Answer
$$
9 y^{\prime \prime}+9 y^{\prime}-4 y=0
$$
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.
Work Step by Step
$$
9 y^{\prime \prime}+9 y^{\prime}-4 y=0 \quad \quad (1)
$$
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.
