Find a linear differential operator that annihilates the given function

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 solve this problem, we are asked to find linear differential operators that annihilate given functions. That means we are to find differential operators L=anDn+an−1Dn−1+…+a0L = a_n D^n + a_{n-1} D^{n-1} + \ldots + a_0, such that L[f(x)]=0L[f(x)] = 0. Here, DD denotes the differential operator ddx\frac{d}{dx}.

Function 1:

f(x)=1+6x−2x3f(x) = 1 + 6x – 2x^3

This is a polynomial of degree 3. We know that the (n+1)(n+1)th derivative of a degree nn polynomial is zero.

D[1+6x−2×3]=6−6x2D[1 + 6x – 2x^3] = 6 – 6x^2
D2[1+6x−2×3]=−12xD^2[1 + 6x – 2x^3] = -12x
D3[1+6x−2×3]=−12D^3[1 + 6x – 2x^3] = -12
D4[1+6x−2×3]=0D^4[1 + 6x – 2x^3] = 0

Thus, the operator D4D^4 annihilates the function:

✅ Answer: D4D^4

Function 2:

f(x)=e−x+2xex−x2exf(x) = e^{-x} + 2x e^x – x^2 e^x

This function is a linear combination of:

e−xe^{-x}
xexx e^x
x2exx^2 e^x

We use a method called annihilator method:

The annihilator of eaxe^{ax} is D−aD – a
The annihilator of xneaxx^n e^{ax} is (D−a)n+1(D – a)^{n+1}

So:

e−x⇒e^{-x} \Rightarrow annihilator: D+1D + 1
xex⇒x e^x \Rightarrow annihilator: (D−1)2(D – 1)^2
x2ex⇒x^2 e^x \Rightarrow annihilator: (D−1)3(D – 1)^3

Now, to annihilate all terms in the function, we must take the least common multiple (LCM) of all these operators:

LCM[(D+1),(D−1)3]=(D+1)(D−1)3\text{LCM}[(D + 1), (D – 1)^3] = (D + 1)(D – 1)^3

✅ Answer: (D+1)(D−1)3(D + 1)(D – 1)^3

✨ Explanation (300+ words)

To solve the problem of finding a linear differential operator that annihilates a given function, we need to understand how different types of functions behave under repeated differentiation.

For the first function:

f(x)=1+6x−2x3f(x) = 1 + 6x – 2x^3
This is a polynomial. The highest power is x3x^3, so its fourth derivative will be zero. This is because:

The first derivative reduces the degree of each term by 1.
Repeating differentiation will eventually remove all terms.

For example:

D[f]=6−6x2D[f] = 6 – 6x^2
D2[f]=−12xD^2[f] = -12x
D3[f]=−12D^3[f] = -12
D4[f]=0D^4[f] = 0

So, D4D^4 is the simplest operator that annihilates this polynomial.

For the second function:

f(x)=e−x+2xex−x2exf(x) = e^{-x} + 2x e^x – x^2 e^x
This is a linear combination of exponential functions and exponential-polynomial products.

To handle such functions, we apply the annihilator method:

The annihilator for eaxe^{ax} is D−aD – a
For xneaxx^n e^{ax}, the annihilator is (D−a)n+1(D – a)^{n+1}

So, analyze each term:

e−xe^{-x}: use D+1D + 1
xexx e^x: use (D−1)2(D – 1)^2
x2exx^2 e^x: use (D−1)3(D – 1)^3

The operator that can annihilate all three terms must handle both e−xe^{-x} and x2exx^2 e^x. This leads to taking the least common multiple:

LCM[(D+1),(D−1)3]=(D+1)(D−1)3\text{LCM}[(D + 1), (D – 1)^3] = (D + 1)(D – 1)^3

This operator, when applied to the function, will give zero. Hence, it’s the required linear differential operator.

✅ Final Answers:

For f(x)=1+6x−2x3f(x) = 1 + 6x – 2x^3: D4D^4
For f(x)=e−x+2xex−x2exf(x) = e^{-x} + 2x e^x – x^2 e^x: (D+1)(D−1)3(D + 1)(D – 1)^3