who else is taking this second session?

6/27/2007 3:28:19 PM

MA341? Which section are you taking?

6/27/2007 5:11:27 PM

lol.. yea ma 341

my bad


i'll be in section 1

6/27/2007 5:39:35 PM

i thought MA 342 was Diff Eq

or is it that calc 3 is 242 and calc 2 is 241?

6/27/2007 6:01:32 PM

^ The latter. The sequence is 141-241-242-341.

FWIW, I'm teaching section 2.

6/27/2007 6:27:18 PM

will you give me an A if I join your section?

6/27/2007 6:42:41 PM

Diff EQ = super easy

6/27/2007 6:50:45 PM

Oops...I taught 341-02 the last two semesters, but I'm scheduled to teach 242-02 this time.

For 341-01, you're in good shape; Rebecca Kalhorn is a grad student who is a very good teacher.

341 is easier than either 241 or 242 for most students. But since it moves quickly, to be over in 5 weeks, just don't get behind.

6/27/2007 10:16:36 PM

^^I hate you

Unless you took Hong, cause that would explain it.

6/27/2007 11:20:54 PM

Hong's final exam sucked

but he did make the class rather easy

[Edited on June 27, 2007 at 11:22 PM. Reason : .]

6/27/2007 11:21:46 PM

^ yea but you dont learn shit....

6/28/2007 7:54:15 AM

nah 341 is cake. I had Pierre Gremaud. He is a pretty funny guy. After a long problem he would write victory on the bored real big. PDE is hard though. Had an A in all the calcs and diff eq. got a C in pde. that shit sucked.

6/28/2007 10:22:40 AM

^^ of course you do

I learned plenty, and I used it a lot in the last couple of years even though I took it when I was a freshman.

6/28/2007 12:10:43 PM

Quote :

"^^I hate you Unless you took Hong, cause that would explain it."

I did take Hong. But his class isn't any easier, he just can actually teach it.

6/29/2007 10:45:16 PM

Oh believe me, his class is easier. Have you taken it with Rodriguez or the Russian caveman guy?

6/30/2007 12:09:05 AM

"Russian caveman guy"

hmmm, this can't be Khyefets since caveman do not wear glasses,...

6/30/2007 2:33:01 AM

^ I believe he's referring to Tsenkov (sp?).

That's who I had for 342. His class wasn't too bad. The tests were all multiple choice from what I remember and there wasn't too much emphasis on actual calculations, more like understanding the actual steps involved. I walked out of there with a low A, so I didn't complain.

6/30/2007 8:22:06 PM

Yeah that was it, lookin at the GDR for that semester I took it (s06) I was surprised, 2/3 are A's, B's, and C's, then I saw there was only 33 total, out of a good 90+ that started the class. I see I wasn't alone in deciding to drop it for Hong.

6/30/2007 11:25:23 PM

I had Tran for MA341 and got an A+. He was the man.

7/1/2007 6:29:59 PM

i'll be there

7/1/2007 8:38:45 PM

hey, do you know when she said she was going to post the h.w?

7/2/2007 1:59:04 PM

she said around noon i got mine off around 4:30 though cause i was at work

7/2/2007 5:55:56 PM

hey.... are you getting # 11 and 13 on the homework


it says yes in the back of the book, but im not getting the same solution they give

7/4/2007 1:29:19 PM

if you gave section numbers we might be able to help.

7/4/2007 6:21:15 PM

ahh yea... section 1.2

7/4/2007 6:59:04 PM

I remember these, you most likely already have the correct solution sitting in front of you, however the algebraic steps needed to massage what you have to what you see in #11 and #13 are far from obvious.

#11 differentiate implicitly, that means y=y(x), note d/dx [ x-1 ] = 1 so,

1 = d/dx [ exp(xy) + y ]
= exp(xy)d/dx[xy] + dy/dx : Chain rule.
= exp(xy)[ y + x*dy/dx ] + dy/dx : Product rule.
= exp(xy)*y + dy/dx [ xexp(xy) + 1 ] : ye old factoring

We want to isolate dy/dx so do so,

dy/dx = [ 1 - yexp(xy) ] / [ xexp(xy) + 1 ]

now one silly step is left. Multiply by one = exp(-xy)/exp(-xy) to obtain

dy/dx = [ exp(-xy) - y ] / [ x + exp(-xy) ]

Behold the book's expression.

7/5/2007 2:21:34 AM

#13 is more fun as I recall,

Given: sin(y) + xy -x^3 = 1

DIfferentiate to get: cos(y)y' + y + xy' -3x^2 = 0 ,lets call this

Diff again: -sin(y)y'y' + cos(y)y'' + y' + y' + xy'' - 6x = 0

For reasons only known to the author of this problem we want to get an expression for y'', so solve
for it,

y'' [ cos(y) + x ] = sin(y)y'y' - 2y' + 6x

y'' = [sin(y)y'y' - 2y' + 6x] / [ cos(y) + x ]

looking at the answer's form we see we want to multiply top and bottom by y'
(multiply by one=y'/y' )

y'' = [sin(y)y'y'y' - 2y'y' + 6xy'] / [ cos(y)y' + xy' ]

Now solve for the denominator of the above

cos(y)y' + xy' = 3x^2 - y

Substitute the last expression and you'll get y'' = the books answer.

stupid isn't it.

7/5/2007 2:30:17 AM

how did you solve for the denominator?

7/5/2007 1:32:35 PM

this class was a total cake-walk.

aslong as you are not retarded, you can at LEAST make a B+.

7/5/2007 6:03:11 PM

cos(y)y' + y + xy' -3x^2 = 0

therefore,

cos(y)y' + xy' = 3x^2 - y

7/5/2007 10:34:04 PM