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Q28E

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. ${y}{\text{'}}{\text{'}}{-}{6}{y}{\text{'}}{+}{10}{y}{=}{{e}}^{3x}{-}{x}$**

${y}_{p}\left(x\right)={c}_{3}+{c}_{4}x+{c}_{5}{e}^{3x}$

The homogeneous of the given equation is

$({D}^{2}-6D+10)\left[y\right]=0$

The solution of the homogeneous is

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

Now ${e}^{3x}-x$ is annihilated by $\left({D}^{3}-3{D}^{2}\right)$

Then, every solution to the given nonhomogeneous equation also satisfies

. $\left({D}^{3}-3{D}^{2}\right)({D}^{2}-6D+10)\left[y\right]=0$

Then

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

is the general solution to this homogeneous equation

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

Comparing (1) & (2)

${y}_{p}\left(x\right)={c}_{3}+{c}_{4}x+{c}_{5}{e}^{3x}$

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