Method of Undetermined Coefficients help

[adamwitt]

Ok, please view the attached image for the Question, and for the given solution.

I need some help understanding the solution.

I can get the Complimentary equation with no problems, I understand how to do that.

However, some questions

1) Why do we first ignore the sin(3x) in our particular solution? so that we only have y'' + 9y = cos(2x) ?
I first attempted to look for a particular solution of the form
y_particular = Acos(2x) + Bsin(2x) + Ccos(3x) + Dsin(3x)

And I managed to solve A = 1/5, but sin(3x) was left over and all the terms making it up had been canceled out. I assume this is what it means by "satisfies the homogenous equation" but I fail to see how the dude who wrote the answer knew that from the beginning.
But my main question really is why are we allowed to look for a particular solution that ignores the sin(3x) ?


2) My second question, is why do attempt to find a particular solution of the form Ax.sin(3x) + Bx.cos(3x) ?
Ie. Why do we suddenly include an 'x' in there? what was our thought process that led us to try this?

Thanks to anyone who helps explain this. Cheers!

 

[Kreizhn]

1) Why do we first ignore the sin(3x) in our particular solution? so that we only have y'' + 9y = cos(2x) ?

The differential equation is linear right? Let's break this into two differential equations. Consider
$$y'' + 9y = \cos(2x)$$
and let the solution to this be $y_{p,1}$. Similarly, let the solution to
$$y'' + 9y = 2\sin(3x)$$
be given by $y_{p,2}$. Now notice that
$$\begin{align*} (y_{p,1} + y_{p_2})'' + 93(y_{p,1} + y_{p_2}) &= [y_{p,1}'' + 9y_{p,1}] + [y_{p,2}'' + 9y_{p,2}] \\ &= \cos(2x) + 2\sin(3x) \end{align*}$$
Thus the particular solution to each part can be solved separately, then added to get the total particular solution. Your method would also work normally, but here we need to be careful.

And I managed to solve A = 1/5, but sin(3x) was left over and all the terms making it up had been canceled out. I assume this is what it means by "satisfies the homogenous equation" but I fail to see how the dude who wrote the answer knew that from the beginning.

Indeed, the homogeneous equation is when you "keep" all the parts that involve y, and set the rest to zero. Namely, the homogeneous equation for this is
$$y'' + 9y = 0$$
Now there are many ways to solve this, and indeed one solution is $\sin(3x)$. The problem now is that when we try to find a particular solution $y_p = A \sin(3x) + B\cos(3x)$ we don't get anything useful out since $y_p'' + 9y_p = 0$. Thus we need to look at something more complicated. In particular, what can we differentiate in order to get a sin(3x) term that isn't sin(3x) itself? Well, how about xsin(3x)?

2) My second question, is why do attempt to find a particular solution of the form Ax.sin(3x) + Bx.cos(3x) ?
Ie. Why do we suddenly include an 'x' in there? what was our thought process that led us to try this?

Hopefully I made this clear from before, but the point is that you can't look for solutions just involving sin(3x) or cos(3x) because they satisfy y'' + 9y = 0, so if you sub them into the DE, they disappear. Thus we want to try terms that when differentiated, give a sin(3x) & cos(3x) term, but are not sin(3x) or cos(3x) themselves. So try xsin(3x). Well

$$\frac{d}{dx} x\sin(3x) = \sin(3x) + 3x\cos(3x)$$

Tada!

 

[adamwitt]

The differential equation is linear right? Let's break this into two differential equations. Consider
$$y'' + 9y = \cos(2x)$$
and let the solution to this be $y_{p,1}$. Similarly, let the solution to
$$y'' + 9y = 2\sin(3x)$$
be given by $y_{p,2}$. Now notice that
$$\begin{align*} (y_{p,1} + y_{p_2})'' + 93(y_{p,1} + y_{p_2}) &= [y_{p,1}'' + 9y_{p,1}] + [y_{p,2}'' + 9y_{p,2}] \\ &= \cos(2x) + 2\sin(3x) \end{align*}$$
Thus the particular solution to each part can be solved separately, then added to get the total particular solution. Your method would also work normally, but here we need to be careful.



Indeed, the homogeneous equation is when you "keep" all the parts that involve y, and set the rest to zero. Namely, the homogeneous equation for this is
$$y'' + 9y = 0$$
Now there are many ways to solve this, and indeed one solution is $\sin(3x)$. The problem now is that when we try to find a particular solution $y_p = A \sin(3x) + B\cos(3x)$ we don't get anything useful out since $y_p'' + 9y_p = 0$. Thus we need to look at something more complicated. In particular, what can we differentiate in order to get a sin(3x) term that isn't sin(3x) itself? Well, how about xsin(3x)?



Hopefully I made this clear from before, but the point is that you can't look for solutions just involving sin(3x) or cos(3x) because they satisfy y'' + 9y = 0, so if you sub them into the DE, they disappear. Thus we want to try terms that when differentiated, give a sin(3x) & cos(3x) term, but are not sin(3x) or cos(3x) themselves. So try xsin(3x). Well

$$\frac{d}{dx} x\sin(3x) = \sin(3x) + 3x\cos(3x)$$

Tada!

Sir, you are of a fine callibre. Thank you greatly, seriously.