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Need help with differential eq. from book

  1. Mar 19, 2009 #1
    On page 11 of Differential Equations Demystified (Krantz), there is an example that goes from this:

    [tex]

    e^{x^2} \cdot y^\prime + e^{x^2} \cdot 2xy = e^{x^2} \cdot x

    [/tex]

    To this:

    [tex]
    \left[ e^{x^2} \cdot y \right]^\prime = x \cdot e^{x^2}
    [/tex]

    Would someone give me a hint as to how they got from the first equation to the second?


    thanks
     
  2. jcsd
  3. Mar 20, 2009 #2
    To solve differential equations in the form

    [tex] y'+P(x)y=Q(x)[/tex]

    it is useful to use an integrating factor defined by

    [tex]\mu=\exp{\left(\int P(x)dx\right)}[/tex]

    We multiply both sides of the equation by this,

    [tex]\mu y'+\mu P(x)y=\mu Q(x)[/tex]

    and if you look closely the left hand side is the product rule for differentiating [itex]y\mu[/itex].

    So then we get that

    [tex]\left(\mu y\right)'=\mu Q(x)[/tex]

    And you'll see if you relate this to what you have, [itex]\mu=e^{x^2}[/itex] and then from the last equation,

    [tex]\left(e^{x^2} y\right)'=e^{x^2}x[/tex]
     
  4. Mar 20, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    More simply, to see that the two expressions are the same, go from
    [tex]\left[ e^{x^2} \cdot y \right]^\prime = x \cdot e^{x^2}[/tex]
    to
    [tex] e^{x^2} \cdot y^\prime + e^{x^2} \cdot 2xy = e^{x^2} \cdot x[/tex]
    by doing that differentiation on the left.
     
  5. Mar 20, 2009 #4
    Thanks, this is what I was looking for but it was not explained in the book.
     
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