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### Messages - Mingdi Xie

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1
##### Term Test 2 / Re: Problem 4 (noon)
« on: November 20, 2019, 03:07:38 PM »
This is my solution.

2
##### Term Test 2 / Re: Problem 3 (noon)
« on: November 20, 2019, 02:54:05 PM »
Solution for part b

3
##### Term Test 2 / Re: Problem 3 (noon)
« on: November 20, 2019, 02:45:02 PM »
this is my solution for part a

4
##### Term Test 2 / Re: Problem 2 (noon)
« on: November 20, 2019, 02:38:35 PM »
Here is my solution

5
##### Term Test 2 / Re: Problem 1 (noon)
« on: November 20, 2019, 02:29:07 PM »
This is my solution

6
##### Term Test 2 / Re: Problem 4 (main sitting)
« on: November 19, 2019, 09:58:57 PM »
Here is my solution

7
##### Term Test 2 / Re: Problem 3 (main sitting)
« on: November 19, 2019, 09:44:19 PM »
There is a small typo in Yiheng Bian's solution, $u_1'$ and $u_2'$ should be alternate.

8
##### Term Test 2 / Re: Problem 3 (main sitting)
« on: November 19, 2019, 09:41:06 PM »
This is my solution

9
##### Term Test 2 / Re: Problem 2 (main sitting)
« on: November 19, 2019, 08:18:51 PM »
This is my solution

10
##### Term Test 2 / Re: Problem 1 (main sitting)
« on: November 19, 2019, 08:02:28 PM »
This is my solutions for problem1.

11
##### Term Test 2 / Re: Problem 3 (morning)
« on: November 19, 2019, 02:47:01 PM »

12
##### Term Test 2 / Re: Problem 2 (morning)
« on: November 19, 2019, 02:14:58 PM »
I think Lan Cheng miss out $c_2$ and $c_3$ term in front of $\cos2t$ and $\sin2t$

13
##### Term Test 2 / Re: Problem 1 (morning)
« on: November 19, 2019, 01:19:13 PM »
Attached is my solution, more detail on the integral parts.

14
##### Quiz-3 / TUT0101 Quiz3
« on: October 12, 2019, 12:22:47 PM »
y’’-6y’+5y=0     y(0)=2    y’(0)=6
r^2-6r+5=0
(r-1)(r-5)=0
r=1   r=5
y(t)=c1e^-3t+c2e^-t
Since y(0)=2
y(0)=c1+c2=2
c2=2-c1                        (1)
y’(t)=c1e^t+5c2e^5t
y’(0)=-1
y’(0)=c1+5c2=6
c2=4
c1=-2
And Wronskian doesn't equal 0
y(t)=-2e^t+4e^5t

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