# Thread: Particular solution to this non-homogen differential equation

1. ## Particular solution to this non-homogen differential equation

Find the particular solution to $y''+9y=2 \cos (2x)+3 \sin (3x)$

I have $y=A \cos (2x)+B \sin (3x)$, but for some reason when I plug in at the end everything cancels, what was wrong? Thanks

2. Originally Posted by tttcomrader
Find the particular solution to $y''+9y=2 \cos (2x)+3 \sin (3x)$

I have $y=A \cos (2x)+B \sin (3x)$, but for some reason when I plug in at the end everything cancels, what was wrong? Thanks
THis is happening because your complimentry solution is

$y_c=c_1\cos(3x)+c_2\sin(3x)$

Since you have repeated factors of $\sin(3x)$ your sine term will look like

$y_p=(Ax^2+Bx+C)\sin(3x)$

Be careful when your complimentry solution and particular have the same factors. Good luck.

3. Originally Posted by tttcomrader
Find the particular solution to $y''+9y=2 \cos (2x)+3 \sin (3x)$

I have $y=A \cos (2x)+B \sin (3x)$, but for some reason when I plug in at the end everything cancels, what was wrong? Thanks
Hi

The term in $B \sin (3x)$ effectively cancels but not the one in $A \cos (2x)$

First derivative
$y'=-2A \sin (2x)+3B \cos (3x)$

Second derivative
$y''=-4A \cos (2x)-9B \sin (3x)$

$y''+9y=5A \cos (2x)$

This means that $y=\frac25 \cos (2x)$ is a particular solution to $y''+9y=2 \cos (2x)$

If you manage to find a particular solution to $y''+9y=3 \sin (3x)$ then by addition you will get a particular solution to $y''+9y=2 \cos (2x)+3 \sin (3x)$

Try $y=Cx \cos (3x)$

EDIT : sorry, too late !

Find the particular solution to: $y''+9y\:=\:2 \cos (2x)+3 \sin (3x)$

I have $y=A \cos (2x)+B \sin (3x)$ .??
but for some reason when I plug in at the end everything cancels.
What was wrong?
Of course, everything cancels out . . .

To get $\cos(2x)$ in the solution, you need: $A\sin(2x) + B\cos(2x)$

And $\sin(3x)$ is a solution to the homogeneous equation.
. . You need: . $Cx\sin(3x) + Dx\cos(3x)$

Try it again with: . $y \;=\;A\sin(2x) + B\cos(2x) + Cx\sin(3x) + Dx\cos(3x)$

You should get: . $A = 0,\;B = \tfrac{2}{5},\;C = 0,\;D = -\tfrac{1}{2}$

5. Thank you so much, I was actually tutoring this course, and embarrassingly don't know how to do this.

Is there a proof for this on the internet? I really want to understand why I do these process. Thanks!

P.S. Why is one of the terms is $A \sin (2x)$, I didn't start with that term in the particular solution, and I didn't find that in the homogeneous solution.

6. Originally Posted by tttcomrader
Thank you so much, I was actually tutoring this course, and embarrassingly don't know how to do this.

Is there a proof for this on the internet? I really want to understand why I do these process. Thanks!

P.S. Why is one of the terms is $A \sin (2x)$, I didn't start with that term in the particular solution, and I didn't find that in the homogeneous solution.
Chris has a section on this here in MHF here is the link

http://www.mathhelpforum.com/math-he...-tutorial.html

The term $A \sin (2x)$ comes from the original driving function and needs to be included in the particular solution.