# I could use a second opinion on the completeness of my solution...

• Jul 2nd 2011, 07:53 AM
Lancet
I could use a second opinion on the completeness of my solution...
I could use a second opinion on my answer to this question:

Determine the values of $\displaystyle a$, if any, for which all solutions to the equation:

$\displaystyle y'' - (2a - 1)y' + a(a - 1)y = 0$

...tend to zero as $\displaystyle t$ approaches infinity.

The general solution comes out to this:

$\displaystyle y = \frac{C_1e^{at}}{e^t} + C_2e^{at}$

So, my answer would be this:

$\displaystyle 0 < a < 1$

if

$\displaystyle C_2 = 0$

$\displaystyle a = 0$

if

$\displaystyle C_2 \neq 0$

Is my answer correct, or did I miss something?
• Jul 2nd 2011, 08:14 AM
FernandoRevilla
Re: I could use a second opinion on the completeness of my solution...
It is not correct. The condition is $\displaystyle \lim_{t\to +\infty}x(t)=0$ for all solutions $\displaystyle x(t)$ . Then, you have to find the values of $\displaystyle a$ such that $\displaystyle \lim_{t\to +\infty}(C_1e^{(a-1)t}+C_2e^{at})=0$ for all $\displaystyle C_1,C_2$
• Jul 2nd 2011, 08:32 AM
Lancet
Re: I could use a second opinion on the completeness of my solution...
I'm not quite sure how to proceed, then.
• Jul 2nd 2011, 08:39 AM
FernandoRevilla
Re: I could use a second opinion on the completeness of my solution...
Quote:

Originally Posted by Lancet
I'm not quite sure how to proceed, then.

Use the following: $\displaystyle \lim_{t\to +\infty}e^{\beta t}=0\Leftrightarrow \beta <0$ .
• Jul 2nd 2011, 09:01 AM
Lancet
Re: I could use a second opinion on the completeness of my solution...
So is it correct that the general answer would be:

$\displaystyle a < 0$
• Jul 2nd 2011, 09:03 AM
FernandoRevilla
Re: I could use a second opinion on the completeness of my solution...
Quote:

Originally Posted by Lancet
So is it correct that the general answer would be: $\displaystyle a < 0$

Right.
• Jul 2nd 2011, 09:09 AM
Lancet
Re: I could use a second opinion on the completeness of my solution...
Quote:

Originally Posted by FernandoRevilla
Right.

Thank you very much for your help!
• Jul 2nd 2011, 09:15 PM
Lancet
Re: I could use a second opinion on the completeness of my solution...
Quick followup:

If a second question for this same problem asked for the values of $\displaystyle a$ for which all nonzero solutions become unbounded as $\displaystyle t$ approaches infinity, would that just mean $\displaystyle a > 0$ ?
• Jul 2nd 2011, 09:43 PM
FernandoRevilla
Re: I could use a second opinion on the completeness of my solution...
Quote:

Originally Posted by Lancet
If a second question for this same problem asked for the values of $\displaystyle a$ for which all nonzero solutions become unbounded as $\displaystyle t$ approaches infinity, would that just mean $\displaystyle a > 0$ ?

It is not right. Choose for instance $\displaystyle C_1=1,C_2=0, a=1/2$ .
• Jul 2nd 2011, 10:02 PM
Lancet
Re: I could use a second opinion on the completeness of my solution...
Quote:

Originally Posted by FernandoRevilla
It is not right. Choose for instance $\displaystyle C_1=1,C_2=0, a=1/2$ .

Ah yes, thank you.

That being the case, $\displaystyle a > 1$ would fill that requirement, would it not?
• Jul 2nd 2011, 10:42 PM
FernandoRevilla
Re: I could use a second opinion on the completeness of my solution...
Quote:

Originally Posted by Lancet
That being the case, $\displaystyle a > 1$ would fill that requirement, would it not?

Yes, if $\displaystyle (C_1,C_2)\neq (0,0)$ and $\displaystyle a>1$ then, $\displaystyle x(t)=e^{(a-1)t}(C_1+C_2e^t)\to (+\infty\;\textrm{or}\; -\infty)$ as $\displaystyle t\to +\infty$ .
• Jul 2nd 2011, 11:01 PM
Lancet
Re: I could use a second opinion on the completeness of my solution...
Thank you again for all your help!