# Thread: Stuck solving for a particular solution

1. ## Stuck solving for a particular solution

$y'' +4y=6xcos(2x)$

I am using the method of undetermined coefficients. I had no trouble solving for the homogeneous solution, but the particular solution has me baffled. I used $y_p=Axcos(2x)+Bxsin(2x)$. After finding the derivatives of this and putting them back into the original equation, I end up with a system of equations that I cannot solve. Have I chosen this function poorly?

2. Hey,

Because of the x out front of the cos you have to choose something along the lines of:

$y_{p}=A\cos{2x} + Bx\cos{2x} + C\sin{2x} + Dx\sin{2x}$

In order to take care of the derivatives where you will have mixed x's on the sines and cosines.
I didn't try to solve it, but I believe this is correct. Give it a shot .

Hope it helps!

3. I tried that and it didn't work.

4. Originally Posted by staevobr
I tried that and it didn't work.
Then please clearly post all the working for what you tried so that it can be reviewed.

5. $y_p=Acos(2x)+Bxsin(2x)+Csin(2x)+Dxcos(2x)$
$y'_p=-2Asin(2x)+Bsin(2x)+2Bxcos(2x)+2Ccos(2x)+Dcos(2x)-2Dxsin(2x)$
$y''_p=-4Acos(2x)+2Bcos(2x)+2Bcos(2x)-4Bxsin(2x)-4Csin(2x)$
$\> -2Dsin(2x)-2Dsin(2x)-4Dxcos(2x)$

$y''_p + 4y = -4Acos(2x)+2Bcos(2x)+2Bcos(2x)-4Bxsin(2x)-4Csin(2x)-2Dsin(2x)$
$-2Dsin(2x)-4Dxcos(2x) + 4Acos(2x) + 4Bxsin(2x) +4Csin(2x) + 4Dxcos(2x)$

$4Bcos(2x)-4Dsin(2x)$
This makes $B=3x/2$, which is not a constant.

6. Originally Posted by staevobr
$y'' +4y=6xcos(2x)$

I am using the method of undetermined coefficients. I had no trouble solving for the homogeneous solution, but the particular solution has me baffled. I used $y_p=Axcos(2x)+Bxsin(2x)$. After finding the derivatives of this and putting them back into the original equation, I end up with a system of equations that I cannot solve. Have I chosen this function poorly?
Since the homogenous solution (which I was hoping you would include in your working because it can be important in deciding what particular solution to try - like in this question) is $y_h = (Ax + B) (\sin (2x) + \cos (2x))$, the 'particular solutions' mentioned so far in this thread will not work (why?).

The particular solution you need to use is (why?):

$y_p = (a_1 x + a_2 x^2) (b_1 \cos (2x) + b_2 \sin (2x))$

7. The homogeneous solution I actually got is

Characteristic equation: $r^2+4=0$
$r= \pm2i$
$y_h=C_1e^{0x}cos(2x)+C_2e^{0x}sin(2x)$
$y_h = C_1cos(2x) + C_2sin(2x)$

8. Originally Posted by staevobr
The homogeneous solution I actually got is

Characteristic equation: $r^2+4=0$
$r= \pm2i$
$y_h=C_1e^{0x}cos(2x)+C_2e^{0x}sin(2x)$
$y_h = C_1cos(2x) + C_2sin(2x)$
Yes, you are of course correct (an aberation on my part). Nevertheless, I think the particular solution I proposed is what you need to use.