Answer
$$
4 y^{\prime \prime}+17 y^{\prime}+4 y=0
$$
The general solution of that equation is given by
$$ y=c_{1} e^{-\frac{1}{4}t} +c_{2} e^{ -4t} $$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
Work Step by Step
$$
4 y^{\prime \prime}+17 y^{\prime}+4 y=0 \quad \quad (1)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
4r^{2}+17 r+4=0,
$$
so its roots are
$$
r_{1,2}=\frac{-17 \pm \sqrt{17^{2}-4 \cdot 4 \cdot 4}}{2 \cdot 4}
$$
Thus the possible values of $r$ are
$r_{1}=-\frac{1}{4} , r_{2}=-4$ ; the general solution of Eq. (1) is
$$ y=c_{1} e^{-\frac{1}{4}t} +c_{2} e^{ -4t} $$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
