}##
    2 - Find a particular solution ##y_1## to the equation. In your case it requires no calculations.
    3 - Add these solutions to have the general solution of the equation
     
  5. Oct 16, 2015 #4
    Hmm, you are right.

    ##x = \frac{1}{2} e^{t^{2}} - 3 e^{-t^{2}} + C##
     
    Last edited: Oct 16, 2015
  6. Oct 16, 2015 #5
    Your answer is incomplete because the solution is not unique unless you have a condition such as ##x(t_0) = y_0##
     
  7. Oct 16, 2015 #6
    Still incomplete, even with ##+C## :smile:. Read post #3
     
  8. Oct 16, 2015 #7

    Ray Vickson

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    Your main error is that initially, ##x## was a solution of ##x' - tx = 0##. Then you put ##x' - tx = 6 t e^{-t^2}## in your expression for ##A'##. That is incorrect: as originally defined, ##A## was a constant, with ##A' = 0##, so you cannot just pull a switcheroo and suddenly make ##A'## non-zero, without at least changing the meaning of ##A##. Go back to tried-and-true methods, as presented in your textbook or (probably) your course notes.
     
  9. Oct 16, 2015 #8

    epenguin

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    I don't think that's quite right, it seems to me

    x = -2 e-t2 + C

    will do it.
     
    Last edited: Oct 16, 2015
  10. Oct 16, 2015 #9
    No, the solutions are the ##x_\lambda(t) = -2e^{-t^2} + \lambda e^\frac{t^2}{2}##, ##\lambda\in\mathbb{R}##
     
  11. Oct 16, 2015 #10

    epenguin

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    :oldtongue::oldtongue:
    I only said #2 mine was A solution. :oldtongue:
     
  12. Oct 17, 2015 #11
    The method I used is the same (I think) as presented in 'Maths Methods For the Physical Sciences' - Mary Boas, on page 346

    It claims that ##y = e^{-I} \int Q e^{I} dx + C e^{-I}## is the general solution to a linear first order differential equation of the form ##y' + Py = Q##
     
  13. Oct 17, 2015 #12
    I appreciate that you have all taken the time to help me here, but I think it's best I go back to square one, as you have highlighted some pretty core misunderstandings.

    Thank you for your help.
     
  14. Oct 17, 2015 #13

    vela

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    The point here is that if you multiply ##x' - tx## by ##e^{-I}##, you turn it into the derivative of ##x e^{-I}##.

    This line would be okay if you didn't say it was equal to A', which is equal to 0 because A is a constant.

    You made a sign error. ##I##, as you defined it above, is equal to ##+\frac{t^2}{2}##, which gives you ##x(t) = -2 e^{-t^2} + c e^{t^2/2}##, which is the correct solution.

    Note that for the general case you cited above, ##I## is defined as ##\int p(x)\,dx##, but the way you did it in this problem, you set ##I = -\int p(x)\,dx##. That's why you have the difference in sign.
     
  15. Oct 17, 2015 #14
    The thing is that you need to be very careful with algebraic manipulations.
    For exemple, a division of your original DE by a function that has got one or several zeros in the interval you are interested in will force you to solve your DE in several sub-intervals and reconnect by continuity and differentiability the solutions found at the bounds.
    If you can't do these connections, your solution set is ##\emptyset## since your ##y## (or ##y'##) is not globally continuous.
    But you will never see that if you just divide and integrate.
     
  16. Oct 17, 2015 #15
    I have worked through the problem again and arrived at the correct solution.

    What I don't understand is why the derivative of the solution is not ## = tx + 6t e^{-t^{2}}##, but rather just ##tx##
     
  17. Oct 17, 2015 #16

    vela

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    You must be making a mistake when differentiating. I find ##x'(t) = 4 e^{-t^2} + c t e^{t^2/2}##. If you subtract ##xt## from that, you're left with ##6te^{-t^2}## as expected.
     
  18. Oct 17, 2015 #17
    You are correct.

    I missed the minus sign in my derivative.

    Thank you.
     
  19. Oct 17, 2015 #18

    Ray Vickson

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    I hope she does not actually say that, because as you have written it, it is either (i) wrong; (ii) inaccurate; (iii) misleading, or (iv) error-prone. What IS true is that the solution of ##y/ + P y = Q## has the form
    [tex] y =c e^{-q(t)} + e^{-q(t)} \int_0^t e^{q(s)} Q(s) \, ds = c e^{-q(t)} + \int_0^t e^{q(s) - q(t)} Q(s) \, ds, [/tex]
    where ##q(t) = \int_0^t Q(s) \, ds##. The correct formula involves specifying function arguments ##q(t)##, plus distinguishing between the variable ##t## in the solution ##y(t)## and the integration dummy variable ##s##. Just writing ##y = e^{-I} \int Q e^{I} dx + C e^{-I}## leaves way too much to the imagination and leaves way too much room for mistakes, etc.
     
  20. Oct 18, 2015 #19
    My mistakes aside, my argument follows fairly closely the steps outlined in the book (see attached).

    Could you recommend a more rigorous text? I am concerned that many of the explanations will be of similar calibre.
     

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  21. Oct 18, 2015 #20
    It is incorrectly explained.
    Just an exemple, when she says around (3.3) second scan "We obtain a solution as follows : ##\frac{dy}{y} = -P dx## ...... ##\ln(y) = ...## ".

    She doesn't know that ##y## does not cancel at some point on the interval of interest. You know it a posteriori, once you have found ##y##.
    So she assumes that the solution has no zero without even knowing it, and she uses divisions by ##y## and uses logarithm of ##y## at all points. It is incorrect !
     
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