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Q21E

Expert-verifiedFound in: Page 337

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.${u}{\text{'}}{\text{'}}{-}{5}{u}{\text{'}}{+}{6}{u}{=}{c}{o}{s}{2}{x}{+}{1}$**

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

The homogeneous of the given equation is

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

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

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

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

Then

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

Hence

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

Then, every solution to the given nonhomogeneous equation also satisfies

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

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)

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

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