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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 method of undetermined coefficients to determine the form of a particular solution for the given equation.${y}{\text{'}}{\text{'}}{\text{'}}{-}{2}{y}{\text{'}}{\text{'}}{-}{5}{y}{\text{'}}{+}{6}{y}{=}{{e}}^{{x}}{+}{{x}}^{{2}}$

${y}_{p}\left(x\right)=-\frac{1}{6}x{e}^{x}+\frac{37}{108}+\frac{5x}{18}+\frac{{x}^{2}}{6}$

See the step by step solution

## Step 1: Find the corresponding auxiliaryequation

Theauxiliary equationof corresponding homogeneous equation

${r}^{3}-2{r}^{2}-5r+6=\left(r-1\right)\left(r-3\right)\left(r+2\right)=0$

The solutions of the auxiliary equation are

$r=-2,r=1andr=3$

## Step 2: Find particular solution

Let the particular solution be

${y}_{p}\left(x\right)=ax{e}^{x}+b+cx+d{x}^{2}$

Then

${y}_{p}^{\text{'}}\left(x\right)=a{e}^{x}+ax{e}^{x}+c+2dx\phantom{\rule{0ex}{0ex}}{y}_{p}^{\text{'}\text{'}}\left(x\right)=2a{e}^{x}+ax{e}^{x}+2d\phantom{\rule{0ex}{0ex}}{y}_{p}^{\text{'}\text{'}\text{'}}\left(x\right)=3a{e}^{x}+ax{e}^{x}\phantom{\rule{0ex}{0ex}}$

Then

${y}_{p}^{\text{'}\text{'}\text{'}}\left(x\right)-2{y}_{p}^{\text{'}\text{'}}\left(x\right)-5{y}_{p}^{\text{'}}\left(x\right)+6{y}_{p}\left(x\right)\phantom{\rule{0ex}{0ex}}=3a{e}^{x}+ax{e}^{x}-4a{e}^{x}-2ax{e}^{x}-4d-5a{e}^{x}-5ax{e}^{x}-5c-10dx+6ax{e}^{x}+6b+6cx+6d{x}^{2}\phantom{\rule{0ex}{0ex}}=-6a{e}^{x}+\left(6b-5c-4d\right)+\left(6c-10d\right)x+6d{x}^{2}$

If $-6a{e}^{x}+\left(6b-5c-4d\right)+\left(6c-10d\right)x+6d{x}^{2}={e}^{x}+{x}^{2}$

Then $-6a=1,6b-5c-4d=0,6c-10d=0,and6d=1$

Then $a=-\frac{1}{6},b=\frac{37}{108},c=\frac{5}{18}andd=\frac{1}{6}$

Hence

${y}_{p}\left(x\right)=-\frac{1}{6}x{e}^{x}+\frac{37}{108}+\frac{5x}{18}+\frac{{x}^{2}}{6}$ ### Want to see more solutions like these? 