Question

In Problems 15–26 find a linear differential operator that annihilates

the given function.

1 + sin x

Short answer

D(D² + 1)

Step-by-step solution

Annihilator Operator

If L is a linear differential operator with constant coefficients and f is a sufficiently differentiable function such that

L(f(x)) = 0

then L is said to be an annihilator of the function.

Using the Differential operators

We have the function,

1 + sin x

And we have to annihilate it using a linear differential operator as the following technique:

xⁿ is being annihilated by the linear differential operator Dⁿ⁺¹ and $\sin \beta \chi$is being annihilated by the linear differential operator $\left(D^2+{\beta }^2\right)$, then we annihilates the components of the given function as,

1 = x^o annihilated by the linear differential operator D¹.

sin x is annihilated by the linear differential operator $\left(D^2+{\beta }^2\right)=\left(D^2+1\right).$

Then we can annihilates the given function as

$D\left(D^2+1\right)\left(1+\sin x\right)=0$

Hence, the used differential operator is D(D² + 1)