# Differential Equation

1. Mar 2, 2008

### suspenc3

1. The problem statement, all variables and given/known data
Write down the complementary solution $$y_c$$ and the correct particular solution $$y_p$$ in terms of its undetermined coefficients.

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

3. The attempt at a solution
I can easily find $$y_c$$ but I dont 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 dont really like the way the book explains it.

Thanks alot.

2. Mar 2, 2008

### HallsofIvy

Staff Emeritus
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.)