Question

In Problems 15–26 find a linear differential operator that annihilates the given function.

${\left(2-e^x\right)}^{\mathbf{2}}$

Short answer

$D\left(D-1\right)\left(D-2\right)$

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

$L\left(f\left(x\right)\right)=0$

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

Using the Differential operators

We have the function,

${\left(2-e^x\right)}^2$

Which is,

$4-4e^x+e^{2x}$

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

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

$4=4x^0$is annihilated by the linear differential operator $D^1$.

$-4e^x$is annihilated by the linear differential operator $\left(D-1\right)$, where $\alpha =1$.

$e^{2x}$is annihilated by the linear differential operator $\left(D-2\right)$, where $\alpha =2.$

Then we can annihilates the given function as

$D\left(D-1\right)\left(D-2\right)\left(4-4e^x+e^{2x}\right)=0$

$D\left(D-1\right)\left(D-2\right){\left(2-e^x\right)}^2=0$

Hence, the used differential operator is$\mathit{D}\left(D-1\right)\left(D-2\right)$.