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Found in: Page 337

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# find a differential operator that annihilates the given function${{x}}^{{2}}{{e}}^{{x}}{-}{x}{s}{i}{n}{4}{x}{+}{{x}}^{{3}}$

${D}^{4}\left(D-1{\right)}^{3}{\left({D}^{2}+16\right)}^{2}$is the differential operator that annihilates the given function.

See the step by step solution

## Step 1: Any nonhomogeneous term of the form f(x)=xkeαxcosβx OR role="math" localid="1663946799201" xkeαxsinβxsatisfiesrole="math" localid="1663946761280" (D-α)2+β2m[f]=0for K=0,1,2,...,m-1

Let the function be $f\left(x\right)={x}^{2}{e}^{x}-x\mathrm{sin}4x+{x}^{3}$

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

Then

$\left(D-1{\right)}^{3}\left[g\right]=0$

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

Then

${\left({D}^{2}+{4}^{2}\right)}^{2}\left[h\right]=0$

Let $i\left(x\right)={x}^{3}$

Then

${D}^{4}\left[i\right]=0$

Hence

$\left(D-1{\right)}^{3}{\left({D}^{2}+16\right)}^{2}{D}^{4}\left[f\right]=0\phantom{\rule{0ex}{0ex}}⇒{D}^{4}\left(D-1{\right)}^{3}{\left({D}^{2}+16\right)}^{2}\left[f\right]=0$

Then ${D}^{4}\left(D-1{\right)}^{3}{\left({D}^{2}+16\right)}^{2}$ is the differential operator that annihilates the given function.