Question

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

the given function

${{\mathbf{e}}^{\mathbf{-}}}^{\mathbf{x}}\mathbf{+}\mathbf{2}\mathit{x}{\mathit{e}}^{\mathbf{x}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{x}}$

Short answer

$\left(D+1\right){\left(D-1\right)}^3$

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,

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

Which is,

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

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

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

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

$\left(2x-x^2\right)e^x$is annihilated by the linear differential operator ${\left(D-1\right)}^3$, where the greater power of the polynomial $\left(2x-x^2\right)$is n = 2.

Then we can annihilates the given function as

$\left(D+1\right){\left(D-1\right)}^3\left(e^{-x}+2x{e}^x-x^2{e}^x\right)=0$

Hence, the used differential operator is localid="1667831002071" $\left(D+1\right){\left(D-1\right)}^{\mathbf{3}}$