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:
L(D)[f(x)]=0
Where D=ddxD = \frac{d}{dx}D=dxd is the differential operator.
Part 1: f(x)=1+6x−2x3f(x) = 1 + 6x – 2x^3f(x)=1+6x−2×3
This is a polynomial of degree 3. Any polynomial of degree nnn can be annihilated by Dn+1D^{n+1}Dn+1.
- Here, the degree is 3.
- So, the annihilator is:
D4\boxed{D^4}D4
Explanation:
- D[1+6x−2×3]=6−6x2D[1 + 6x – 2x^3] = 6 – 6x^2D[1+6x−2×3]=6−6×2
- D2[1+6x−2×3]=−12xD^2[1 + 6x – 2x^3] = -12xD2[1+6x−2×3]=−12x
- D3[1+6x−2×3]=−12D^3[1 + 6x – 2x^3] = -12D3[1+6x−2×3]=−12
- D4[1+6x−2×3]=0D^4[1 + 6x – 2x^3] = 0D4[1+6x−2×3]=0
Thus, the operator D4D^4D4 annihilates the polynomial function.
Part 2: f(x)=e−x+2xex−x2exf(x) = e^{-x} + 2x e^x – x^2 e^xf(x)=e−x+2xex−x2ex
This is a combination of two terms:
- e−xe^{-x}e−x
- (2x−x2)ex(2x – x^2)e^x(2x−x2)ex, which is a polynomial times exe^xex
We can treat each separately and take the least common multiple (LCM) of their annihilators.
For e−xe^{-x}e−x:
- D+1D + 1D+1 annihilates e−xe^{-x}e−x because: (D+1)[e−x]=D[e−x]+e−x=−e−x+e−x=0(D + 1)[e^{-x}] = D[e^{-x}] + e^{-x} = -e^{-x} + e^{-x} = 0(D+1)[e−x]=D[e−x]+e−x=−e−x+e−x=0
For (2x−x2)ex(2x – x^2)e^x(2x−x2)ex:
This is a polynomial of degree 2 multiplied by exe^xex.
- A polynomial of degree 2 times exe^xex is annihilated by (D−1)3(D – 1)^3(D−1)3
Why?
Because:
- If f(x)=P(x)exf(x) = P(x)e^xf(x)=P(x)ex, and P(x)P(x)P(x) is degree nnn, then the annihilator is (D−1)n+1(D – 1)^{n+1}(D−1)n+1
So:
- (D−1)3(D – 1)^3(D−1)3 annihilates (2x−x2)ex(2x – x^2)e^x(2x−x2)ex
Now take the LCM of the annihilators:
- For e−xe^{-x}e−x: D+1D + 1D+1
- For (2x−x2)ex(2x – x^2)e^x(2x−x2)ex: (D−1)3(D – 1)^3(D−1)3
So the overall annihilator is: (D+1)(D−1)3\boxed{(D + 1)(D – 1)^3}(D+1)(D−1)3
✅ Final Answers:
- For 1+6x−2×31 + 6x – 2x^31+6x−2×3