Question

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

the given function.

$\mathbf{1}\mathbf{+}\mathbf{6}\mathit{x}\mathbf{-}\mathbf{2}{\mathit{x}}^{\mathbf{3}}$

Short answer

D⁴

Step-by-step solution

Annihilator Operator

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

$\mathrm{L}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=0$

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

Using the Differential operators

We have the polynomial function,

$1+6\mathrm{x}-2{\mathrm{x}}^3=0$

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

${\mathrm{x}}^{\mathrm{n}}$is being annihilated by the linear differential operator ${\mathrm{D}}^{\mathrm{n}+1}$, then we annihilates the components of the given function as,

$1={\mathrm{x}}^0$is annihilated by the linear differential operator ${\mathrm{D}}^1$.

$6\mathrm{x}$is annihilated by the linear differential operator ${\mathrm{D}}^2$.

$-2{\mathrm{x}}^3$ is annihilated by the linear differential operator ${\mathrm{D}}^4$.

Since the component of the highest power for the polynomial function is $-2{\mathrm{x}}^3$, then

we have

${\mathrm{D}}^4(1+6\mathrm{x}-2{\mathrm{x}}^3)=0$

Hence, the used differential operator islocalid="1667897931758" ${\mathit{D}}^{\mathbf{4}}$.