Finding the particular solution

I was thinking about the problem that says:

Suppose $y_{p}=x\cos(2x)$ is a particular solution of $y^{n}+\alpha y=-4\sin(2x),$ where by $y^{n}$, i mean $$y^{n}=\underbrace{\frac{d}{dx}\frac{d}{dx}\cdots\frac{d}{dx}}_{n\text{ times}}\,y.$$ Then the constant $\alpha$ equals to

(a) $-4$
(b) $-2$
(c) $2$
(d) $4$

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since this is labeled (differential-equations), do you mean \begin{align} y^n =y\cdot y\cdot ... \text{ or } y^{[n]} = \frac{d}{dx} \frac{d}{dx}\frac{d}{dx} ... (y) \end{align} Furthermore are there some restrictions for $n$ ? – k1next Dec 14 '12 at 11:09
Sir, i mean $y^{n}=\frac{d}{dx}\frac{d}{dx}\frac{d}{dx}\frac{d}{dx}.....(y)$.There is no restrictions on $n$. – user52976 Dec 14 '12 at 11:13
First thing to do is work out ${d^n\over dx^n}(x\cos2x)$. – Gerry Myerson Dec 14 '12 at 12:10
I see that $D^n(x cos(2x))=2^{n-1}\{2x cos(2x+\frac{n\pi}{2})+n cos(2x+\frac{(n-1)\pi}{2})\}$. I know i have to put the value of $y_{p}$ in the given D.E. but i could not do the trick in the calculation to get the desired result. – user52976 Dec 14 '12 at 12:26

Imagine solving this linear DE with constant coefficients. We would probably find the general solution of the homogeneous DE $y^{(n)}+\alpha y=0$, then find a particular solution of the inhomogeneous DE. Adding, we get the general solution of the inhomogeneous DE.
When we solve $y^{(n)}+\alpha y=0$, we form the characteristic equation $t^n+\alpha=0$, and solve it. If the roots $r_1,\dots,r_n$ are distinct, the general solution has shape $\sum_1^n C_i e^{r_i x}$. The $r_i$ are (in general) complex, and therefore the complex exponentials can be expressed in terms of sines and cosines.
In this case, the roots are distinct. For $\alpha$ cannot be $0$, since $x\cos 2x$ is not a solution of $y^{(n)}=-4\sin 2x$. It is also fairly easy to see that $n$ cannot be $1$.
If $\alpha\ne 4$, to find a particular solution we look for constants $A$ and $B$ such that $A\cos\sqrt[n]{-\alpha} x+B\sin \sqrt[n]{-\alpha} x$ is a solution, and there are such constants. But $x\cos 2x$ cannot be expressed as a linear combination of exponentials (including sines and cosines). So we must have $\alpha=4$ (and $n=2$).
Remark: This is a multiple choice question, so one does not need the above exposition, which despite its length has missing details. You would be expected to answer the question based on experience with simple situations. You probably have experience solving some similar DE, and have noticed that the particular solution involves $x$ times trigonometric function only in the special cases when the homogeneous equation has solutions in sines and cosines that have the same period as the trigonometric function on the right-hand side. So picking out $4$, if one does not have to write a justification, is fairly quick.