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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

# use the annihilator method to determinethe form of a particular solution for the given equation.${u}{\text{'}}{\text{'}}{-}{5}{u}{\text{'}}{+}{6}{u}{=}{c}{o}{s}{2}{x}{+}{1}$

${u}_{p}\left(x\right)={c}_{3}+{c}_{4}\mathrm{sin}2x+{c}_{5}\mathrm{cos}2x$

See the step by step solution

## Step 1: Solve the homogeneous of the given equation

The homogeneous of the given equation is

$\left({D}^{2}-5D+6\right)\left[u\right]=0\phantom{\rule{0ex}{0ex}}⇒\left(D-2\right)\left(D-3\right)\left[u\right]=0$

The solution of the homogeneous is

${u}_{h}\left(x\right)={c}_{1}{e}^{2x}+{c}_{2}{e}^{3x}$ (1)

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

Then

$\left({D}^{2}+{2}^{2}\right)\left[g\right]=0\phantom{\rule{0ex}{0ex}}⇒\left({D}^{2}+4\right)\left[g\right]=0$

Let $h\left(x\right)=1$

Then

$D\left[h\right]=0$

Hence

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

Then, every solution to the given nonhomogeneous equation also satisfies

$D\left({D}^{2}+4\right)\left(D-2\right)\left(D-3\right)\left[u\right]=0$

Then

$u\left(x\right)={c}_{1}{e}^{2x}+{c}_{2}{e}^{3x}+{c}_{3}+{c}_{4}\mathrm{sin}2x+{c}_{5}\mathrm{cos}2x$ (2)

is the general solution to this homogeneous equation

We know $u\left(x\right)={u}_{h}+{u}_{p}$

Comparing (1) & (2)

${u}_{p}\left(x\right)={c}_{3}+{c}_{4}\mathrm{sin}2x+{c}_{5}\mathrm{cos}2x$