Understanding Undetermined Coefficients in Differential Equations

[suspenc3]

Homework Statement

Write down the complementary solution $$y_c$$ and the correct particular solution $$y_p$$ in terms of its undetermined coefficients.

$$y^{'''}+y^{''}+3y^{'} -5y=x^2e^{x}+x^{2}e^{-x}cos(2x)+xe^{-x}sin(2x)$$

The Attempt at a Solution

I can easily find $$y_c$$ but I don't understand $$y_p$$.

The solution is:TRY:$$(Ax^2+Bx+C)e^x+e^{-x}[(Dx^2+Ex+F)cos(2x)+(Gx^2+Hx+I)sin(2x)]$$. And then you would adjust the particular solution. How come The $$Gx^2+Hx+I$$ term is a quadratic? If I didnt know the solution I would have put is as Gx+H.

If you had a good site explaining undetermined coefficients, that would be awesome, I don't really like the way the book explains it.

Thanks a lot.

 

[HallsofIvy]

Why aren't you asking about the "Dx²+ Ex + F" as well? The answer is exactly the same:
Since e^-x cos(2x) and e^-xsin(2x) are already solutions to the homogeneous equation, using (Dx+ C)cos(2x) and (Gx+ Y)sin(2x) would only give you "e^-xcos(2x)" and "e^-xsin(2x)", not "xe^-xcos(2x)" and "x^-xsin(2x)". Whenever a function already satisfies the homogeneous equation, you need to multiply whatever you would "normally try" by x.

(You don't actually need Dx²+ Ex+ F or Gx²+ Hx+ I. What you need is (Dx²+ Ex)e^-xcos(2x) and (Gx²+ Hx)e^-xsin(2x). Try those and see what happens.)