Answer
$$
y^{\prime \prime}-2 y^{\prime}+10 y=0
$$
The general solution of that equation is given by
$$
y(t) =e^{t}( c_{1}\cos 3t+ c_{2} \sin 3t)
$$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
Work Step by Step
$$
y^{\prime \prime}-2 y^{\prime}+10 y=0 \quad \quad (1)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}- 2r+10=0,$$
so its roots are
$$r_{1}=1+3i ,\quad r_{2}=1-3i.
$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(t) = e^{1+3i } t=e^{t}(\cos 3t + i \sin 3t)
$$
and
$$
y_{2}(t) = e^{1-3i } t=e^{t}( \cos 3t - i \sin 3t)
$$
Thus the general solution of the differential equation is
$$
y(t) =e^{t}( c_{1}\cos 3t+ c_{2} \sin 3t)
$$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
