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Messages - Chengyin Ye

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1
Quiz 2 / TUT 0601 Quiz2
« on: January 30, 2020, 03:40:13 PM »
Question: Find the limit of the function at the given point or explain why it does not exist.
f(z)=(z3-8i)/(z+2i)
(z≠-2i) at z0=-2i
Solution:
f(z)=(z3-8i)/(z+2i)
f(z)=(z3+(2i)3)/(z+2i)
f(z)=((z+2i)(z2-2iz+(2i)2)/(z+2i)
f(z)=(z+2i)(z2-2iz+(2i)2
limz → -2i f(z) =limz → -2i (z+2i)(z2-2iz+(2i)2
=(-2i)2-2i(-2i)+(2i)2=12i2=-12

2
Final Exam / Re: FE-P6
« on: December 16, 2018, 07:37:50 PM »
dx/dt=2x2y+2y3+8y
dy/dt=-2x3-2xy2+32x
so,dx/dy=2x2y+2y3+8y/-2x3-2xy2+32x
so,(2x3+2xy2-32x)dx+(2x2y+2y3+8y)dy=0
Let M=2x3+2xy2-32x,N=2x2y+2y3+8y
My=4xy,Nx=4xy
My=Nx, so Exact.
There exists a 𝐻(𝑥,𝑦) s.t. 𝐻x(𝑥,𝑦)=M,𝐻y(𝑥,𝑦)=N
𝐻(𝑥,𝑦)=1/2x4+x2y2-16x2+h(y)
𝐻y(𝑥,𝑦)=2x2y+ℎ′(𝑦)
so,ℎ′(𝑦)=2y3+8y
h(y)=1/2y4+4y2+C
Therefore, 𝐻(𝑥,𝑦)=1/2x4+x2y2-16x21/2y4+4y2=C

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Final Exam / Re: FE-P3
« on: December 14, 2018, 12:52:25 PM »
I think that instead of writing e-t(2et-e2t+3 -2ln|et+1| +C2), we should write e-t(2et -2ln|et+1| +C2) because e-t(-e2t) and 3et  are in the homogeneous solution.

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Quiz-4 / Re: Q4 TUt 0401
« on: October 26, 2018, 06:22:39 PM »
Here is my solution.

5
Term Test 1 / Re: TT1 Problem 4 (afternoon)
« on: October 16, 2018, 09:20:55 AM »
Here is my solution.

6
Thanksgiving Bonus / Re: Thanksgiving bonus 7
« on: October 05, 2018, 11:09:52 PM »
Here is my solution for Question 7.

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