Find the relationship between $x_0$ and $y_0$ that ensures $x(t) → 0$ as $t → ∞$ If the roots of the auxiliary equation of a second order linear homogeneous
ODE are $k_1 > 0$ and $−k_2 < 0$ then the solution is
$$x(t) = Ae^{k_1t}+ Be^{−k_2t}.$$
For most choices of initial conditions
$x(0) = x_0$, $\dot x(0) = y_0$
we will have that $x(t) → ±∞$ as $t → ∞$. However, there are some special
initial conditions for which $x(t) → 0$ as $t → ∞$. Find the relationship between
$x_0$ and $y_0$ that ensures this
My working so far is $x_0 = A + B$ and $y_0 = Ak_1 - Bk_2$ and I'm really unsure of how to tackle this question further

If
$x(t) \to 0 \; \text{as} \; x \to \infty, \tag 1$
we have
$A = 0, \tag 2$
lest the term
$Ae^{k_1t} \to \infty \tag 3$
and dominate
$x(t) = Ae^{k_1t} + Be^{-k_2t} \tag 4$
as $t$ increases without bound.  (We note here that
$Be^{-k_2t} \to 0 \; \text{as} \; x \to \infty.) \tag 5$
In light of (2),
$x(0) = x_0 = B \tag 6$
and
$\dot x(0) = y_0 = -Bk_2; \tag 7$
thus the relationship 'twixt $x_0$ and $y_0$ is
$x_0 = B = -\dfrac{y_0}{k_2}, \tag 8$
that is,
$y_0 = -k_2 x_0. \tag 9$
It is also easy to see that this equation forces $A = 0$, for
$x_0 = A + B \Longrightarrow k_1A - k_2B = y_0 = -k_2x_0 = -k_2A - k_2B \Longrightarrow k_1A = -k_2A$
$\Longrightarrow (k_1 + k_2)A \Longrightarrow A = 0, \tag{10}$
since
$k_1 + k_2 > 0; \tag{11}$
thus, (9) implies
$x(t) \to 0 \; \text{as} \; t \to \infty. \tag{12}$
