Match each function name with its equation.

/ Exams & Certification / By

Match each function name with its equation.

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 a differential operator L(D)L(D), written in terms of D=ddxD = \frac{d}{dx}, such that: L(D)[f(x)]=0L(D)[f(x)] = 0

We are given two separate functions to analyze:

Function 1: f(x)=1+6x−2x3f(x) = 1 + 6x – 2x^3

This is a polynomial function of degree 3.

Step 1: Recall the rule

Any polynomial of degree nn is annihilated by the differential operator Dn+1D^{n+1}. So, for a polynomial of degree 3: D4[1+6x−2×3]=0D^4[1 + 6x – 2x^3] = 0

✅ Answer:

The linear differential operator that annihilates this function is: D4\boxed{D^4}

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

This is a linear combination of exponential and exponential-polynomial terms. Let’s break it down:

  • e−xe^{-x}: Annihilated by (D+1)(D + 1)
  • 2xex2x e^x: Annihilated by (D−1)2(D – 1)^2
  • −x2ex-x^2 e^x: Annihilated by (D−1)3(D – 1)^3

In general, xneaxx^n e^{ax} is annihilated by (D−a)n+1(D – a)^{n+1}

So, we take the least common multiple (LCM) of all individual annihilators:

  • (D+1)(D + 1) from e−xe^{-x}
  • (D−1)3(D – 1)^3 from 2xex2x e^x and −x2ex-x^2 e^x

✅ Answer:

The full differential operator is: (D+1)(D−1)3\boxed{(D + 1)(D – 1)^3}

✅ Final Answers:

  1. For 1+6x−2×31 + 6x – 2x^3:                                                                                   

D4\boxed{D^4}

  1. For e−x+2xex−x2exe^{-x} + 2x e^x – x^2 e^x:                                                     

(D+1)(D−1)3\boxed{(D + 1)(D – 1)^3}

✍️ Explanation

In differential equations, a linear differential operator like D=ddxD = \frac{d}{dx} acts on a function to produce its derivative. An operator annihilates a function when applying it results in zero. To find such an operator, we look at the structure of the function.

For the first function, 1+6x−2×31 + 6x – 2x^3, it’s a polynomial of degree 3. Each differentiation reduces the degree by 1. After four derivatives, the result is zero. Hence, D4D^4 is the smallest differential operator that annihilates it.

For the second function, e−x+2xex−x2exe^{-x} + 2x e^x – x^2 e^x, each term is exponential or exponential times a polynomial. The term e−xe^{-x} is annihilated by D+1D + 1 because the derivative of e−xe^{-x} is −e−x-e^{-x}. The terms 2xex2x e^x and −x2ex-x^2 e^x are of the form xnexx^n e^x, which are annihilated by (D−1)n+1(D – 1)^{n+1}. Specifically, xexx e^x is annihilated by (D−1)2(D – 1)^2, and x2exx^2 e^x by (D−1)3(D – 1)^3.

To annihilate the whole function, we need a common operator that eliminates all terms. That means we need to take the product of their annihilators: (D+1)(D + 1) and (D−1)3(D – 1)^3. This product ensures that each term individually is sent to zero.

Therefore, the second function is annihilated by (D+1)(D−1)3(D + 1)(D – 1)^3.

Related Posts

Scroll to Top