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?

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 $\displaystyle C_1,C_2$

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

I'm not quite sure how to proceed, then.

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: $\displaystyle \lim_{t\to +\infty}e^{\beta t}=0\Leftrightarrow \beta <0$ .

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

So is it correct that the general answer would be:

$\displaystyle a < 0$

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.

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!

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 $\displaystyle a > 0$ ?

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 $\displaystyle C_1=1,C_2=0, a=1/2$ .

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?

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 $\displaystyle t\to +\infty$ .

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

Thank you again for all your help!