### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Qian Li

Pages: [1]
1
##### MAT244 Misc / Office Hour
« on: December 11, 2013, 01:07:27 AM »
What is the office hour befor final?
Is it the same as that was in regular semester?
Thanks a lot!

2
##### MAT244 Misc / Coverage about quiz 5 (day section)
« on: November 16, 2013, 01:00:12 PM »
The coverage of quiz 5 my instructor posted on blackboard is 7.5 7.6 and 7.7, but the forum says it  only includes 7.5 and 7.6. So what would be covered on quiz 5 for day section?

3
##### Quiz 1 / Re: Q1, P2--Day section
« on: October 05, 2013, 03:40:39 PM »
3 from section 2.6, i.e.
$$(3x^2-2xy+2)+(6y^2-x^2+3)y'=0$$
We already know it is exact, partial derivative of $(3x^2-2xy+2)$ with respect to $y$ is $-2x$, while partial derivative of
$(6y^2-x^2+3)$ with respect to $x$ is also $-2x$.

So just assume the original function is  $F(x,y)$. $P(x,y)=3x^2-2xy+2$ is obtained from $F(x,y)$ by taking patial derivative with respect to $x$, and $Q(x,y)=(6y^2-x^2+3)$ with respect to $y$.

Now integrate $P(x,y)$ with respect to $x$, and we can get $F(x,y)=x^3-x^2y+2x+h(y)$. Take partial derivative with respect to $y$, and we get $Q(x,y)= -x^2+h'(y)$. ERROR!!!

If compare this $Q(x,y)$ with the $6y^2-x^2+3$,  we get $h'(y)=-6y^2+3$, which means $h(y)=2y^3+3y+C$, where $C$ is any arbitrary constant.

In conclusion, $F(x,y)=x^3-x^2y+2x+2y^3+3y+C$, or we can write it as $x^3-x^2y+2x+2y^3+3y=C$.

Do not hijack topics--this was about Q1, P1--so I split it

4
##### MAT244 Misc / What is the time for the quizzes?
« on: September 14, 2013, 01:59:49 PM »
I am wondering what is the date for the 6 quizzes of the day section (L0101)? It is not mentioned on the website.
Thanks!

Pages: [1]