# Nonhomogenious - method of undetermined coefficients

• May 21st 2008, 09:35 AM
taurus
Nonhomogenious - method of undetermined coefficients
How to find a particular solution for this using method of undetermined coefficients:
y'' - y = e^(-x) cos( 4 x )

I have got the homogenious equation as:
y = A*e^(1*x) + B*e^(-1*x)

But now am not sure how to get a particular solution with the right side?

thanks
• May 21st 2008, 10:45 AM
topsquark
Quote:

Originally Posted by taurus
How to find a particular solution for this using method of undetermined coefficients:
y'' - y = e^(-x) cos( 4 x )

I have got the homogenious equation as:
y = A*e^(1*x) + B*e^(-1*x)

But now am not sure how to get a particular solution with the right side?

thanks

Trial functions:
$e^{-x}cos(4x),~e^{-x}sin(4x)$

-Dan
• May 21st 2008, 10:56 AM
taurus
why choose those?
• May 21st 2008, 11:04 AM
topsquark
Quote:

Originally Posted by taurus
why choose those?

Because they are similar in form to the non-homogeneous part and have derivatives that are also similar in form to the non-homogeneous part.

In addition since they are not solutions to the homogeneous equation we don't have to worry about messy terms like
$xe^{-x}cos(4x)$
etc.

-Dan
• May 21st 2008, 11:51 AM
taurus
could you explain that? How do you choose a particular solution. I am completly confused
• May 21st 2008, 12:19 PM
galactus
Sometimes a DE book will have a list of trial particular solutions.

In this case, we use $y_{p}=Ae^{-x}cos(4x)+Be^{-x}sin(4x)$

Because our problem is of that form. Suppose we would've had

$xe^{3x}cos(4x)$, then we would use

$(ax+B)e^{3x}cos(4x)+(Cx+E)e^{3x}sin(4x)$

Find the second derivative:

$y''_{p}=(-15A-8B)e^{-x}cos(4x)+(8A-15B)e^{-x}sin(4x)$

Sub them into the equation and get:

$y''_{p}-y_{p}=(-16A-8B)e^{-x}cos(4x)+(8A-16B)e^{-x}sin(4x)=e^{-x}cos(4x)$

Now, equate coefficients:

There is no sin on the right so that is 0:

$-16A-8B=1$

$8A-16B=0$

$A=\frac{-1}{20}, \;\ B=\frac{-1}{40}$

A particular solutions is:

$\frac{-1}{20}e^{-x}cos(4x)-\frac{1}{40}e^{-x}sin(4x)$

A general solution is:

$\frac{-1}{20}e^{-x}cos(4x)-\frac{1}{40}e^{-x}sin(4x)+C_{1}e^{x}+C_{2}e^{-x}$
• May 21st 2008, 12:35 PM
taurus
would you have a list of the trials?
• May 21st 2008, 12:37 PM
galactus
No, I'm sorry. You can probably google some.
• May 21st 2008, 12:44 PM
taurus
Quote:

Originally Posted by galactus
In this case, we use $y_{p}=Ae^{-x}cos(4x)+Be^{-x}sin(4x)$

How would i differentiate that? with respect to what do i differentiate?
• May 21st 2008, 01:42 PM
topsquark
Quote:

Originally Posted by taurus
How would i differentiate that? with respect to what do i differentiate?

The derivatives in your differential equation are with respect to x, so take the derivatives with respect to x.
$\frac{dy_p}{dx} = A(-e^{-x})cos(4x) + Ae^{-x}(-4~sin(4x)) + B(-e^{-x})sin(4x) + Be^{-x}(4~cos(x))$
by the usual product rule.

etc.

-Dan