The purpose of this page is to allow students in Math 23 to submit and answer questions concerning the course or homework at their convenience.
The page is quite easy to use. Click on Post to post your question, or to answer to a previously posted question. I suggest answering previously posted questions using the page rather than blitz; that way everyone can share the wealth.
FROM: Stumped
DATE: Tue Jan 20 10:26:54 EST 1998
e-mail: e-mail not given
Does anyone know if there's a way to solve part D of 2.5 #9 without using a graphing calculator?
FROM: Erik Guentner
DATE: Sat Jan 24 21:29:16 EST 1998
e-mail: guentner@dartmouth.edu
Yes Stumped, there is a way to solve this problem.
Based on your question I assume you can get to the point
of solving
exp(40r)-2exp(20r)+1=0.
Then substitute x=exp(10r) to reduce to finding roots of
x^4-2x^3+1=0.
By inspection, x=1 is a root (corresponding to r=0) and
if we factor
x^4-2x^3+1=(x-1)(x^3-x^2-x-1)
we are reduced to finding the roots of the cubic
polynomial on the right hand side of this equation.
There is a seldom used formula similar to the quadratic
formula for finding roots of cubic polynomials which you
need to look up in a book. It will give you the answer
in the form of a fraction (for typesetting purposes write
a=root of 11 and b=root of 3 and c=19+3ab):
c^{2/3} + 4 + c^{1/3}
x= ------------------------
3c^{1/3}
Not very satisfactory, I know. I'd stick with the
graphing calculator myself.
That's the best I can do
FROM: trent
DATE: Sun Jan 25 04:08:33 EST 1998
e-mail: trent@hell.com
in section 3.1, problem 30, what is the question asking for? i don't know what it's trying to explain either. hmm..
FROM: Erik Guentner
DATE: Sun Jan 25 11:37:59 EST 1998
e-mail: guentner@dartmouth.edu
Trent, you should solve the given equation using the
substitution v=y'. Then v'=y'' and the equation becomes
v' + tv^2 = 0.
This separable equation should be straightforward to
solve for v. Once you've got v, integrate to obtain y.
Hope this helps.
BTW, this problem is assigned because this substitution
is very common; we'll encounter it again later.
FROM: ???
DATE: Mon Jan 26 17:28:46 EST 1998
e-mail: e-mail not given
In sectrion 2.10, problem #15, does anyone know the method to get the thing started? Is it separable, exact, etc.?
FROM: Justin May
DATE: Tue Jan 27 19:41:56 EST 1998
e-mail: Beowulf@Dartmouth.EDU
I'm doing section 3.4. Easy stufff right? Sure is if you remember how to do derivatives. Question on section 3.4 #18... I've got a nice cancellation taking place I get y=exp(-2t) C1=1/(cost+2*Sint) C2= 2/(cost+2sint) when I put these constants into the general equation: y=C1exp(-2t)cost + C2exp(-2t)sint I am getting the above answer. this doesn;t agree with the back of the book. Anyone else out there? Don't be shy.. This is the information age.
FROM: Chris Holden
DATE: Tue Jan 27 20:59:05 EST 1998
e-mail: cph@dartmouth.edu
Justin, In doing #18 I get roots being -2(+-)i. Thus y=c1e^(-2t)cost + c2e(-2t)sint y prime=(-2c1+c2)e^(-2t)cost + (-c1-2c2)e^(-2t)sint So, y(0)=1 therefore c1=1 y prime(0)=-2c1+c2=0 thus c2=2 Final answer: y=e^(-2t)(cost+2sint) Hope this helps...
FROM: anonymous
DATE: Thu Jan 29 13:45:29 EST 1998
e-mail: e-mail not given
whenever, i try to solve a reduction of order, i end up with v'/v'' = something, how do you solve this
FROM: even more anonymous
DATE: Thu Jan 29 17:33:35 EST 1998
e-mail: e-mail not given
anonymous, when you get v'/v'' try making the sustitution u=v' and then you get a separable equation with the variables u and x (or t). Solve for u and then sub back in for v. That should work
FROM: Confused
DATE: Sat Jan 31 14:42:54 EST 1998
e-mail: e-mail not given
In section 3.6, #3 and #22a, I don't understand why the answers have an extra t in the particular solution. For example, in #3, I found a particular solution of the form Y=Ate^(-t) but the answer in the back of the book wants something that looks like Y=Ate^(-t) + B(t^2)e^(-t). Does anybody have a clue why this is?
FROM: ?
DATE: Tue Feb 3 23:32:50 EST 1998
e-mail: e-mail not given
number three is probably just a little algebra mistake. You should get, if you just use y = Ate^-t, -4A=-3t. This can't be true, since A must be just a constant. Therefore, you have to make Y=At^2e^-t + Bte^-t I have basically the same question for 20a and 22a, though. I know that if you actually go through the algebra, you can find out if you need to multiply by t again. Is there any way you can see that without going through the algebra? It seems as if in number 20 and 22 for the answers they know to guess a higher power of t just by looking at the question. If there is a way to do this, it would save a lot of time on the other problems too
FROM: fixed up
DATE: Tue Feb 3 23:36:03 EST 1998
e-mail: e-mail not given
number three is probably just a little algebra mistake. You should get, if you just use y = Ate^-t, -4A=-3t. This can't be true, since A must be just a constant. Therefore, you have to make Y=At^2e^-t + Bte^-t I have basically the same question for 20a and 22a, though. I know that if you actually go through the algebra, you can find out if you need to multiply by t again. Is there any way you can see that without going through the algebra? It seems as if in number 20 and 22 for the answers they know to guess a higher power of t just by looking at the question. If there is a way to do this, it would save a lot of time on the other problems too
FROM: for fixed up
DATE: Fri Feb 6 14:06:11 EST 1998
e-mail: e-mail not given
for #20a and #22a, they don't just look at the question, they are looking at the solution of the corresponding homogeneous equation, seeing that this matches the first guess you would make, and so they multiply by t in the right places. Other than this I think you need to just go through the algebra...which isn't fun...