# Homework Help: Differential Equation Question

1. May 19, 2014

### Calu

I have a differential equation

y'' + y' -2y = 3e-2x + 5cosx

y = yc + yp

I found

yc = Ae-2x + Bex

for A, B arb. Const.

Then when selecting a trial function to find the particular integral, yp I came up with:

yp = ae-2x + bcosx + csinx

However the correct trial function to choose is:

yp = axe-2x + bcosx + csinx

Where the ae-2x has been multiplied by x. I was wondering in what instance I would have to multiply the term in the yp equation by x or x2.

2. May 19, 2014

### LCKurtz

When you plug in your expression for $y_p$, you know you need to get out a $3e^{-2x}$ term. But if you plug $ae^{-2x}$ into the equation, you know you will get zero because it is in $y_c$. That's when you want to try a term $y_p=axe^{-2x}$. There's more to that story but that's the short answer.

3. May 19, 2014

### Zondrina

You want the terms in $y_p$ to be linearly independent of the terms in $y_c$. Note that $f(x) = 3e^{-2x} +5cos(x)$.

Your original guess for $y_p$ includes a multiple of one of your homogeneous solutions, namely the $Ae^{-2x}$ term, so it cannot be linearly independent of $y_c$. Multiplying by $x$ is a quick shortcut to obtaining a linearly independent $y_p$ guess.

This method was developed by D'alembert, you can find out more by checking out "reduction order".

4. May 20, 2014

### Calu

These were both great answers! Thanks a lot!