Consider the following example.
$$\frac{\partial^2 u}{\partial^2 x}=\left(\frac{1}{R(t)}\frac{\partial }{\partial t}\right)^2u$$
Given that:
$$\left( \frac{1}{R(t)}\frac{\partial }{\partial t}\frac{\partial}{\partial x}-\frac{\partial}{\partial x}\frac{1}{R(t)}\frac{\partial }{\partial t}\right)u(x,t)=0,$$
we notice that it follows that
$$\left(\sqrt{\frac{\partial^2}{\partial x^2}}\right)u^*=\left(\pm\sqrt{\left(\frac{1}{R(t)}\frac{\partial }{\partial t}\right)^2}\right)u^*$$
where $\left\lbrace u^*\right\rbrace \subset \left\lbrace u\right\rbrace$.
The resulting equations for $u^*$ are much simpler to solve as they are first order and separable. So we solve them by the method of characteristics to get
$$\left\lbrace u^*\right\rbrace = \left\lbrace F_{1}\left(x-\int R(t)dt\right), F_{2}\left(x+\int R(t)dt\right)\right\rbrace $$
Where $F_{1}$ and $F_{2}$ are arbitrary functions. Since the original equation is linear, we then argue that a linear combination of all the elements of $\left\lbrace u^*\right\rbrace$ must produce all of the elements of $\left\lbrace u\right\rbrace$.
$$ u = F_{1}\left(x-\int R(t)dt\right) + F_{2}\left(x+\int R(t)dt\right)$$
So, the second order problem is thus algebraically reduced to a much simpler problem and the solutions are preserved.
Question:
Is this a universal property of differential operators and differential equations? And How can we prove it?