Find a linear differential operator that annihilates the given function. (Use D for the differential operator.)
For, 1+6x – 2×3 and, e-x + 2xex – x2ex
The Correct Answer and Explanation is:
To find a linear differential operator that annihilates a given function means to find an operator $L(D)$, expressed in terms of the differential operator $D = \frac{d}{dx}$, such that:
$$
L(D)[f(x)] = 0
$$
We will analyze each function separately.
Part 1: Function $f(x) = 1 + 6x – 2x^3$
This is a polynomial function. A general rule is:
- A polynomial of degree $n$ is annihilated by the $(n+1)^\text{th}$ derivative.
Here, the highest degree term is $x^3$, so:
$$
D^4 [1 + 6x – 2x^3] = 0
$$
Therefore, the linear differential operator is:
$$
\boxed{D^4}
$$
Part 2: Function $g(x) = e^{-x} + 2x e^x – x^2 e^x$
Break it into parts:
- $e^{-x}$: exponential
- $2x e^x$, $x^2 e^x$: polynomial × exponential
Step 1: Annihilate $e^{-x}$
The function $e^{-x}$ is annihilated by:
$$
(D + 1)
\quad \text{because} \quad (D + 1)[e^{-x}] = De^{-x} + e^{-x} = -e^{-x} + e^{-x} = 0
$$
Step 2: Annihilate $2x e^x – x^2 e^x$
This expression is a linear combination of:
- $x e^x$ and $x^2 e^x$
Let’s denote a general form:
- $e^x$, $x e^x$, $x^2 e^x$ ⇒ annihilated by $(D – 1)^3$
This is because:
- $(D – 1)[e^x] = 0$
- $(D – 1)^2[x e^x] = 0$
- $(D – 1)^3[x^2 e^x] = 0$
Thus, $(D – 1)^3$ annihilates the entire $(2x – x^2) e^x$ part.
Step 3: Annihilator for entire function
To annihilate the full expression:
$$
e^{-x} + 2x e^x – x^2 e^x
$$
We combine annihilators using the least common multiple (LCM) of the two operators:
- $(D + 1)$ and $(D – 1)^3$
Since these are distinct, the combined annihilator is:
$$
\boxed{(D + 1)(D – 1)^3}
$$
✅ Final Answers:
- For $f(x) = 1 + 6x – 2x^3$
Annihilator: $\boxed{D^4}$ - For $g(x) = e^{-x} + 2x e^x – x^2 e^x$
Annihilator: $\boxed{(D + 1)(D – 1)^3}$
📘 Explanation (Approx. 300 words):
In differential equations, a linear differential operator is an operator involving derivatives that acts linearly on functions. For example, $D = \frac{d}{dx}$, $D^2 = \frac{d^2}{dx^2}$, etc. If a differential operator $L(D)$ applied to a function $f(x)$ yields zero, we say that $L(D)$ annihilates $f(x)$.
In the first function $f(x) = 1 + 6x – 2x^3$, we have a polynomial of degree 3. It’s a known rule that a polynomial of degree $n$ is annihilated by $D^{n+1}$. Here, $D^4 f(x) = 0$, so $D^4$ is the annihilator.
In the second function $g(x) = e^{-x} + 2x e^x – x^2 e^x$, we deal with exponentials and polynomial-exponential terms. The function $e^{-x}$ is annihilated by $D + 1$, because differentiating $e^{-x}$ gives $-e^{-x}$, and adding back $e^{-x}$ yields zero.
For terms like $x e^x$, $x^2 e^x$, we use a standard rule: $x^n e^{ax}$ is annihilated by $(D – a)^{n+1}$. So $x^2 e^x$ is annihilated by $(D – 1)^3$, which also annihilates $x e^x$ and $e^x$.
To annihilate the full function, we combine these operators. Since they operate on independent parts of the function, their least common multiple gives the full annihilator: $(D + 1)(D – 1)^3$.
This method is fundamental in solving differential equations using the method of annihilators, especially when handling nonhomogeneous equations
.