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# Differential Equations watch

1. 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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2. (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?
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.
3. (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.
Isn't the limit for the second term 0 though? Subbing in a large value, I get back 0.
4. (Original post by ChrisLorient)
Isn't the limit for the second term 0 though? Subbing in a large value, I get back 0.
Yes. You can also write it as 1/e^x. So, as x goes to infinity, e^-x goes to zero.
5. (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.
OK. But the other term is infinity, no? So isn't that giving that 0 = infinity?
6. (Original post by ChrisLorient)
OK. But the other term is infinity, no? So isn't that giving that 0 = infinity?
Well, there's one particular value of A for which it doesn't go to infinity.
7. (Original post by ghostwalker)
Well, there's one particular value of A for which it doesn't go to infinity.
0?

Wait, A and B are both 0?
8. (Original post by ChrisLorient)
0?
Yes, so A=0

Wait, A and B are both 0?
No - why would you think that?
9. (Original post by ghostwalker)
Yes, so A=0

No - why would you think that?
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.

So I incorrectly assumed B = 0.
10. (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).
Yes.

So the e^(-x) is hugely negative when infinity is approached, but I need B to make it 0.
No. You've already covered this. See posts 3 and 4.
11. (Original post by ghostwalker)
Yes.

No. You've already covered this. See posts 3 and 4.
OK, so the e term will always give 0. Does that mean B could take any value?
12. (Original post by ChrisLorient)
OK, so the e term will always give 0. Does that mean B could take any value?
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.
13. (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.
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
14. (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
Yep. Done.
15. (Original post by ghostwalker)
Yep. Done.
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
16. (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
Yes.
17. (Original post by ghostwalker)
Yes.
Thinking about this again, why have I not found a particular integral? Is it because it's homogenous?
18. (Original post by ChrisLorient)
Thinking about this again, why have I not found a particular integral? Is it because it's homogenous?
Looking at the RHS of your DE, if there was a PI, then it's clearly 0, hence none.

Edit: That is treating the PI as something separate to the complementary functions.

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