Find a linear differential operator that annihilates the given function

/ Exams & Certification / By

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, we look for a differential operator L=p(D)L = p(D) such that L[f(x)]=0L[f(x)] = 0, where D=ddxD = \frac{d}{dx}.

Given Functions:

  1. f1(x)=1+6x−2x3f_1(x) = 1 + 6x – 2x^3
  2. f2(x)=e−x+2xex−x2exf_2(x) = e^{-x} + 2x e^x – x^2 e^x

Step 1: Annihilator for f1(x)=1+6x−2x3f_1(x) = 1 + 6x – 2x^3

This is a polynomial of degree 3.

A standard result:

The annihilator of a polynomial of degree nn is Dn+1D^{n+1}, because successive derivatives of a polynomial eventually reduce it to zero.

Here, f1(x)f_1(x) is degree 3. So its annihilator is: D4\boxed{D^4}

Because:

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

Step 2: Annihilator for f2(x)=e−x+2xex−x2exf_2(x) = e^{-x} + 2x e^x – x^2 e^x

Let’s break it down:

  • e−xe^{-x} → annihilator: D+1D + 1
  • 2xex2x e^x, x2exx^2 e^x → both are of the form P(x)exP(x)e^x, where P(x)P(x) is a polynomial.

A general rule:

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

So:

  • 2xex2x e^x → annihilator: (D−1)2(D – 1)^2
  • x2exx^2 e^x → annihilator: (D−1)3(D – 1)^3

So for the full function:

  • e−x+(2x−x2)exe^{-x} + (2x – x^2) e^x

The total annihilator is the least common multiple (LCM) of:

  • D+1D + 1 (for e−xe^{-x})
  • (D−1)3(D – 1)^3 (for x2exx^2 e^x)

So the annihilator is: (D+1)(D−1)3\boxed{(D + 1)(D – 1)^3}

Final Answer:

  • Annihilator of 1+6x−2×31 + 6x – 2x^3:                     D4\boxed{D^4}
  • Annihilator of e−x+2xex−x2exe^{-x} + 2x e^x – x^2 e^x:   (D+1)(D−1)3\boxed{(D + 1)(D – 1)^3}

Explanation

Annihilators are differential operators that send a function to zero. These are especially useful in solving differential equations using the method of annihilators.

The first function, f1(x)=1+6x−2x3f_1(x) = 1 + 6x – 2x^3, is a polynomial of degree 3. Derivatives of a polynomial reduce its degree by 1 at each step. Thus, applying DD (i.e., taking the derivative) four times will reduce this function to zero. Therefore, D4D^4 is the minimal annihilator. That is, D4[f1(x)]=0D^4[f_1(x)] = 0.

The second function, f2(x)=e−x+2xex−x2exf_2(x) = e^{-x} + 2x e^x – x^2 e^x, is a combination of exponential terms, some of which are multiplied by polynomials. For such functions, the annihilator depends on both the exponential part and the polynomial degree.

  • For e−xe^{-x}, the annihilator is D+1D + 1, since D[e−x]=−e−xD[e^{-x}] = -e^{-x}.
  • For xnexx^n e^x, the annihilator is (D−1)n+1(D – 1)^{n+1}. Hence:
    • 2xex2x e^x is annihilated by (D−1)2(D – 1)^2
    • x2exx^2 e^x is annihilated by (D−1)3(D – 1)^3

Since the function contains both e−xe^{-x} and terms with exe^x, we need a differential operator that annihilates all components. The least common multiple (LCM) of all the individual annihilators gives the annihilator for the entire function.

Thus, the annihilator of f2(x)f_2(x) is (D+1)(D−1)3(D + 1)(D – 1)^3, which when applied to f2(x)f_2(x), gives zero.

Related Posts

Scroll to Top