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Method of undetermined coefficients

  1. Oct 31, 2006 #1
    hello everyone, i need some huge help here. heres the equation :

    y''+y'-6y=10e^2x-18e^3x-6x-11.

    complementary solution:c1e^2x+c2e^-3x
    s1={e^2x}
    s2={e^3x}
    s3=(x,1}
    ok since e^2x exists in the complimentary solution, it is therefore a solution, so i multiply it by x to get s'1{xe^2x}, so now i have that and my s2 and s3.

    so i end up with this:

    yp=Axe^2x+Be^3x+Cx+D

    heres where i get confused do i use the product rule for Axe^2x, if so do i end up with these two derivatives: y'p=Axe^2x+2Ae^2x+...etc, y''p=Axe^2x+4Ae^2x+...etc. i really get confused with this product rule thing.
     
  2. jcsd
  3. Oct 31, 2006 #2

    arildno

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    "heres where i get confused do i use the product rule for Axe^2x, if so do i end up with these two derivatives: y'p=Axe^2x+2Ae^2x+...etc, y''p=Axe^2x+4Ae^2x+...etc. i really get confused with this product rule thing."

    Certainly you are to differentiate the whole thing!
    Those derivatives of xe^2x that doesn't differentiate the x-factor, but effectively only differentiates the exponential, will disappear.
     
  4. Oct 31, 2006 #3
    why does'nt it differentiate the x factor?
     
  5. Oct 31, 2006 #4

    arildno

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    h(x)=f(x)g(x)
    h'(x)=f(x)g'(x)+(f'(x)g(x))
    h''(x)=f(x)g''(x)+(2f'(x)g'(x)+f''(x))
    The paranthesized terms differentiaties f, the other terms regard f effectively as a constant multiple of the g-function.
    Assuming some diff.eq:
    [tex]\alpha{h}''+\beta{h}'+\gamma{h}=L(x)[/tex]
    we may write this as:
    [tex]f(x)(\alpha{g}''+\beta{g}'+\gamma{g})+(\alpha(2f'(x)g'(x)+f''(x)g)+\beta{f}'g)=L(x)[/tex]
    Thus, if g is a solution of the associated homogenous problem, the first term vanishes identically.
     
    Last edited: Oct 31, 2006
  6. Nov 1, 2006 #5

    HallsofIvy

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    You misinterpreted this. arildno meant "in those terms where, using the product rule, the x itself was not differentiated".
    For example, differentiating xf(x) using the product rule, I get f(x)+ xf '(x). The term "xf '(x)" is the one arildno was referring to.
     
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