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, we must find a differential operator L(D)L(D) such that: L(D)[f(x)]=0L(D)[f(x)] = 0
We’ll treat each function separately.
1. Function: f1(x)=1+6x−2x3f_1(x) = 1 + 6x – 2x^3
This is a polynomial function of degree 3. For polynomial functions:
- The (n+1)th derivative of a polynomial of degree n is 0.
Since f1(x)f_1(x) is a degree-3 polynomial, the fourth derivative will annihilate it.
Define the operator: L1(D)=D4L_1(D) = D^4
Then: D4[f1(x)]=0D^4[f_1(x)] = 0
✅ Annihilator: D4\boxed{D^4}
2. Function: f2(x)=e−x+2xex−x2exf_2(x) = e^{-x} + 2xe^x – x^2e^x
This is a linear combination of terms involving exponentials multiplied by polynomials.
Break it into parts:
- e−xe^{-x}
- 2xex2xe^x
- −x2ex-x^2e^x
Let’s identify annihilators of each term:
- e−xe^{-x}: annihilated by (D+1)(D + 1)
- xexxe^x: annihilated by (D−1)2(D – 1)^2
- x2exx^2e^x: annihilated by (D−1)3(D – 1)^3
To annihilate the full function, we need the least common multiple (LCM) of the annihilators:
- LCM of (D+1),(D−1)2,(D−1)3(D + 1), (D – 1)^2, (D – 1)^3 is:
(D+1)(D−1)3(D + 1)(D – 1)^3
✅ Annihilator: (D+1)(D−1)3\boxed{(D + 1)(D – 1)^3}
📘 Explanation)
A linear differential operator is an operator involving derivatives, often denoted using D=ddxD = \frac{d}{dx}. The process of annihilation means applying this operator to a function such that the result is zero. An annihilator effectively removes the function—much like applying a fourth derivative to a cubic polynomial yields zero.
For f1(x)=1+6x−2x3f_1(x) = 1 + 6x – 2x^3:
This is a polynomial of degree 3. The derivative of a constant is zero; each differentiation lowers the degree by one. Hence, after four differentiations:
- D[f1(x)]=6−6x2D[f_1(x)] = 6 – 6x^2
- D2[f1(x)]=−12xD^2[f_1(x)] = -12x
- D3[f1(x)]=−12D^3[f_1(x)] = -12
- D4[f1(x)]=0D^4[f_1(x)] = 0
So, D4D^4 annihilates this polynomial.
For f2(x)=e−x+2xex−x2exf_2(x) = e^{-x} + 2xe^x – x^2e^x:
This function includes terms of the form xneaxx^n e^{ax}, common in differential equations. There is a systematic rule:
- eaxe^{ax}: annihilated by (D−a)(D – a)
- xneaxx^n e^{ax}: annihilated by (D−a)n+1(D – a)^{n+1}
So:
- e−xe^{-x} is annihilated by (D+1)(D + 1)
- xexxe^x needs (D−1)2(D – 1)^2
- x2exx^2e^x needs (D−1)3(D – 1)^3
To annihilate the entire sum, we take the least common multiple: Annihilator=(D+1)(D−1)3\text{Annihilator} = (D + 1)(D – 1)^3
This operator guarantees that all three terms vanish when the operator is applied.
✅ Final Answers:
- f1(x):D4f_1(x): \boxed{D^4}
- f2(x):(D+1)(D−1)3f_2(x): \boxed{(D + 1)(D – 1)^3}
