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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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    (continued) You will get something like what Didier Piau wrote. You get $k$ times some constant is equal to $F_0$. If you multiply $F_0$ by something, the $k$ that works must be multiplied by the same thing. Easy. that's what you were supposed to show. There is a big difference between steady state solution and particular solution, though maybe not to a physicist. Which is older, your book or the (correct) web book you linked me to?2011-08-18

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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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    @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.2011-08-17
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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?2015-06-28