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

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

See the step by step solution

Step 1: Differentiate the function continuously

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

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

Then

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

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

Then

$h\text{'}\left(x\right)=-6{e}^{x}=h\left(x\right)\phantom{\rule{0ex}{0ex}}⇒h\text{'}\left(x\right)-h\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(D-1\right)\left[h\right]=0$

Then

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

Then

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

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

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