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

${y}_{p}\left(x\right)=-\frac{4}{25}x{e}^{x}+\frac{1}{10}{x}^{2}{e}^{x}-\frac{1}{2}$

See the step by step solution

## Step 1: Find the corresponding auxiliaryequation

Theauxiliary equationof corresponding homogeneous equation

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

The solutions of the auxiliary equation are

$r=-1+i,r=-1-i,r=1$

## Step 2: Find particular solution

Let the particular solution be

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

Then

${y}_{p}^{\text{'}}\left(x\right)=a{e}^{x}+\left(a+2b\right)x{e}^{x}+b{x}^{2}{e}^{x}\phantom{\rule{0ex}{0ex}}{y}_{p}^{\text{'}\text{'}}\left(x\right)=\left(2a+2b\right){e}^{x}+\left(a+4b\right)x{e}^{x}+b{x}^{2}{e}^{x}\phantom{\rule{0ex}{0ex}}{y}_{p}^{\text{'}\text{'}\text{'}}\left(x\right)=\left(3a+6b\right){e}^{x}+\left(a+6b\right)x{e}^{x}+b{x}^{2}{e}^{x}$

Then

${y}_{p}^{\text{'}\text{'}\text{'}}\left(x\right)+{y}_{p}^{\text{'}\text{'}}\left(x\right)-2{y}_{p}\left(x\right)\phantom{\rule{0ex}{0ex}}=\left(3a+6b\right){e}^{x}+\left(a+6b\right)x{e}^{x}+b{x}^{2}{e}^{x}+\left(2a+2b\right){e}^{x}+\left(a+4b\right)x{e}^{x}+b{x}^{2}{e}^{x}-2ax{e}^{x}-2b{x}^{2}{e}^{x}-2c\phantom{\rule{0ex}{0ex}}=\left(5a+8b\right){e}^{x}+10bx{e}^{x}-2c$

If $\left(5a+8b\right){e}^{x}+10bx{e}^{x}-2c=x{e}^{x}+1$

Then $5a+8b=0,10b=1and-2c=1$

Then $a=-\frac{4}{25},b=\frac{1}{10}andc=-\frac{1}{2}$

Hence ${y}_{p}\left(x\right)=-\frac{4}{25}x{e}^{x}+\frac{1}{10}{x}^{2}{e}^{x}-\frac{1}{2}$