Question

Can Someone please help!
Find a particular solution yp of the differential equation

6y''+6y'+8y=(cos^2(x))
using the Method of Undetermined Coefficients. Primes denote derivatives with respect to x.

Answer 1

We can use this:
cos^2(x)=[1+cos(2x)]/2
So we are looking for a yp like this:
yp=A+B*cos(2x)+C*sin(2x)
Now we find the derivatives:
yp'=-2B*sin(2x)+2C*cos(2x)
yp''=-4B*cos(2x)-4C*sin(2x)
Now we plug that into the equation:
6*(-4B*cos(2x)-4C*sin(2x))+6(-2B*sin(2x)+2C*cos(2x))+8*(A+B*cos(2x)+C*sin(2x))=
=cos(2x)*[-24B+12C+8B] +sin(2x)*[-24C-12B+8C]+8A
=[1+cos(2x)]/2
So then we get this:
8A=1/2
-24C-12B+8C=0
-24B+12C+8B=1/2
A=1/16
Now we have to solve for B and C

Answer 2

-12B+16C=0
-16B+12C=1/2
I tihnk that you can solve this, right?

Answer 3

so you'd get B=0 and C=(1/24) right

Answer 4

but then that would be incorrect for the end solution

Answer 5

I got B=-1/14
C=-3/56

Answer 6

So then the solution is: yp=1/8-(1/14)*cos(2x)-(3/56)*sin(2x)

Answer 7

How did you solve for B?

Answer 8

We can solve like this:
-12B+16C=0
-16B+12C=1/2
Solve for B on the first one.
B=(16/12)*C
then plug that into the second one:
-16*(16C/12)+12C=1/2
There we can solve for C:
-256C/12+12C=1/2
(144-256)C/12=1/2
-112C/12=1/2
C=-12/(2*112)
C=-3/56
Now we go back to the first one:
-12B+16*(-3/56)=0
And now we solve for B
-12B=48/56
B=-1/14

Answer 9

ok thank you so much!!

Answer 10

Thanks to you! You are welcome! :)

Answer 11

wait shouldnt the solution be 1/16-(1/14)*cos(2x)-(3/56)*sin(2x)

Answer 12

Yes, you are right! :)
Sorry about my confusion.

Answer 13

unfortunately thats still wrong! It's part of thisonline web homework and its saying its wrong

Answer 14

A sec please.

Answer 15

-24C-12B+8C=0
-24B+12C+8B=1/2
I found my mystake:
The equation that we have to solve is this one:
-16C-12B=0
-16B+12C=1/2
B=-1/50
C=3/200
Sorry.

Answer 16

and A is 1/16?

Answer 17

Yes.

Answer 18

Is that correct now?

Answer 19

yes it is! thanks a bunch!!!

Answer 20

You are welcome! :)

Answer 21

:D