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DE:y' + 2ty = 5t

  1. Jun 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve DE:y' + 2ty = 5t

    3. The attempt at a solution
    For yh: y' + 2ty = 0
    y' = -2ty
    Thus int(y-1,y) = int(-2t,t)
    ln(y) = -t2 + C1
    y = e-t2 + C1 = eC1e-t2 = Ce-t2

    To find yp there are two ways to do it. With an integrating factor exp(-A(t)) where A'(t) = 2t or by saying yp = v(t)yh(t). Both ways will result in int(e-t2,t) which is unsolvable.

    So I checked up the answer which read 5/2 + Cet2, so yp(t)=5/2. So I checked it by entering 5/2 in the DE and of course it was right.

    Now my question is, how would you arrive at that answer? When would you make the assumption yp' = 0 and check if it is right? Isn't that too trivial? And when you would find an answer for yp' = 0, would you stop searching for another answer?
     
    Last edited: Jun 24, 2011
  2. jcsd
  3. Jun 24, 2011 #2

    micromass

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    Hi luitzen! :smile:

    Can't you just argue that

    [tex]y^\prime+2ty=5t[/tex]

    is equivalent with

    [tex]y^\prime=t(5-2y)[/tex]

    and thus

    [tex]\frac{y^\prime}{5-2y}=t[/tex]
     
  4. Jun 24, 2011 #3
    Yes, of course. The question was: Check if the following equations are separable equations and if yes, give the solution.

    So, if you have a separable equation, you don't need to find yh and yp, am I right?

    If that's not the case, you try to find yh and after that yp, right?

    Thank you very much:)
     
  5. Jun 24, 2011 #4

    micromass

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    Right! Seperable equations are easy!

    Indeed, if the equation is not seperable, then you'll need to do other tricks. For example, working with yh and yp...
     
  6. Jun 24, 2011 #5

    LCKurtz

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    In my opinion, the best way to have worked that problem is to observe that it is a linear first order so you can multiply by [itex]e^{t^2}[/itex] making an exact derivative.
     
  7. Jun 24, 2011 #6
    How would you integrate et2?
     
  8. Jun 24, 2011 #7

    vela

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    You shouldn't have to. How are you coming up with that integral?
     
  9. Jun 24, 2011 #8

    LCKurtz

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    You wouldn't:
    [tex]e^{t^2}y'+2te^{t^2}y=5te^{t^2}[/tex]
    [tex](ye^{t^2})' =5te^{t^2}[/tex]
     
  10. Jun 25, 2011 #9
    Ok this is what I did:
    [itex]2ty-5t+y'=0[/itex]

    [itex]e^{t^{2}}\left(2ty-5t\right)+e^{t^{2}}y'=0[/itex]

    [itex]\dfrac{\partial F\left(y,t\right)}{\partial t}=e^{t^{2}}\left(2ty-5t\right)[/itex]

    [itex]\dfrac{\partial F\left(y,t\right)}{\partial y}=e^{t^{2}}[/itex]
     
  11. Jun 25, 2011 #10
    If I follow that route, I'll get this:

    [itex]\left(ye^{t^{2}}\right)'=5te^{t^{2}}[/itex]

    [itex]ye^{t^{2}}=\int5te^{t^{2}}dt[/itex]

    [itex]\int5te^{t^{2}}dt=5t\int e^{t^{2}}dt-5\int\int e^{t^{2}}dtdt[/itex]

    [itex]y=5e^{t^{-2}}\left(t\int e^{t^{2}}dt-\int\int e^{t^{2}}dtdt\right)[/itex]
     
  12. Jun 25, 2011 #11

    HallsofIvy

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    No, no, no!

    Let [itex]u= t^2[/itex]. Then du= 2t dt, tdt= du/2.
     
  13. Jun 25, 2011 #12
    lol, I think I use integration by parts too quickly.

    It's not too hard, but I still don't get it why it's easier though.
     
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