Question on nonhomogeneous ODE 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?
 A: 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. 
A: 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, $\lambda=\omega_0^2-w_d^2-\text{i}w_d\Gamma$.
