Find the relationship between $x_0$ and $y_0$ that ensures $x(t) → 0$ as $t → ∞$

Question

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

Answer 1

The second term always converges to zero for $t\to\infty$. The first term always diverges at the same time if $A\ne 0$. Thus you need $A=0$, elimination of $B$ gives the desired relation.

Answer 2

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

$x(t) \to 0 \; \text{as} \; t \to \infty. \tag{12}$