Is it always okay to substitute $a^2$ in place of $D^2$ in the inverse differential operator of cosh ax?

Question

In class, we were told that it was okay to make the following substitution while trying to solve a particular integral of a nonhomogeneous linear ODE, if $f(D)$ contains only even powers of $D$, and $f(a) \neq 0$ $$\frac{1}{f(D)}\cosh ax = \frac{1}{f(a)}\cosh ax$$

But in some examples, we only replaced $D^2$ with $a^2$, while keeping odd powers. For example, the general solution of the following equation $$\left(D^2-2D-3\right)y=2\cosh 3x$$ is $$y=c_1e^{3x}+c_2e^{-x}+\frac{xe^{3x}}{4}+\frac{1}{12}e^{-3x}$$

But our instructor replaced $D^2$ with $3^2$ while he was solving for the particular solution, which seems to give an inaccurate result $$y_p=\frac{1}{\left(D^2-2D-3\right)}2\cosh 3x = \frac{-1}{2}\frac{1}{\left(D-3\right)}\left(e^{3x}+e^{-3x}\right) = \frac{-xe^{3x}}{2}+\frac{e^{-3x}}{12}$$

So, my questions is, when is it really okay to make this substitution without getting an erroneous result? and does the same go for replacing $D^2$ with $-a^2$ if the image of the linear differential operator is $\sin ax$ or $\cos ax$?

Answer 1

You are allowed to substitute $D^2$ with $a^2$ in $\frac{1}{f(D)}\cosh ax$ and $\frac{1}{f(D)}\sinh ax$ or substitute $D^2$ with $-a^2$ in $\frac{1}{f(D)}\cos ax$ and $\frac{1}{f(D)}\sin ax$ if $f(a) \ne 0$.

In the cases that $f(a)$ becomes zero, you need to factor the term that makes $f(a)=0$, i.e. $f(D)=(D-a)^ng(D)$. And then, you can substitute $D^2$ with $\pm a^2$ in $g(D)$.

For example in your case,$$\frac{1}{\left(D^2-2D-3\right)}2\cosh 3x=\frac{1}{\left(D+1)(D-3\right)}2\cosh 3x=\frac{D-1}{\left(D^2-1)(D-3\right)}2\cosh 3x=\frac{D-1}{\left(8)(D-3\right)}2\cosh 3x$$

Or, in another example $$\frac{1}{(D^3-1)}\cosh x=\frac{1}{(D-1)(D^2+D+1)}\cosh x=\frac{1}{(D-1)(D+2)}\cosh x=\frac{D-2}{(D-1)(D^2-4)}\cosh x=\frac{D-2}{(D-1)(-3)}\cosh x=\frac{-1}{3}(\frac{1}{D-1}\sinh x-\frac{2}{D-1}\cosh x)$$

Answer 2

As $cos h(ax)$ is the linear combination of $e^{ax}$ and $e^{-ax}$ so you need to take care of their presence in your homogeneous solution and the partial substitution for only even powers of $D$ will give you erroneous result. On the other hand, if the RHS of differential equation is $sin(ax)$ which is the linear combination of $e^{iax}$ and $e^{-iax}$, you can go for partial substitution like $D^2=-a^2$ to find $y_p$ unless your homogeneous solution contains the linear combination of $sin(ax)$ and $cos(ax)$ (in which case you again rely on expressing $\frac{1}{f(D)}sin(ax)=\frac{1}{2i}\frac{1}{f(D)}(e^{iax}-e^{-iax})$, etc.)
For the sake of completeness, the link of related question you asked earlier has the answer here: https://math.stackexchange.com/a/4683737/246221