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Q28E

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

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

See the step by step solution

Step 1: Solve the homogeneous of the given equation

The homogeneous of the given equation is

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

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

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

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)

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