Text: Advanced Calculus for Applications, Second Edition by Francis B. Hildebrand
My inquiry today involves the following problem:
Given the notation:
$$y(x) = \frac{1}{D-r}h(x)$$
used to indicate that $y(x)$ satisfies the equation $(D-r)y=h$, where $D=\frac{d}{dx}$ and $r \in \mathbb{R}$, verify that: \begin{align}y &=\Big[\frac{1}{r_1-r_2}\Big(\frac{1}{D-r_1} - \frac{1}{D-r_2}\Big)\Big]h(x) \\ &= \frac{1}{r_1 - r_2}\Big[\frac{1}{D-r_1}h(x) - \frac{1}{D - r_2}h(x)\Big] \end{align} satisfies the equation $(D-r_1)(D-r_2)y=h$, where $r_1 \neq r_2$.
The strategy I embarked on was to reduce the expression to the form:
$$y = \frac{1}{(D-r_2)(D-r_1)}h(x) = \frac{1}{D-r_2}\Big[\frac{1}{D-r_1}h(x)\Big]$$
since it was proven in a previous problem that such a $y$ would satisfy the aforementioned equation. I proceeded as follows:
\begin{align}y &=\Big[\frac{1}{r_1-r_2}\Big(\frac{1}{D-r_1} - \frac{1}{D-r_2}\Big)\Big]h(x) \\ &= \Big[\frac{1}{r_1-r_2}\Big(\frac{1}{D-r_1}\color{red}{\Big(\frac{D-r_2}{D-r_2}\Big)} - \frac{1}{D-r_2} \color{red}{ \Big(\frac{D-r_1}{D-r_1}\Big)}\Big)\Big]h(x) \\ &=\Big[\frac{1}{r_1-r_2}\Big(\frac{D-r_2-D+r_1}{(D-r_1)(D-r_2)}\Big)\Big]h(x) \\ &=\Big[\frac{1}{r_1-r_2}\Big(\frac{r_1-r_2}{(D-r_1)(D-r_2)}\Big)\Big]h(x) \\ &=\frac{1}{(D-r_1)(D-r_2)}h(x) \\ \end{align}
Would the steps colored in red be legal? That is, is it fair to say that:
$$\frac{D-r}{D-r} = 1$$
If the cancellation law fails to hold, what exactly is the justification, and what would be the appropriate manner in going about this proof?