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 function means to find a differential operator $L(D)$, where $D = \frac{d}{dx}$, such that:
$$
L(D)[f(x)] = 0
$$
We are given two functions:
- $f_1(x) = 1 + 6x – 2x^3$
- $f_2(x) = e^{-x} + 2x e^x – x^2 e^x$
We will find the annihilator for each function separately and then combine them.
🔹 Part 1: Polynomial Function
Function:
$$
f_1(x) = 1 + 6x – 2x^3
$$
This is a polynomial of degree 3. The annihilator for any polynomial of degree $n$ is:
$$
D^{n+1}
$$
So here, since the highest power is $x^3$, we use:
$$
D^4
$$
Why? Because differentiating a cubic polynomial four times gives zero:
$$
D[1 + 6x – 2x^3] = 6 – 6x^2,\quad D^2 = -12x,\quad D^3 = -12,\quad D^4 = 0
$$
Thus, the annihilator is:
$$
L_1(D) = D^4
$$
🔹 Part 2: Exponential Function
Function:
$$
f_2(x) = e^{-x} + 2x e^x – x^2 e^x
$$
This is a linear combination of:
- $e^{-x}$
- $x e^x$
- $x^2 e^x$
✅ Step 1: Annihilator of $e^{-x}$
Annihilator is:
$$
(D + 1)
$$
✅ Step 2: Annihilator of $x^2 e^x$
The annihilator for $x^n e^{ax}$ is:
$$
(D – a)^{n+1}
$$
So for $x^2 e^x$, it is:
$$
(D – 1)^3
$$
✅ Combine Both
To annihilate all terms in $f_2(x)$, we take the least common multiple (LCM) of the individual annihilators:
$$
L_2(D) = (D + 1)(D – 1)^3
$$
✅ Final Answer
To annihilate the full expression:
$$
f(x) = f_1(x) + f_2(x) = 1 + 6x – 2x^3 + e^{-x} + 2x e^x – x^2 e^x
$$
We combine the two operators:
- From Part 1: $D^4$
- From Part 2: $(D + 1)(D – 1)^3$
So the final annihilator is:
$$
\boxed{D^4 (D + 1)(D – 1)^3}
$$
This is the differential operator that, when applied to the full function, gives zero.