Why does treating the linear differential equation as algebraic equation works? I was reading about solving second order differential equations here which says that you can treat them as algebraic equation to solve them. I understand the method but why does this work at all this way?
 A: Well, because we don't even need to make that replacement! Namely, we think of the action of taking the derivative as a linear function $d_x:C^{2}(\mathbb{R})\to C^{2}(\mathbb{R})$. Then we can also make sense of $1:C^{2}(\mathbb{R})\to C^{2}(\mathbb{R})$ and $d_x-1$ likewise. This leads to actual equations $(d_x-1)(d_x+1)=d_x^2-1$ in the space of functions on $C^{2}(\mathbb{R})$. This just expresses $(d_x-1)(d_x+1)f=(d_x^2-1)f$ for all functions. Thus solutions to $y''-y=(d_x^2-1)y=0$ or elements of $\ker(d_x^2-1)$, are the same as $\ker(d_x-1)(d_x+1)=(d_x-1)^{-1}(\ker(d_x-1))$. Now in general $A^{-1}(\ker(B))\neq span(\ker(A), \ker(B))$, but we always have $span(\ker(A), \ker(B))\subset A^{-1}(\ker(B))$, and in the case above, we note that $dim((d_x-1)^{-1}(\ker(d_x-1)))=dim(\ker(d_x-1)(d_x+1))\le 2$, since second order differential equations have atmost two independant solutions, and we have $dim(span(\ker(d_x-1), \ker(d_x+1)))=2$, so that we must have $span(\ker(d_x-1), \ker(d_x+1))=(d_x-1)^{-1}(\ker(d_x-1))=\ker(d_x-1)(d_x+1)$, and writing out what this means gives the results in the blog.
Its probably best to explore a case where this doesn't work, as Andre pointed out in the comments. If we proceed the same way with $y''-2y'+y=0$, we get $d_x^2-2d_x+1=(d_x-1)^2$, so that we are looking at $\ker((d_x-1)^2)$, but this is not equal to $\ker(d_x-1)$, since it contains the function $xe^x$, which is not in $\ker(d_x-1)$.
