We have these 2 theorems/definitions.
*For each natural number n, $(D-m)^n e^{mx} y = e^{mx}D^ny$
*If f(D) is a polynomial in D with constant coefficients then $e^{mx}f(D)y=f(D-m)e^{mx}y$$\,\,\,\,\,\,\,\,$ where $D^n y= {d^ny\over dx^n}$
Now, we have this example, $ (D-2)^3y=0$ and this how my teacher solved it, $$ \begin{align}(D-2)^3y=0 \\ e^{-2x}(D-2)^3y=0 \\ D^3(e^{-2x}y)=0 \\ D^2(e^{-2x}y)= c_1 \\ D(e^{-2x}y = c_1x+ c_2 \\ e^{-2x}y = {c_1x^2 \over 2} + c_2x + c_3 \\ y= ({c_1x^2 \over 2} + c_2x + c_3)e^{2x} \end{align} $$
My question is how come she multiplied the second line by $e^{-2x}$ and not by $e^{2x}$. If we look at the above theorems/definitions and $m=2$ from the example, we have to multiply the second line by $e^{2x}$ and not by $e^{-2x}$ to use that theorem above. Please enlighten me! Thanks!
$\endgroup$1 Answer
$\begingroup$Note that $e^{-2x} (D-2)^3 y \neq (D-2)^3 e^{-2x} y$, since $(D-m)^n$ is a differential operator which acts on $y$ or $e^{-2x} y$.
Probably, sometimes it is better to write $(D-2)^3 (e^{-2x} y)$ to reflect this action and not to be confused.
(For example: $e^{-x} Dy \neq D(e^{-x} y)$, since $D(e^{-x}y) = -e^{-x} y + e^{-x} Dy$)
To understand how to apply the theorem above to the equation $(D-2)^3y=0$, I would suggest you to make a simple change of variables: let $y = e^{2x} u$.
Then, you will get $$ 0 = (D-2)^3 (e^{2x} u) \underbrace{=}_\text{by theorem} e^{2x} D^3 u. $$ Hence, $D^3 u = 0$, and therefore $$ u = \frac{c_1}{2}x^2 + c_2 x + c_3. $$ Finally, $y = \left(\frac{c_1}{2}x^2 + c_2 x + c_3\right) e^{2x}$.
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