## Calculus: Early Transcendentals 8th Edition

y = $C_{1}$ + $C_{2}e^{\frac{4}{3}x}$
Question: 3y"=4y' We write: $3y"- 4y'=0$ We know: y = $e^{rx}$ y' = $re^{rx}$ y" = $r^{2}e^{rx}$ This results in $(3r^{2}-4r)e^{rx}$=0 The corresponding characteristic equation is: $3r^{2}-4r=0$ Solving the equation gives: $r(3r-4)=0$ $r=0$ or $r=\frac{4}{3}$ Adding the constants gives the following general solution: y = $C_{1}e^{0}$ + $C_{2}e^{\frac{4}{3}x}$ $e^{0} = 1$ so y = $C_{1}$ + $C_{2}e^{\frac{4}{3}x}$