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Given $$\frac{d^2z}{dt^2}+\Gamma\frac{dz}{dt}+\omega_0^2 \;z = \mathcal{F}(t)$$

where $\mathcal{F}(t) = F_0 e^{-iw_d t}$

The question says

"Without finding the particular solution explicitly, prove just using linerarity that the particular solution is proportional to $F_0$"

I can find the particular solution explicitly, but I don't understand the linearity property. If $z_1$ and $z_2$ are solutions, then $z_0 = z_1+z_2$ is a solution not of the original differential equation but of : $$\frac{d^2z_0}{dt^2}+\Gamma\frac{dz_0}{dt}+\omega_0^2 \;z_0 = 2\mathcal{F}(t)$$

What am I doing wrong, or am I misunderstanding the question?

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It seems like an ill-posed question. I think they mean something to do with the fact the particular solution scales linearly with $\mathcal{F}$ on the right side. – anon Aug 17 '11 at 21:25
If you look at the homogeneous equation (put right-hand side equal to $0$), solutions will add nicely. What is puzzling about the question is the reference to "the particular solution" since there are many. The problem may refer to the particular solution produced by the technique you were taught. I can deal with that if you tell me the name of the technique, or even if you give a very brief description, since there are not many, and I am likely to recognize it. – André Nicolas Aug 17 '11 at 23:51
@André My source is: Howard Georgi's Physics of Waves Page 51, Problem 2.2 (in the linked online edition, different in print edition) – kuch nahi Aug 18 '11 at 3:12
@kuch nahi what is $\Gamma$? Is it a constant? – user38268 Aug 18 '11 at 4:01
@D Lim yes, it is a real constant. $\omega_0$ and $\omega_d$ are also real constants. – kuch nahi Aug 18 '11 at 4:02
up vote 2 down vote accepted

I also find the question puzzling, for reasons that will be described in the comment at the end. But the following is certainly true.

Let $z_a$ be a particular solution of the equation that has $F_0=a\ne 0$.

Let $z_b=(b/a)z_a$. Then $z_b$ is a particular solution of the equation that has $F_0=b$.

Verification is straightforward. Substitute $(b/a)z_a$ for $z_b$ into the left-hand side.

In general the derivative of $kf(t)$, where $k$ is a constant, is $kf'(t)$, and the second derivative is $kf''(t)$. Thus $$\frac{d^2z_b}{dt^2}+\Gamma\frac{dz_b}{dt}+\omega_0^2z_b=(b/a)\left(\frac{d^2z_a}{dt^2}+\Gamma\frac{dz_a}{dt}+\omega_0^2z_a\right).$$

Note that linearity was essential. The argument could not be pushed through with a nonlinear equation like $\frac{d^2y}{dt^2}+y\frac{dy}{dt}+y=F(t)$.

Comment: The wording of the problem is imprecise. We have shown that given a particular solution $z_1$ for $F_0=1$, we can find a particular solution $z_c$ for $F_0=c$ by just multiplying $z_1$ by $c$. But not all particular solutions arise in this way.

I will illustrate this with the simpler equation $$\frac{dy}{dt} +y =5.$$

One particular solution is $y=5$ (the constant function $5$).

Now look at the equation $$\frac{dy}{dt}+y=105.$$

We can certainly get a particular solution of this equation by multiplying $5$ by $(105/5)$.

But the equation $\frac{dy}{dt}+y=105$ has plenty of other particular solutions, like $y=e^{-t}+105$. This particular solution cannot be obtained by multiplying the particular solution $5$ by a proportionality constant!

Note that the ratio $z_b/z_a$ is exactly $b/a$, which is the ratio of two values of $F_0$. This gives the required proportionality.

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They might have intended the reverse direction: scaling the particular solution necessarily scales the RHS. Very bad wording. – anon Aug 17 '11 at 22:35
@anon: The scaling argument works both ways, so it is hard to know what was in the person's mind. Here is a guess. The students have been given one of the usual explicit algorithms that produces a particular solution for this sort of equation. "The" particular solution may refer to the solution produced by this algorithm. I agree that the wording is painfully wanting. – André Nicolas Aug 17 '11 at 22:53

A first remark is that the LHS for $z=\mathcal F$ equals $\lambda\mathcal F$ for a given nonzero complex number $\lambda$. A second remark is that the LHS for $\alpha z$ equals $\alpha$ times the LHS for $z$, for every complex number $\alpha$ (this is where linearity is used). Hence the LHS for $z=\lambda^{-1}\mathcal F$ equals $\mathcal F$. This means that $z=\lambda^{-1}\mathcal F$ is a particular solution. Numerically, $\lambda=\omega_0^2-w_d^2-\text{i}w_d\Gamma$.

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@NonStandard Fully rigorous, thank you (note that the fact that the word "remark" appears in an answer does not mean that the answer is a remark). Which step do you have trouble to follow? – Did Jun 28 '15 at 20:50

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