Condition for Solutions to Constant Coefficient Homogeneous LSOODE to tend to Zero

Theorem
Let:
 * $(1): \quad y'' + p y' + q y = 0$

be a constant coefficient homogeneous linear second order ODE.

Let the general solution to $(1)$ be $\map y {x, C_1, C_2}$.

Then:
 * $\displaystyle \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$


 * $p$ and $q$ are both strictly positive.

Proof
By Solution of Constant Coefficient Homogeneous LSOODE, $y$ is in one of the following forms:
 * $y = \begin{cases}

C_1 e^{m_1 x} + C_2 e^{m_2 x} & : p^2 > 4 q \\ & \\ C_1 e^{m_1 x} + C_2 x e^{m_2 x} & : p^2 = 4 q \\ & \\ e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} & : p^2 < 4 q \end{cases}$ where:
 * $m_1$ and $m_2$ are the roots of the auxiliary equation $m^2 + p m + q = 0$
 * $a + i b = m_1$
 * $a - i b = m_2$

Sufficient Condition
Let:
 * $\displaystyle \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$

Let $y$ be of the form:
 * $y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$

Then it follows that $m_1 < 0$ and $m_2 < 0$.

From Sum of Roots of Quadratic Equation:
 * $p = -\paren {m_1 + m_2}$

from which it follows that:
 * $p > 0$

From Product of Roots of Quadratic Equation:
 * $q = m_1 m_2$

from which it follows that:
 * $q > 0$

Let $y$ be of the form:
 * $y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$

From Limit at Infinity of Polynomial over Complex Exponential:
 * $\displaystyle \lim_{x \mathop \to \infty} C_2 x e^{m_1 x} = 0$

$m_1 < 0$.

Thus it follows that $m_1 < 0$.

Again from Sum of Roots of Quadratic Equation:
 * $p = -\paren {2 m_1}$

from which it follows that:
 * $p > 0$

From Product of Roots of Quadratic Equation:
 * $q = m_1^2$

from which it follows that:
 * $q > 0$

Let $y$ be of the form:
 * $y = e^{a x} \paren {C_1 \sin b x + C_2 \cos b x}$

Thus it follows that $a < 0$.

From Sum of Roots of Quadratic Equation:

Then:

Thus it is seen that in all three cases:
 * $\displaystyle \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$

implies that $p$ and $q$ are both strictly positive.

Necessary Condition
Let $p$ and $q$ both be strictly positive.

Let $p^2 > 4 q$.

Then both $m_1$ and $m_2$ are real and unequal.

Thus from Solution of Constant Coefficient Homogeneous LSOODE:
 * $y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$

From Product of Roots of Quadratic Equation:
 * $m_1 m_2 = q > 0$

and so either $m_1$ and $m_2$ are either both strictly positive or strictly negative.

It follows that $m_1 + m_2$ is also either strictly positive or strictly negative.

From Sum of Roots of Quadratic Equation:
 * $-\paren {m_1 + m_2} = p > 0$

and so $m_1 + m_2 < 0$

Hence both $m_1$ and $m_2$ are strictly negative.

It follows that:
 * $\displaystyle \lim_{x \mathop \to \infty} C_1 e^{m_1 x} + C_2 e^{m_2 x} = 0$

Let $p^2 = 4 q$.

Then from Solution to Quadratic Equation with Real Coefficients:
 * $m_1 = m_2 = -\frac p 2$

and so:
 * $y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$

From Limit at Infinity of Polynomial over Complex Exponential:
 * $\displaystyle \lim_{x \mathop \to \infty} C_2 x e^{m_1 x} = 0$

$m_1 < 0$.

It follows that:
 * $\displaystyle \lim_{x \mathop \to \infty} C_1 e^{m_1 x} + C_2 x e^{m_1 x} = 0$

Let $p^2 < 4 q$.

Then both $m_1$ and $m_2$ are complex and unequal:
 * $m_1 = a + i b = -\dfrac p 2 + i \dfrac {\sqrt {4 q - p^2} } 2$
 * $m_2 = a + i b = -\dfrac p 2 - i \dfrac {\sqrt {4 q - p^2} } 2$

Then from Solution of Constant Coefficient Homogeneous LSOODE:
 * $y = e^{a x} \paren {C_1 \sin b x + C_2 \cos b x}$

As $p > 0$ it follows that $a < 0$.

Thus it follows that:
 * $\displaystyle \lim_{x \mathop \to \infty} e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} = 0$

Thus in all cases:
 * $\displaystyle \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$