# How to show that all solutions of $ay''+by'+cy=0$ approach $0$ as $t \rightarrow \infty$

Given $ay''+by'+cy=0$ and assuming that $a, b, c > 0$ show that all solutions approach $0$ as $t\rightarrow\infty$

I was able to begin by seperating the problem into three cases:

Case 1: Repeated roots of the characteristic equation

I was able to find the limit as $t\rightarrow\infty$ for the general solution to Case 1 using l'Hospital's Rule: Fairly straightforward.

Case 2: Imaginary roots of the characteristic equation

I was able to find the limit as $t\rightarrow\infty$ for the general solution to Case 2 using the squeeze theorem: Also fairly straightforward.

Case 3: Two distinct real roots of the characteristic equation

In this case the two roots are: $$r_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ $$r_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$

and the general solution is: $$y=c_1e^{r_1t}+c_2e^{r_2t}$$

The only way I see that $\lim_{t\to\infty} y=0$ is if $r_1,r_2<0$. However if the only information I am given is that $a,b,c>0$ how can I prove that $r_1,r_2<0$?

The characteristic polynomial is $aX^2+bX+c$ where $a,b,c>0$. For $r \geq 0$, $ar^2+br+c>0$ so any real root is negative. This argument works even in the higher degree case.

You know $r_2 \lt 0$ because both terms in the numerator are negative and the denominator is positive. Next you need to prove $r_1 \lt 0$, which boils down to the condition that $b > \sqrt{b^2-4ac}$.

• But how can it be shown that $b>\sqrt{b^2-4ac}$ if I'm only given that all coefficients are positive? Mar 11, 2015 at 1:20
• Clearly $b^2>b^2-4ac$ since $4ac$ is positive. So $b>\sqrt{b^2-4ac}$.
– ET93
Mar 11, 2015 at 1:38

In fact you can see that for the cases

$b^2-4ac = 0\, \quad \rm and\quad b^2 - 4ac < 0$

the limit goes to $0$. Try to study the other case.