Question

. find a differential operator that annihilates the given function.

$x{e}^{-2x}+x{e}^{-5x}sin3x$

Short answer

$(D+2{)}^2{\left(D^2+10D+34\right)}^2$is the differential operator that annihilates the given function.

Step-by-step solution

Any nonhomogeneous term of the form f(x)=xkeαxcosβx  OR xkeαxsinβxsatisfies (D-α)2+β2m[f]=0for k=0,1,..,m-1

Let the function be$f\left(x\right)=x{e}^{-2x}+x{e}^{-5x}\sin 3x$

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

Then

$(D+2{)}^2\left[g\right]=0$

Let$h\left(x\right)=x{e}^{-5x}\sin 3x$

Then

${\left[{(D+5)}^2+3^2\right]}^2\left[g\right]=0$

Hence$(D+2{)}^2{\left[{(D+5)}^2+3^2\right]}^2\left[f\right]=0$

Then

$$\begin{gathered} (D+2{)}^2{\left[{(D+5)}^2+3^2\right]}^2 \\ =(D+2{)}^2{\left(D^2+10D+25+9\right)}^2 \\ =(D+2{)}^2{\left(D^2+10D+34\right)}^2 \end{gathered}$$

is the differential operator that annihilates the given function.