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Integrating xe[itex]^{x}[/itex] without integration by parts

  1. Jun 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Use the method of separation of variables or an integrating factor to find a particular solution of the differential equation that satisfies the given initial condition.

    y'=x-y+2 ; y(0)=4


    2. The attempt at a solution
    I've used an integrating factor of e[itex]^{x}[/itex] to obtain the following from y'=x-y+2:

    [itex]\frac{d}{dx}[/itex]e[itex]^{x}[/itex]y=xe[itex]^{x}[/itex]+2e[itex]^{x}[/itex]

    I know that I know have to integrate both sides of the equation. However, this is an issue since the equation contains the term xe[itex]^{x}[/itex]. This book hasn't yet taught integration by parts which is commonly used to integrate xe[itex]^{x}[/itex]. I don't think that separation of variables can be used to solve integrate this either.

    Therefore, my question is: is there a way to integrate xe[itex]^{x}[/itex] without using integration by parts?
     
  2. jcsd
  3. Jun 24, 2013 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    hi cmajor47! :smile:
    hint: can you integrate (x + 1)ex without using integration by parts? :wink:
     
  4. Jun 24, 2013 #3

    Mark44

    Staff: Mentor

    This doesn't look right to me. Can you show what you did to get it?
     
  5. Jun 24, 2013 #4

    lurflurf

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    Homework Helper

    ^There problem is in your look.
    we desire
    (u y)'=u(y+y')
    u' y+u y'=u y + u y'
    u=u'
    u=e^x
    so
    (e^x y)'=e^x (x+2)

    to solve (e^x y)'=e^x(x+2)
    we have a few choices all similar
    1)Integration by parts
    You want to avoid this one, but it is just a use of the product rule
    (u v)'=u' v + u v'
    u v'=(u v)'-u' v
    suppose u=x+2 v'=e^x
    then u'=1 v=e^x
    and
    u v'=(u v)'-u' v
    becomes
    (x+2)e^x=((x+2)e^x)'-e^x

    2)Undetermined coefficients
    assume
    ((a x+b)e^x)'=(x+2)e^x
    determine a,b

    3)repeated differentiation
    (e^x y)'=e^x (x+2)
    (e^-x (e^x y)')''=(x+2)''=0
    this just reproduces the undetermined coefficients

    4)mixed
    As tiny-tim points out we can rewrite the equation in terms of
    (x e^x)'
     
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