Elementary Differential Equations and Boundary Value Problems 9th Edition

$$9 y^{\prime \prime}+9 y^{\prime}-4 y=0$$ The general solution of that equation is given by $$y=c_{1} e^{ \frac{1}{3} t} +c_{2} e^{ -\frac{4}{3}t}$$ where $c_{1}$and $c_{2}$ are arbitrary constants.
$$9 y^{\prime \prime}+9 y^{\prime}-4 y=0 \quad \quad (1)$$ We assume that $y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation $$9r^{2}+9 r-4=0,$$ so its roots are $$r_{1,\:2}=\frac{-9\pm \sqrt{9^2-4\cdot \:9\left(-4\right)}}{2\cdot \:9}$$ Thus the possible values of $r$ are $$r_{1}= \frac{1}{3} ,\quad r_{2}=-\frac{4}{3} .$$ the general solution of Eq. (1) is $$y=c_{1} e^{ \frac{1}{3} t} +c_{2} e^{ -\frac{4}{3}t}$$ where $c_{1}$and $c_{2}$ are arbitrary constants.