# Thread: 2nd order ODE with a catch

1. ## 2nd order ODE with a catch

I have an odd ODE question that I could use a hand with...

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$y'' - y' - 2y = 0$

$y(0) = a$

$y'(0) = 2$

Find "a" so that the solution approaches 0 as t approaches infinity.

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So here's what I've done:

$r^2 - r - 2 = 0$

$r = -1, 2$

$y = C_{1}e^{-t} + C_{2}e^{2t}$

$a = C_{1} + C_{2}$

$C_{2} = a - C_{1}$

$y' = -C_{1}e^{-t} + 2C_{2}e^{2t}$

$2 = -C_{1} + 2C_{2}$

$C_{1} = 2C_{2} - 2$

$C_{1} = 2a - 2C_{1} - 2$

$C_{1} = \frac{2a - 2}{3}$

$y = \frac{C_{1}}{e^t} + C_{2}e^{2t}$

So, unless I've made a mistake somewhere, I've done most of the work. But I don't understand how I can get an "a" that will cause the solution to approach 0 as t approaches infinity. Since the second term is not part of a fraction, I'm not sure how this can be done.

Can someone help me to understand what I'm missing?

2. ## Re: 2nd order ODE with a catch

Originally Posted by Lancet
I have an odd ODE question that I could use a hand with...

----------------

$y'' - y' - 2y = 0$

$y(0) = a$

$y'(0) = 2$

Find "a" so that the solution approaches 0 as t approaches infinity.

----------------

So here's what I've done:

$r^2 - r - 2 = 0$

$r = -1, 2$

$y = C_{1}e^{-t} + C_{2}e^{2t}$

$a = C_{1} + C_{2}$

$C_{2} = a - C_{1}$

$y' = -C_{1}e^{-t} + 2C_{2}e^{2t}$

$2 = -C_{1} + 2C_{2}$

$C_{1} = 2C_{2} - 2$

$C_{1} = 2a - 2C_{1} - 2$

$C_{1} = \frac{2a - 2}{3}$

$y = \frac{C_{1}}{e^t} + C_{2}e^{2t}$

So, unless I've made a mistake somewhere, I've done most of the work. But I don't understand how I can get an "a" that will cause the solution to approach 0 as t approaches infinity. Since the second term is not part of a fraction, I'm not sure how this can be done.

Can someone help me to understand what I'm missing?
If y --> 0 as t --> +oo then C2 has to equal zero.

So you have y = C1 e^(-t) and since y(0) = a, C1 = a.

So y = a e^(-t).

Now use y'(0) = 2 to get a.

3. ## Re: 2nd order ODE with a catch

Originally Posted by mr fantastic
If y --> 0 as t --> +oo then C2 has to equal zero.

So you have y = C1 e^(-t) and since y(0) = a, C1 = a.

So y = a e^(-t).

Now use y'(0) = 2 to get a.
Thank you so much for that - I just couldn't see what I was missing. In retrospect, it seems obvious, and worth kicking myself over.

Thanks again!

4. ## Re: 2nd order ODE with a catch

No, no, don't kick yourself. Let us do it!