# the general solution of the equation $y^{\prime \prime} + Py^{\prime} +Qy=0$, approaches zero as $x$ approaches $\infty$

Show that the general solution of the equation $$y^{\prime \prime} + Py^{\prime} +Qy=0$$ where P and Q are constants, approaches zero as $x$ approaches $\infty$ if and only if P ,Q are both positive. I have no idea on how to prove this. Anyone has any idea?

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You could solve the equation. Recall that one uses the polynomial $r^2+Pr+Q=0$. Write down the roots. At least if the roots are distinct, the (complex) general solution is $Ae^{r_1 t}+Be^{r_2 t}$ where $r_1$ and $r_2$ are the roots. (If there is a double root $r_1$, the general solution is $Ae^{r_1 t}+Bte^{r_1 t}$.) – André Nicolas Feb 18 '13 at 7:15

Hint: The general solution is $$y = c_1e^{ax} + c_2e^{bx}$$ What is the relationship between $a, b$ and $P, Q$?

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The solution is only for real roots right? Then how about complex roots? – Idonknow Feb 18 '13 at 7:21
@Idonknow: This is the solution for complex roots as well, you just have to interpret the exponential as a complex function. The only case this doesn't cover is a repeated real root. – Jim Feb 18 '13 at 7:27

Here is one way.

Let $x_1 = y', x_2 = y$. Then the equation is equivalent to $\dot{x} = \begin{bmatrix} -P & -Q \\ 1 & 0\end{bmatrix} x = Ax$. By looking at the Jordan form of $A$, we see that all solutions converge to $0$ iff all eigenvalues of $A$ have negative real parts.

A quick calculation shows that the eigenvalues are given by solutions to $\lambda^2+P \lambda + Q=0$, that is, $\lambda = \frac{1}{2} (-P \pm \sqrt{P^2-4Q})$, hence you need to find conditions on $P,Q$ that are equivalent to $\text{Re}( -P \pm \sqrt{P^2-4Q}) < 0$.

It may help to split the condition into $P^2 < 4Q$ and $P^2 \geq 4Q$.

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