Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
    
homework should not be used as a standalone tag; see tag-wiki and meta. –  Martin Sleziak Feb 27 '13 at 10:54

2 Answers 2

up vote 2 down vote accepted

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

share|improve this answer
    
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 $P^2 \geq 4Q$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.