Differential Equations

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.

$\displaystyle \lim_{x\to\infty} y(x)=\lim_{x\to\infty}(Ae^{2x}+Be^{-x})=0$

So,

$\displaystyle 0 =\lim_{x\to\infty}(Ae^{2x})+\lim_{x\to\infty}(Be^{-x})$

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.

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?

Well, there's one particular value of A for which it doesn't go to infinity.

0?

Wait, A and B are both 0?

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.

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?

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

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

Yes.

Thinking about this again, why have I not found a particular integral? Is it because it's homogenous?