Method of undetermined coefficients

[badman]

hello everyone, i need some huge help here. here's 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.

 

[arildno]

"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.

 

[badman]

why doesn't it differentiate the x factor?

 

[arildno]

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:
$$\alpha{h}''+\beta{h}'+\gamma{h}=L(x)$$
we may write this as:
$$f(x)(\alpha{g}''+\beta{g}'+\gamma{g})+(\alpha(2f'(x)g'(x)+f''(x)g)+\beta{f}'g)=L(x)$$
Thus, if g is a solution of the associated homogenous problem, the first term vanishes identically.

 

[HallsofIvy]

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.