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Q16E

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

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

See the step by step solution

## Step 1: Differentiate the function continuously

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

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

Then

$g\text{'}\left(x\right)=2x\phantom{\rule{0ex}{0ex}}g\text{'}\text{'}\left(x\right)=2\phantom{\rule{0ex}{0ex}}g\text{'}\text{'}\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}{D}^{3}\left[g\right]=0$

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

Then

$h\text{'}\left(x\right)=-{e}^{x}\phantom{\rule{0ex}{0ex}}h\text{'}\left(x\right)-h\left(x\right)=0\phantom{\rule{0ex}{0ex}}\left(D-1\right)\left[h\right]=0$

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

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