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differential equation help

Hey,
Could someone please explain to me why when Q tends to a finite limit as t tends to infinity, B=0.
I don't understand this.
I have attached the question and ms for reference.
(edited 12 months ago)
Reply 1
Think about how exponentials behave: Be^t normally blows up (i.e. tends to infinity) as t tends to infinity, unless B is a very specific value.
Original post by tonyiptony
Think about how exponentials behave: Be^t normally blows up (i.e. tends to infinity) as t tends to infinity, unless B is a very specific value.

So Ae^-3t will tend to 0 and Be^5t will tend to infinity. But how does that mean that B=0?
Reply 3
Original post by action123
So Ae^-3t will tend to 0 and Be^5t will tend to infinity. But how does that mean that B=0?


So we are given Q tends to a finite limit, which is the sum of the A term and the B term (let's call them that).
If A tends to a finite limit (0 is finite), then the B term better tend to a finite limit as well - and is precisely "limit of Q - limit of A".

Then follow the argument in #2.
(edited 12 months ago)
Reply 4
Maybe as a cautionary tale, it is worth emphasizing that "infinity minus infinity" is NOT zero. In fact it could be anything - finite or infinite! So be careful when dealing with limits.
so Be^5t = finite limit
and if this is the case, B=0?
Is there a specific rule for this that I can write down in my notes?
Reply 6
Original post by action123
so Be^5t = finite limit
and if this is the case, B=0?
Is there a specific rule for this that I can write down in my notes?


It's less of a "rule" but more obvious if you think about what exponentials normally look like (think here means using pen and paper and curve sketch). They just grow and grow and grow. If you want to stop an exponential from growing by multiplying a constant to it, you don't have much choice.

That said, if you think you should add into your notes, go for it. For me, this would be on the very low end of the "do I need this in my notes" scale.
(edited 12 months ago)

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