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: :$q > 0$ {{qed|lemma}} 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: :$\displaystyle \lim_{x \mathop \to \infty} C_2 x e^{m_1 x} = 0$ {{iff}} $m_1 < 0$. 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: :$q > 0$ {{qed|lemma}} 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: {{begin-eqn}} {{eqn | l = p       | r = -\paren {a + i b + a - i b}       | c = Sum of Roots of Quadratic Equation }} {{eqn | r = -2 a       | c =  }} {{eqn | ll= \leadsto       | l = p       | o = >       | r = 0       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = q       | r = \paren {a + i b} \paren {a - i b}       | c = Product of Roots of Quadratic Equation }} {{eqn | r = a^2 + b^2       | c =  }} {{eqn | ll= \leadsto       | l = q       | o = >       | r = 0       | c = whatever $a$ and $b$ are }} {{end-eqn}} {{qed|lemma}} 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:  :$\displaystyle \lim_{x \mathop \to \infty} C_1 e^{m_1 x} + C_2 e^{m_2 x} = 0$ {{qed|lemma}} 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: :$\displaystyle \lim_{x \mathop \to \infty} C_2 x e^{m_1 x} = 0$ {{iff}} $m_1 < 0$. It follows that:  :$\displaystyle \lim_{x \mathop \to \infty} C_1 e^{m_1 x} + C_2 x e^{m_1 x} = 0$ {{qed|lemma}} 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:  :$\displaystyle \lim_{x \mathop \to \infty} e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} = 0$ {{qed|lemma}} Thus in all cases: :$\displaystyle \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$ {{qed}}