 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q28E

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

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

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)

${y}_{p}\left(x\right)={c}_{3}+{c}_{4}x+{c}_{5}{e}^{3x}$ ### Want to see more solutions like these? 