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Anonymous Wed 04 Sep 00:01:24 2019 No.10941589 View Reply Original Report

Quoted By: >>10942944 >>10944617 >>10944645 >>10944677

I need to solve the homogenous second order ODE shown

b and c are negative unknown constants.

Therefore, the exponent of the first term is always positive and the exponent of the second term can be either positive or negative.

The boundary limits provided are:

y(0) = y0(positive contant)
y(infinity) = 0

I'm trying to find the values of the constants C1 and C2. However, I can't find how to use the infinity boundary limit to obtain the second equation. What am I missing here?

Anonymous

Anonymous Wed 04 Sep 2019 11:21:50 No.10942944 Report

Quoted By:

>>10941589
you might try taking the laplace transform and sending s->0. this is equal to sending t->infinity. its called the final value theorem. mind you it only works where y does NOT go to infinity. alternatively just look at the sign of the real component of the exponent. negative goes to zero, positive goes to infinity, 1 is a steady sine wave. for you that depends on the magnitudes and signs of a and b

Anonymous

Anonymous Thu 05 Sep 2019 00:01:19 No.10944617 Report

Quoted By:

>>10941589
in the general solution, plug in infinity for x. b and c are negative, so RHS is C1e^(infinity) and C2e^{0,-infinity, or infinity}. In any case, C1 must be 0 or -C2. This will depend on specific choice of b and c. So your IVP gives 2 unique solutions, which makes sense, since it is a 2nd order eq, you have 2 conditions, and 2 unknowns in the diffeq.

Anonymous

Anonymous Thu 05 Sep 2019 00:10:43 No.10944645 Report

Quoted By:

>>10941589
C1 = 0, C2 = y0

Anonymous

Anonymous Thu 05 Sep 2019 00:23:36 No.10944677 Report

Quoted By: >>10944686

>>10941589
You're pretty much finished OP. I would just clear up the text though.
Define:
$r_{+} = \frac{-b + \sqrt{b^2 - 4c}}{2}$
and
$r_{-} = \frac{-b - \sqrt{b^2 - 4c}}{2}$

Then write your solution as such.
$y(x) = A e^{r_{+} x} + B e^{r_{-} x}$
When if bother b and c are negative constants, then you know that it' is impossible to get complex numbers for any of the r's. Because both therms in inside the square root are positive. That means A has to be zero ... or your solution would blow up to infinity.

Anonymous

Anonymous Thu 05 Sep 2019 00:26:43 No.10944686 Report

Quoted By:

>>10944677
that means you're solution is
$y(x) = y_0 e^{r_{-} x}$

Where the r(-) is defined in my previous post.

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