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Iffy diffy q

  1. Jun 22, 2004 #1

    Math Is Hard

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    I am trying to solve a linear equation and getting stuck.

    [tex]y' + 2xy = x^2 [/tex]

    I am using [tex]e^{x^2} [/tex] as my integrating factor and multiplying that to both sides.

    Afterwards, I am able to wrap up the LHS as [tex][y e^{x^2}]' [/tex]

    and I have [tex][y e^{x^2}]' = x^2 e^{x^2} [/tex]

    Now all I need to do is integrate both sides and I and home free, but I haven't found a way to integrate [tex] x^2 e^{x^2} [/tex].
    Using integration by parts just makes things more and more complicated. :grumpy:
    I am letting u = [tex]e^{x^2} [/tex] and dV = [tex]x^2 dx[/tex]
    du = [tex]e^{x^2} 2x dx[/tex] and V = [tex](x^3)/3[/tex]
    I don't think I have any other choice for this.

    Am I missing something really obvious or have I made a mistake along the way? Or is there another technique I can apply?

    Thanks in advance for your responses.
     
  2. jcsd
  3. Jun 22, 2004 #2

    HallsofIvy

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    You are doing integration by parts THE WRONG WAY!

    let dv= exdx so that v= ex and
    u= x2 so that du= 2x dx. Any time you have x to a power as a factor of an integrand, consider using that as the "u" since differentiating reduces the power.
    Repeating that will eventually reduce it to a constant.
     
  4. Jun 22, 2004 #3

    Gokul43201

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    Here's a useful aid for Integration by parts : Remember ILATE, a mnemonic for "inverse trig - log - algebraic - trig - exp." When you have 2 functions, often the right way to pick u and dv is in the order they appear in ILATE.

    In your case, you have an Algebraic term, and an Exponential term. In the mnemonic, A comes before E. So, you pick u and dv as the A term and the E term, respectively, as HallsofIvy has shown.
     
  5. Jun 22, 2004 #4

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    Thank you for your responses. But I am unclear on why dV should be
    [tex]e^{x} dx [/tex] instead of [tex]e^{x^2} dx [/tex]

    And thanks, for the tip, Gokul! I had not heard of this mnemonic. That's very cool. :smile:
     
  6. Jun 22, 2004 #5
    It should be [itex]e^{x^2}[/itex]. The integral doesn't turn out nicely.

    cookiemonster
     
  7. Jun 22, 2004 #6

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    Thank you for looking at this, Cookie. Here's the solution for y from my textbook by the way:

    [tex]y = x/2 + Ce^{-x^2} - e^{-x^2} {\int \ (1/2) e^{x^2}\, dx}[/tex]

    sorry about the less than perfect latex. Still learning.
     
  8. Jun 22, 2004 #7
    To get that form, use u = x and [itex]dv = xe^{x^2}dx[/itex].

    cookiemonster
     
  9. Jun 23, 2004 #8

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    whoa! I hadn't even thought about that! interesting split!

    oh, well. Thank goodness that one didn't show up on the final tonight.
     
  10. Jun 23, 2004 #9
    To figure it out, I just noticed that letting u = e^(x^2) didn't accomplish anything, and I couldn't integrate it when u = x^2 because then dv = e^(x^2)dx, which I notice can't integrate without an x in front. But I just so happened to have an x in the u term, so I moved it over and voila.

    cookiemonster
     
  11. Jun 23, 2004 #10

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    You are one smart cookie!!!

    Here - have a bag of Oreos. It's on the house.
    :biggrin:
     
  12. Jun 23, 2004 #11

    HallsofIvy

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    Oops, I missed that it was e^(x^2) rather than e^x!
     
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