# Differential Equation Question

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.

LCKurtz
Homework Helper
Gold Member
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.

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.

1 person
STEMucator
Homework Helper
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".

1 person
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.

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".

These were both great answers! Thanks a lot!