Solved it for the first initial conditions, but I would appreciate a check.
What is the situation with the underlined new initial condition though?
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ChrisLorient
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- 17-10-2017 07:41
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ghostwalker
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- 17-10-2017 08:10
(Original post by ChrisLorient)
Solved it for the first initial conditions, but I would appreciate a check.
What is the situation with the underlined new initial condition though?
So,
What's the limit for the second term?
Then what must A be for the whole thing to equal 0?
Edit: First one looks fine, though to finish off, I would restate y(x)= ... with the values of A,B filled in.Last edited by ghostwalker; 17-10-2017 at 11:13. -
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- 18-10-2017 07:33
(Original post by ghostwalker)
I presume it's just the limit criterion that's causing the problem.
So,
What's the limit for the second term?
Then what must A be for the whole thing to equal 0?
Edit: First one looks fine, though to finish off, I would restate y(x)= ... with the values of A,B filled in. -
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- 18-10-2017 09:05
(Original post by ChrisLorient)
Isn't the limit for the second term 0 though? Subbing in a large value, I get back 0. -
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- 18-10-2017 13:33
(Original post by ghostwalker)
Yes. You can also write it as 1/e^x. So, as x goes to infinity, e^-x goes to zero. -
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- 18-10-2017 14:13
(Original post by ChrisLorient)
OK. But the other term is infinity, no? So isn't that giving that 0 = infinity? -
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- 18-10-2017 15:57
(Original post by ghostwalker)
Well, there's one particular value of A for which it doesn't go to infinity.
Wait, A and B are both 0?Last edited by ChrisLorient; 18-10-2017 at 15:58. -
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- 18-10-2017 16:05
(Original post by ChrisLorient)
0?
Wait, A and B are both 0? -
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- 18-10-2017 16:17
Therefore, RHS = 0
If the first term when e^(2x) approaches infinity is infinity, I need A = 0, as nothing else gives a non-positive result (that I know).
So the e^(-x) is hugely negative when infinity is approached, but I need B to make it 0.
So I incorrectly assumed B = 0. -
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- 18-10-2017 17:11
(Original post by ChrisLorient)
LHS = 0
Therefore, RHS = 0
If the first term when e^(2x) approaches infinity is infinity, I need A = 0, as nothing else gives a non-positive result (that I know).
So the e^(-x) is hugely negative when infinity is approached, but I need B to make it 0. -
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- 18-10-2017 17:17
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- 18-10-2017 17:20
(Original post by ChrisLorient)
OK, so the e term will always give 0. Does that mean B could take any value? -
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- 18-10-2017 17:34
(Original post by ghostwalker)
Based on the limit criterion, yes, B could by anything. But you also have another condition, and you can use that to now determine B.
y(x) = Ae^(2x) + Be^(-x)
So...
1 = Ae^(2*0) + Be^(-0)
1 = A + B
B = 1 - A
If A is 0, then B = 1 -
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- 18-10-2017 19:20
(Original post by ChrisLorient)
My other condition is that y(0) = 1 .
y(x) = Ae^(2x) + Be^(-x)
So...
1 = Ae^(2*0) + Be^(-0)
1 = A + B
B = 1 - A
If A is 0, then B = 1 -
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- 18-10-2017 19:38
(Original post by ghostwalker)
Yep. Done.
y(x) = e^(-x)
Having subbed in A = 0 and B = 1 -
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- 18-10-2017 19:58
(Original post by ChrisLorient)
One last question on this problem. It would be written in its final form as...
y(x) = e^(-x)
Having subbed in A = 0 and B = 1 -
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- 30-10-2017 12:40
(Original post by ghostwalker)
Yes.Last edited by ChrisLorient; 30-10-2017 at 14:33. -
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- 30-10-2017 14:35
(Original post by ChrisLorient)
Thinking about this again, why have I not found a particular integral? Is it because it's homogenous?
Edit: That is treating the PI as something separate to the complementary functions.Last edited by ghostwalker; 30-10-2017 at 14:42.
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