When solving a linear differential equation by factoring the operator, how does one guarantee no solutions are lost? I think the best way to make this question clear is with an example.
Lets say we want to solve the differential equation $(\Delta^2 - \lambda^4)\phi=0,$ calculations are greatly simplified if  we factor the operator as 
$(\Delta -  \lambda^2)(\Delta + \lambda^2) \phi=0.$ Since the factors commute, it is clear that we can find solutions by solving each of the equations
$$ (\Delta -  \lambda^2)\phi_1 =0\\(\Delta +  \lambda^2)\phi_2=0  $$ separately. 
However it is not obvious that any solution to the original problem can be given in the form $\phi = \phi_1 + \phi_2,$ which is what texts concerning vibrations of plates claim. For example the book from Leissa, Vibrating Plates, NASA. I have not found yet any reference that addresses this issue. 
How can one guarantee no solutions are lost by solving the equation through this method? That is, how can one guarantee there are no other solutions $\phi$ that can't be written as $\phi=\phi_1+\phi_2$?
I would greatly appreciate help with this problem. Thanks in advance. 
 A: 
Lets say we want to solve the differential equation $(\Delta^2 - \lambda^4)\phi=0.$

Like Tomás said in the comment, if we are dealing with a bounded domain (a vibrating plate), then boundary condition is necessary for us to look a meaningful solution. From the routine of looking for proper boundary condition imposed for second order problem (Poisson, Helmholtz, please refer to Evans), we can use the Green's formula (integration by parts):
$$
\int_{\Omega} (\Delta^2 u )\,v = \int_{\Omega}\Delta u\Delta v + \int_{\partial \Omega} v\frac{\partial \Delta u}{\partial n} dS - \int_{\partial \Omega} \frac{\partial v}{\partial n} \Delta u\,dS,
$$
where can set the test function $v=0$ on boundary, and the true solution's Laplacian vanishes on the boundary to make the boundary term vanish, so that to make $\Delta^2$ a positive definite operator, thus to have a bounded inverse. Therefore the Dirichlet-type homogeneous boundary condition for the fourth order problem should be:
$$
\phi = 0,\;\text{and }\; \Delta \phi = 0\;\text{ on }\partial \Omega.\tag{1}
$$
Once we have these two boundary conditions, for $(\Delta -  \lambda^2)(\Delta + \lambda^2) \phi=0$, let $\psi = (\Delta +\lambda^2)\phi$. The equation turns into 
$$(\Delta -  \lambda^2)\psi = 0  \quad\text{in }\Omega\\
\psi= 0 \quad \text{ on }\partial \Omega.$$ 
For $(-\Delta +\lambda^2)$ is positive definite, you can either argue by maximum principle or energy method that $\psi = 0$ identically in $\Omega$. This implies 
$$
(\Delta +\lambda^2)\phi = 0, 
$$
and $\phi$ is an eigenfunction of $-\Delta$ with eigenvalue $\lambda^2$. In this case, we can't see $\phi_1$, and $\phi = \phi_2$ with boundary condition (1). 
I just wanna say: boundary condition does matter. Now let's back to your main question.


How can one guarantee no solutions are lost by solving the equation through this method? That is, how can one guarantee there are no other solutions $\phi$ that can't be written as $\phi=\phi_1+\phi_2$?

This is more of an algebraic result: if we already know that under certain properly imposed boundary condition (Dirichlet or Neumann), the equation has a solution $\phi$ with certain regularities, then
$$
\phi = \underbrace{\frac{1}{2\lambda^2}(\Delta + \lambda^2)\phi}_{\phi_1} + \underbrace{\frac{1}{2\lambda^2}(-\Delta + \lambda^2)\phi}_{\phi_2}.
$$
We can check:
$$
(\Delta-\lambda^2)\phi_1 = 0,\;\text{ and }\; (\Delta+\lambda^2)\phi_2=0.
$$
