## Calculus: Early Transcendentals 8th Edition

y = $C_{1}e^{x}$ + $C_{2}e^{-x}$
Question: y=y" We write: y"-y=0 We know: y = $e^{rx}$ y' = $re^{rx}$ y" = $r^{2}e^{rx}$ This results in $(r^{2}-1)e^{rx}=0$ The corresponding characteristic equation is: $r^{2}-1=0$ Solving the equation gives: $r^{2}=1$ $r=\frac{+}{-}1$ Adding the constants gives the following general solution: y = $C_{1}e^{x}$ + $C_{2}e^{-x}$