Method of undetermined coefficients

  • Thread starter badman
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  • #1
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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.
 

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  • #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.
 
  • #3
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why does'nt it differentiate the x factor?
 
  • #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.
 
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  • #5
HallsofIvy
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arildno said:
Those derivatives of xe^2x that don't differentiate the x-factor, but effectively only differentiates the exponential, will disappear.
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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