## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 9 - Section 9.1 - Modeling with Differential Equations - 9.1 Exercises - Page 590: 3

#### Answer

(a) $r = \frac{1}{2}, -1$ (b) Every member of this family of solutions $y = ae^{r_1x}+be^{r_2x}$ is also a solution.

#### Work Step by Step

(a) $y = e^{rx}$ We can find the values of $r$: $2y''+y'-y=0$ $2r^2~e^{rx}+r~e^{rx}-e^{rx}=0$ $2r^2+r-1=0$ $(2r-1)(r+1) = 0$ $r = \frac{1}{2}, -1$ (b) $y = ae^{r_1x}+be^{r_2x}$ We can evaluate the differential equation: $2y''+y'-y$ $= 2(ar_1^2e^{r_1x}+br_2^2e^{r_2x})+(ar_1e^{r_1x}+br_2e^{r_2x})-(ae^{r_1x}+be^{r_2x})$ $= (2ar_1^2e^{r_1x}+ ar_1e^{r_1x}-ae^{r_1x}) +(2br_2^2e^{r_2x}+br_2e^{r_2x}-be^{r_2x})$ $= a(2r_1^2e^{r_1x}+ r_1e^{r_1x}-e^{r_1x}) +b(2r_2^2e^{r_2x}+r_2e^{r_2x}-e^{r_2x})$ $= a(0) +b(0)$ $= 0$ Then every member of this family of solutions $y = ae^{r_1x}+be^{r_2x}$ is also a solution.

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