关于含微分算子D的ODE公式及性质的推导请求

Hey there! Let’s walk through exactly why this differential operator identity holds—it’s all about solving a simple linear ODE, so let’s break it down step by step.

First, let’s clarify what the notation means:

• The symbol D is the differential operator, so D = \frac{d}{dx} (it just means "take the derivative with respect to x").
• The expression $\frac{1}{D - a}f(x)$ isn’t regular division—it’s shorthand for finding a particular solution to the differential equation $(D - a)y = f(x)$. Translating that to standard ODE form, we’re solving:
  $$y' - a y = f(x)$$

Now let’s solve this first-order linear nonhomogeneous ODE using the integrating factor method—this is the key to deriving the identity:

1. Find the integrating factor:
   For a linear ODE of the form $y' + P(x)y = Q(x)$, the integrating factor is $\mu(x) = e^{\int P(x)dx}$. Here, P(x) = -a, so:
   $$\mu(x) = e^{\int -a dx} = e^{-ax}$$

2. Multiply the entire ODE by the integrating factor:
   $$e^{-ax}y' - a e^{-ax}y = e^{-ax}f(x)$$
   Notice that the left-hand side is the derivative of a product: $\frac{d}{dx}\left(e^{-ax}y\right)$ (you can double-check this with the product rule if you want!). So we rewrite the equation as:
   $$\frac{d}{dx}\left(e^{-ax}y\right) = e^{-ax}f(x)$$

3. Integrate both sides with respect to x:
   Integrate the left side to undo the derivative, and integrate the right side as-is:
   $$e^{-ax}y = \int e^{-ax}f(x)dx + C$$
   (We can ignore the constant C since we only need a particular solution, which is what $\frac{1}{D - a}f(x)$ refers to.)

4. Solve for y:
   Multiply both sides by $e^{ax}$ to isolate y:
   $$y = e^{ax}\int e^{-ax}f(x)dx$$

That’s exactly the identity you were asking about! Let’s test it with a quick example to confirm it works:

• Let f(x) = x and a = 2. Then $\frac{1}{D - 2}x = e^{2x}\int e^{-2x}x dx$.
• Calculating the integral using integration by parts gives $\int x e^{-2x}dx = -\frac{x e^{-2x}}{2} - \frac{e^{-2x}}{4} + C$.
• Multiply by $e^{2x}$ and drop the constant: $y = -\frac{x}{2} - \frac{1}{4}$.
• Now check if $(D - 2)y = y' - 2y$ equals x:
  $y' = -\frac{1}{2}$, so $y' - 2y = -\frac{1}{2} - 2\left(-\frac{x}{2} - \frac{1}{4}\right) = -\frac{1}{2} + x + \frac{1}{2} = x$—perfect, it matches!

备注：内容来源于stack exchange，提问作者Shashank j