Question

find a differential operator that annihilates the given function.

$x^2-e^x$

Short answer

$D^4-D^3$is the differential operator that annihilates the given function.

Step-by-step solution

Differentiate the function continuously

Let the function be$f\left(x\right)=x^2-e^x$

Let$g\left(x\right)=x^2$

Then

$$\begin{gathered} g\text{'}\left(x\right)=2x \\ g\text{'}\text{'}\left(x\right)=2 \\ g\text{'}\text{'}\text{'}\left(x\right)=0 \\ {D}^3\left[g\right]=0 \end{gathered}$$

Let$h\left(x\right)=-e^x$

Then

$$\begin{gathered} h\text{'}\left(x\right)=-e^x \\ h\text{'}\left(x\right)-h\left(x\right)=0 \\ (D-1)\left[h\right]=0 \end{gathered}$$

Hence$D^3(D-1)\left[f\right]=0$

Then $D^4-D^3$is the differential operator that annihilates the given function.