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 ( L(D) ) that annihilates a given function, we look for a differential equation of the form:
[
L(D) y = 0
]
where ( L(D) ) is a polynomial in the differentiation operator ( D ) (where ( D = \frac{d}{dx} )), and ( y(x) ) is the given function.
1st Function: ( f(x) = 1 + 6x – 2x^3 )
This is a polynomial function. The general rule is that a polynomial of degree ( n ) is annihilated by ( D^{(n+1)} ). Here, the highest degree term is ( -2x^3 ), so we use:
[
D^{4} (1 + 6x – 2x^3) = 0
]
Thus, the annihilating operator is:
[
L(D) = D^4
]
2nd Function: ( g(x) = e^{-x} + 2x e^x – x^2 e^x )
This function consists of terms involving exponentials and polynomials. The general rule is:
- ( e^{\lambda x} ) is annihilated by ( D – \lambda ).
- ( x^n e^{\lambda x} ) is annihilated by ( (D – \lambda)^{n+1} ).
In our case:
- ( e^{-x} ) is annihilated by ( D + 1 ).
- ( 2x e^x ) is annihilated by ( (D – 1)^2 ).
- ( -x^2 e^x ) is annihilated by ( (D – 1)^3 ).
Thus, the minimal annihilating operator for ( g(x) ) is the least common multiple (LCM) of these:
[
L(D) = (D + 1)(D – 1)^3
]
Explanation:
A differential operator annihilates a function when applying the operator results in zero. For polynomials, the order of the derivative required is one more than the polynomial degree. For exponentials, a first-order differential operator suffices, but if multiplied by ( x^n ), we need ( (D – \lambda)^{n+1} ). By finding the LCM of all required factors, we determine the minimal operator.