Adding up the solutions of a two different linear differential equation I read a paper in which there is a linear differential equation as follows:

You can see two constants on the right side of the equation omega3 and omega1. The authors have solved the equation for three cases:


*

*$\Omega_3 = 10$ and $\Omega_1 = 0$

*$\Omega_3 = 0$ and $\Omega_1 = 10$

*$\Omega_3 = 10$ and $\Omega_1 = 10$


And they say that since the equation is linear the solution of 3. is a sum of 1. and 2.
I did not understand this because 1. and 2. are different differential equations. They are not the solutions of the same equation. So they cannot be added up to form a new solution. 
Can someone explain to me what's going on here?
 A: First lets consider the left hand side of the equation. To make the notation easier we will define a linear operator $L$ whose action on theta results in the LHS. 
$$L(\theta)=\ddot{\theta}+2\zeta\omega_n\dot{\theta}+(\omega_n^2+\dot{\gamma}^2)\theta$$
$L$ is certainly linear in $\theta$.
Now lets consider the solution to situation (1) indicated in your question. In this case the solution, lets call it $\theta_1$, must satisfy the following condition, 
$$ L(\theta_1)=2\dot{\gamma}\Omega_3\cos(\gamma),$$
Similarly we have $\theta_2$ which satisfies the following condition.
$$ L(\theta_2)=2\dot{\gamma}\Omega_1\sin(\gamma)$$
Now what happens when $L$ acts on the sum of $\theta_1$ and $\theta_2$? Well since $L$ is linear the result will be the sum of $L$ applied to each function.
$$ L(\theta_1+\theta_2) = L(\theta_1)+L(\theta_2) = 2\dot{\gamma}\Omega_3\cos(\gamma)+2\dot{\gamma}\Omega_1\sin(\gamma).$$
This of course means that the sum satisfies the differential equation when the third condition is true. 

To establish the linearity of $L$ we need to prove that the following properties hold.


*

*$ L(\theta_a + \theta_b) = L(\theta_a)+L(\theta_b)$

*$ L(k \theta) = k L(\theta)$  ($k$ is a constant)


For the purposes of this particular answer we only need to establish the first property. 
$$L(\theta_a+\theta_b)=\color{red}{\frac{\mathrm{d}^2}{\mathrm{d}t^2} \left(\theta_a+\theta_b\right)}\color{blue}{+2\zeta\omega_n\frac{\mathrm{d}}{\mathrm{d}t}\left( \theta_a+\theta_b\right)}\color{green}{+(\omega_n^2+\dot{\gamma}^2)\left(\theta_a+\theta_b\right)}
$$
Now we use the properties of derivatives to distribute them onto each of the $\theta$'s. 
$$= \color{red}{\frac{\mathrm{d}^2}{\mathrm{d}t^2}\theta_a+\frac{\mathrm{d}^2}{\mathrm{d}t^2}\theta_b}\color{blue}{+2\zeta\omega_n\frac{\mathrm{d}}{\mathrm{d}t}\theta_a+2\zeta\omega_n\frac{\mathrm{d}}{\mathrm{d}t}\theta_b}\color{green}{+(\omega_n^2+\dot{\gamma}^2)\theta_a+(\omega_n^2+\dot{\gamma}^2)\theta_b}$$
$$= \left[ \color{red}{\frac{\mathrm{d}^2}{\mathrm{d}t^2}\theta_a}\color{blue}{+2\zeta\omega_n\frac{\mathrm{d}}{\mathrm{d}t}\theta_a}\color{green}{+(\omega_n^2+\dot{\gamma}^2)\theta_a} \right]+\left[ \color{red}{\frac{\mathrm{d}^2}{\mathrm{d}t^2}\theta_b}\color{blue}{+2\zeta\omega_n\frac{\mathrm{d}}{\mathrm{d}t}\theta_b}\color{green}{+(\omega_n^2+\dot{\gamma}^2)\theta_b} \right]$$
$$=L(\theta_a)+L(\theta_b)$$
