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anonymous

  • 4 years ago

Solve u'-a*t*u=sin(t).

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  1. anonymous
    • 4 years ago
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    Assume a is a constant and u is a function of t. I'm using an integrating factor (e^((-at^2)/2), which is causing me problems. Specifically, I end up needing to integrate and integral that involves the error function as a solution, which I feel must be wrong.

  2. myininaya
    • 4 years ago
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    Integrating factor is \[v=e^{\int\limits_{}^{}-a t dt }=e^{-a \frac{t^2}{2}} \text{ we didn't need the +constant here}\] So we have \[e^{-a \frac{t^2}{2}} u'-e^{\frac{-a t^2}{2}} a t u=e^{\frac{-a t^2}{2}} \sin(t)\]

  3. myininaya
    • 4 years ago
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    so we can write \[(u e^{-a \frac{t^2}{2}})'=e^{-\frac{a t^2}{2}} \sin(t)\]

  4. myininaya
    • 4 years ago
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    Now integrate both sides

  5. anonymous
    • 4 years ago
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    This makes sense so far, but I don't know how to integrate the right side. I've tried integrating by parts, but that's not working.

  6. anonymous
    • 4 years ago
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    Well for an equation: \[u'(t)+p(t)u(t)=g(t)\] The solution is given by: \[u(t)=e^{\int\limits p(t)dt} \int\limits e^{-\int\limits p(t)dt}g(t)dt+Ce^{-\int\limits p(t)dt}\]

  7. anonymous
    • 4 years ago
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    If I remember correctly D:

  8. myininaya
    • 4 years ago
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    \[u e^{\frac{-a t^2}{2}}=\int\limits_{}^{}e^{\frac{-a t^2}{2}} \sin(t) dt +C\]

  9. myininaya
    • 4 years ago
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    so lets look at that part we haven't integrated

  10. myininaya
    • 4 years ago
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    \[\int\limits_{}^{}e^{\frac{-a t^2}{2}} \sin(t) dt=e^{\frac{-a t^2}{2}}[-\cos(t)]-\int\limits_{}^{}[-at] e^{\frac{-a t^2}{2}} [-\cos(t)] dt\]

  11. myininaya
    • 4 years ago
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    so far I have performed one round on integration by parts

  12. anonymous
    • 4 years ago
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    Okay, so this was integration by parts once?

  13. anonymous
    • 4 years ago
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    And you must integrate by parts again then on the remaining integral?

  14. myininaya
    • 4 years ago
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    \[\int\limits_{}^{}e^{-\frac{at^2}{2}} \sin(t) dt=-\cos(t) e^{-\frac{at^2}{2}}-a \int\limits_{}^{} t e^{-\frac{a t^2}{2}} \cos(t) dt\]

  15. myininaya
    • 4 years ago
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    i just wanted to clean some stuff up yes we need to do it again lol

  16. myininaya
    • 4 years ago
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    this is nasty looking but yes we need to do it again

  17. anonymous
    • 4 years ago
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    Alright this is making sense, I think I gave up too easy after a single round of integration by parts was looking like it was going nowhere.

  18. myininaya
    • 4 years ago
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    \[\int\limits\limits_{}^{}e^{-\frac{at^2}{2}} \sin(t) dt=-\cos(t) e^{-\frac{at^2}{2}}-a \int\limits\limits_{}^{} t e^{-\frac{a t^2}{2}} \cos(t) dt \] \[=- \cos(t) e^{-\frac{at^2}{2}}-a[\sin(t) t e^{-\frac{a t^2}{2}}-\int\limits_{}^{}\sin(t)[e^{-\frac{a t^2}{2}}+t (-at )e^{\frac{-a t^2}{2}} ] dt]\]

  19. myininaya
    • 4 years ago
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    I'm sorry this getting to messey to do for me on here. I think I will do this on paper and scan it for you.

  20. anonymous
    • 4 years ago
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    Thank you for going through all this work. I might be able to do it myself if it's only one more round of integration by parts.

  21. myininaya
    • 4 years ago
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    no it is not just one more

  22. myininaya
    • 4 years ago
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    this thing looks more beastly to me

  23. anonymous
    • 4 years ago
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    Yeah, it's pretty bad. It's supposed to be ODE review in a PDE class.

  24. anonymous
    • 4 years ago
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    I took ODE's over 2 years ago, so I'm a little rusty. I thought maybe some other method besides integration by parts may work better.

  25. myininaya
    • 4 years ago
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    i'm starting to think this isn't gonna work

  26. anonymous
    • 4 years ago
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    I think it should integrate by some means, since I'm also supposed to sketch several members of the family of solutions to the ODE.

  27. anonymous
    • 4 years ago
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    I don't have a problem sketching solutions, but I think that the integral must go away for me to be able to sketch solutions.

  28. myininaya
    • 4 years ago
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    oh you know what?

  29. myininaya
    • 4 years ago
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    this isn't an elementary integral

  30. myininaya
    • 4 years ago
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    so we can't use our elementary ways

  31. myininaya
    • 4 years ago
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    and that is all i know

  32. anonymous
    • 4 years ago
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    I'm not sure I know of any non-elementary ways to integrate.

  33. myininaya
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=integrate%28e%5E%28-a+t%5E2%2F2%29+sin%28t%29%2Ct%29

  34. anonymous
    • 4 years ago
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    Okay, well thank you for all of your time and help. I'll check out your link.

  35. asnaseer
    • 4 years ago
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    I think you need to use an integrating factor here of:\[e^{\int{atdt}}\]

  36. anonymous
    • 4 years ago
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    Oh no, the error function again. This is what I originally was running into problems with.

  37. myininaya
    • 4 years ago
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    hey asnaseer thats not where we were having the problem

  38. asnaseer
    • 4 years ago
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    sorry - missed a minus sign

  39. myininaya
    • 4 years ago
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    yes you need that lol

  40. myininaya
    • 4 years ago
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    \[\int\limits\limits_{}^{}e^{\frac{-a t^2}{2}} \sin(t) dt= \]

  41. myininaya
    • 4 years ago
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    this is the part we got stuck on

  42. myininaya
    • 4 years ago
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    and ike robin says we are getting the error function

  43. anonymous
    • 4 years ago
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    I think this method is typically used for second order ODEs, but could we find the homogeneous and particular solutions?

  44. anonymous
    • 4 years ago
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    I know the result should not involve the error function, sice II need to graph a family of curves, and I don't think the error function can be graphed.

  45. asnaseer
    • 4 years ago
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    ah - ok - I see what you mean now - I've followed through the discusion above. Let me see if there is another approach that can be used here...

  46. myininaya
    • 4 years ago
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    by the way thanks for coming asanaseer

  47. asnaseer
    • 4 years ago
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    np

  48. anonymous
    • 4 years ago
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    Thank you both for helping.

  49. asnaseer
    • 4 years ago
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    yw - I'm off to do some research on this interesting problem - I'll be back if I find a solution

  50. anonymous
    • 4 years ago
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    Good luck, I've been working on this for hours!

  51. JamesJ
    • 4 years ago
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    The integral on the RHS doesn't have an expression in terms of elementary functions. If you want to deal with it all and get it beyond the form \[ \int e^{-at^2/2} \sin t \ dt \ \ \ \ --(*)\] then write \[ \sin t = \frac{1}{2i}( e^{it} - e^{-it} ) \] complete the square, and you're left with error functions involving complex numbers. I'm not convinced that's a gain. I would probably just leave the integral (*) as it is.

  52. asnaseer
    • 4 years ago
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    yes - I tried that approach as well and ended up with the error function. I was just wondering if there are other approaches to solving equations like this?

  53. JamesJ
    • 4 years ago
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    No, not really. This integral isn't too bad. Part of dealing with ODEs and PDEs is learning what sorts of expressions can and cannot be integrated. This one can't.

  54. asnaseer
    • 4 years ago
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    ok - thx for the clarification - really appreciated.

  55. JamesJ
    • 4 years ago
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    For what it's worth, even if you tried a different technique here such as Laplace transform, when you went to invert the solution, you'd find you had an integral equivalent to this one.

  56. asnaseer
    • 4 years ago
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    Yes - I tried that as well :)

  57. anonymous
    • 4 years ago
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    For anyone who comes across a problem like this in the future, I found a way to solve the differential equation, without having to integrate ∫e−at2/2sint dt. General solution of differential equation = homogenous solution + particular solution. Find homogenous solution: \[u'-a*t*u = 0\] This is separable. \[du/dt =a*t*u\] \[(1/u) du = (a*t) dt\] \[\int\limits_{}^{}(1/u) du = \int\limits_{}^{} (at) dt\] \[\ln \left| u \right|+c = (1/2) a *t^{2}+c\] \[\left| u \right|=\exp((1/2)a*t ^{2}) + c\] \[u=\pm\exp((1/2)a*t ^{2}) + c\] Find the particular solution, using undetermined coefficients: \[p(t) = A \sin t + B \cos t\] \[p'(t) = A \cos t - B \sin t\] \[u'-a*t*u = \sin t\] \[A \cos t - B \sin t - a * t* (A \sin t + B \cos t) = \sin t\] Group like terms and factor to get system of equations: \[A - a*t*B = 0\] \[-B - a*t*A =1\] Solve for A and B. \[A = (at)/(1-a ^{2}t ^{2})\] \[B = 1/ (1-a ^{2}t ^{2}\] Plug back A and B into particular solution, add homogenous solution to get final solution. \[u =\pm e^{(1/2)at ^{2}} +(at)/(1-a ^{2}t ^{2})(\sin t) + 1/(1-a ^{2}t ^{2})(\cos t)+c\] Please correct any mistakes I made.

  58. JamesJ
    • 4 years ago
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    No, that's definitely not right because (a) this solution doesn't satisfy the differential equation (b) it falls apart when take the derivative. If the coefficients A and B are functions of t, then they must have non zero derivative, vs. the way you treat them above when you write down p'(t)

  59. anonymous
    • 4 years ago
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    Ah, I wasn't sure if undetermined coefficients would work or not in this case. I see why it doesn't work now. Is there another way to find the particular solution then without using undetermined coefficients?

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