Question

find a differential operator that annihilates the given function.

$e^{2x}-6e^x$

Short answer

$D^2-3D+2$is the differential operator that annihilates the given function.

Step-by-step solution

Differentiate the function continuously

$h\left(x\right)=-6e^x$Let the function be$f\left(x\right)=e^{2x}-6e^x$

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

Then

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

Let$h\left(x\right)=-6e^x$

Then

$$\begin{gathered} h\text{'}\left(x\right)=-6e^x=h\left(x\right) \\ \Rightarrow h\text{'}\left(x\right)-h\left(x\right)=0 \\ \Rightarrow (D-1)\left[h\right]=0 \end{gathered}$$

Then

$(D-2)\left[f\right]\left(x\right)=(D-2)[g+h]\left(x\right)=(D-2)\left[g\right]\left(x\right)+(D-2)\left[h\right]\left(x\right)=(D-2)\left[h\right]\left(x\right)=6e^x=-h\left(x\right)$

Then

$(D-1)\left[\right(D-2\left)\right[f\left]\right]\left(x\right)=(D-1)[-h]\left(x\right)=-(D-1)\left[h\right]\left(x\right)=0$

Hence$(D-1)(D-2)\left[f\right]=0$

Then$D^2-3D+2$is the differential operator that annihilates the given function.