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Expert-verified 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.${y}{\text{'}}{\text{'}}{\text{'}}{+}{2}{y}{\text{'}}{\text{'}}{-}{y}{\text{'}}{-}{2}{y}{=}{{e}}^{{x}}{-}{1}$

${y}_{p}\left(x\right)={c}_{1}+{c}_{5}x{e}^{x}$

See the step by step solution

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

The homogeneous of the given equation is

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

The solution of the homogeneous is

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

Now ${e}^{x}-1$ is annihilated by $\left({D}^{2}-D\right)$

Then, every solution to the given nonhomogeneous equation also satisfies

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

Then

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

is the general solution to this homogeneous equation

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

Comparing (1) & (2)

${y}_{p}\left(x\right)={c}_{1}+{c}_{5}x{e}^{x}$ ### Want to see more solutions like these? 