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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 ${\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{5}}{\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{6}}{\mathbit{y}}{\mathbf{=}}{{\mathbit{e}}}^{\mathbf{3}\mathbf{x}}{\mathbf{-}}{{\mathbit{x}}}^{{\mathbf{2}}}$

$y\text{'}\text{'}-5y\text{'}+6y={e}^{3x}-{x}^{2}$

See the step by step solution

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

The homogeneous of the given equation is

$\left({D}^{2}-5D+6\right)\left[y\right]=\left(D-2\right)\left(D-3\right)\left[y\right]=0$

The solution of the homogeneous is

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

Let $g\left(x\right)={e}^{3x}$

Then

$\left(D-3\right)\left[g\right]=0$

Let $h\left(x\right)={x}^{2}$

Then

${D}^{3}\left[h\right]=0$

Hence

${D}^{3}\left(D-3\right)\left[g-h\right]=0$

Then, every solution to the given nonhomogeneous equation also satisfies

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

Then

$y\left(x\right)={c}_{1}{e}^{2x}+{c}_{2}{e}^{3x}+{c}_{3}x{e}^{3x}+{c}_{4}+{c}_{5}x+{c}_{6}{x}^{2}$ (2)

is the general solution to this homogeneous equation

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

Comparing (1) & (2)

${y}_{p}\left(x\right)={c}_{3}x{e}^{3x}+{c}_{4}+{c}_{5}x+{c}_{6}{x}^{2}$

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