Elementary Differential Equations and Boundary Value Problems 9th Edition

$y=8+8e^{\frac{-1}{2}t}-18e^{\frac{-1}{3}t}$
Let $\;\;\;\;\;y=e^{rt}\\\\$ $6{y}'''+5{y}''+{y}'=0 \;\;\;\;\Rightarrow \;\;\;\; 6r^3e^{rt}+5r^2e^{rt}+re^{rt}=0\\\\$ $6r^3+5r^2+r=r(2r+1)(3r+1)=0$$\rightarrow\;\;\;\;\; r_{1}= 0\;\;\;\;\;or\;\;\;r_{2}=\frac{-1}{2}\;\;\;or\;\;\;\;r_{3}=\frac{-1}{3} \;\;\;\;\;\\\\$ So the 3 roots is: $\;\;\;r_{1}=0 \;\;\;,\;\;r_{2}=\frac{-1}{2} \;\;\;,\;\;\;r_{3}=\frac{-1}{3}$ $y= C_{1}+C_{2}e^{\frac{-1}{2}t}+C_{3}e^{\frac{-1}{3}t}$ Derivatives of the general solution; ${y}'=-\frac{1}{2}C_{2}e^{\frac{1}{2}t}-\frac{1}{3} C_{3}e^{\frac{-1}{3}t}$ ${y}''=\frac{1}{4}C_{2}e^{\frac{-1}{2}t}+\frac{1}{9} C_{3}e^{\frac{-1}{3}t}$ At; $y(0)=C_{1}+C_{2}+C_{3}=-2$ ${y}'(0)=-\frac{1}{2}C_{2}-\frac{1}{3} C_{3}=2$ ${y}''(0)=\frac{1}{4}C_{2}+\frac{1}{9} C_{3}=0$ $\;\;\;\;\;\Rightarrow \;\;\;\;C_{1}=8\;\;\;\;\;\;\;C_{2}=8\;\;\;\;\;\;C_{3}=-18 \;\;\;\;\;\;$ $\therefore y=8+8e^{\frac{-1}{2}t}-18e^{\frac{-1}{3}t}$