## Calculus: Early Transcendentals 8th Edition

y = $C_1e^{3x}$ + $C_2xe^{3x}$
Question: y"-6y'+9y=0 We know: y = $e^{rx}$ y' = $re^{rx}$ y" = $r^{2}e^{rx}$ This results in $(r^{2}-6r+9)e^{rx}$=0 The corresponding characteristic equation is: $r^{2}-6r+9=0$ Factoring gives: $(r-3)(r-3) = 0$ r=3 and r=3 Because we have the same solution twice we need to add an extra x to one of the answers. Adding the constants gives the following general solution: y = $C_1e^{3x}$ + $C_2xe^{3x}$