Thread: Nonhomogenious - method of undetermined coefficients

1. 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

2. 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:
$\displaystyle e^{-x}cos(4x),~e^{-x}sin(4x)$

-Dan

3. why choose those?

4. 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
$\displaystyle xe^{-x}cos(4x)$
etc.

-Dan

5. could you explain that? How do you choose a particular solution. I am completly confused

6. Sometimes a DE book will have a list of trial particular solutions.

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

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

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

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

Find the second derivative:

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

Sub them into the equation and get:

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

Now, equate coefficients:

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

$\displaystyle -16A-8B=1$

$\displaystyle 8A-16B=0$

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

A particular solutions is:

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

A general solution is:

$\displaystyle \frac{-1}{20}e^{-x}cos(4x)-\frac{1}{40}e^{-x}sin(4x)+C_{1}e^{x}+C_{2}e^{-x}$

7. would you have a list of the trials?

8. No, I'm sorry. You can probably google some.

9. Originally Posted by galactus
In this case, we use $\displaystyle y_{p}=Ae^{-x}cos(4x)+Be^{-x}sin(4x)$
How would i differentiate that? with respect to what do i differentiate?

10. 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.
$\displaystyle \frac{dy_p}{dx} = A(-e^{-x})cos(4x) + Ae^{-x}(-4~sin(4x)) + B(-e^{-x})sin(4x) + Be^{-x}(4~cos(x))$
by the usual product rule.

etc.

-Dan