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.
