Question

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

the given function.

$\mathbf{1}\mathbf{+}\mathbf{7}{\mathbf{e}}^{\mathbf{2}\mathbf{x}}$

Short answer

$\mathbf{D}\mathbf{(}\mathbf{D}\mathbf{-}\mathbf{2}\mathbf{)}$

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,

$1+7{\mathrm{e}}^{2\mathrm{x}}=0$

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 ${\mathrm{D}}^{\mathrm{n}+1}\mathrm{and}{\mathrm{e}}^{\mathrm{\alpha x}}$is annihilated by $(\mathrm{D}-\mathrm{\alpha })$, then we annihilates the components of the given function as,

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

$7{\mathrm{e}}^{2\mathrm{x}}$is annihilated by the linear differential operator $(\mathrm{D}-2).$

Then we can annihilates the given function as

$\mathrm{D}(\mathrm{D}-2)(1+7{\mathrm{e}}^{2\mathrm{x}})=0$

Hence, the used differential operator is $\mathrm{D}(\mathrm{D}-2).$