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### Messages - Jingze Wang

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1
##### Final Exam / Re: FE-P6
« on: December 17, 2018, 05:50:10 AM »
I think (-4,0) and (4,0) are maximum and (0,0) is minimum

2
##### Final Exam / Re: FE-P6
« on: December 17, 2018, 05:44:04 AM »
Since system is integrable, so just saddle and center as in linear system

3
##### Final Exam / Re: FE-P6
« on: December 16, 2018, 08:11:04 PM »
Can anyone tell me why Professor said my answer is wrong? Also, to avoid confusion, I use 0.5 instead of 1/2.

4
##### Final Exam / Re: FE-P6
« on: December 16, 2018, 01:29:15 PM »
Thanks, all corrected, finally

5
##### Final Exam / Re: FE-P6
« on: December 15, 2018, 07:12:23 PM »

6
##### Final Exam / Re: FE-P5
« on: December 14, 2018, 10:42:58 AM »
This is the computer generated global phase portrait.

We already know that Wolfram Alpha provides rather crappy pictures here. V.I.

7
##### Final Exam / Re: FE-P1
« on: December 14, 2018, 10:18:57 AM »
Sorry I have not finished my typed solution at that time, so you just saw part of my solution, but we get the same answer finally

8
##### Final Exam / Re: FE-P1
« on: December 14, 2018, 09:53:16 AM »
Let $M=2x\sin(y)+1, N=4x^2\cos(y)+3x\cot(y)+5\sin(2y)$
$M_y=2x\cos(y), N_x=8x\cos(y)+3\cot(y)$
Check and find this is not exact
Then try to find integrating factor
$N_x-M_y=8x\cos(y)+3\cot(y)-2x\cos(y)=6x\cos(y)+3\cot(y)$
By observation, $\frac{N_x-M_y}{M}=3\cot(y)$
Therefore, the integrating factor is $\sin^{3}(y)$
$M'=2x\sin^4(y)+\sin^3(y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+h(y)$
$\psi_y=4x^2\sin^3(y)\cos(y) + x \sin^2(y)\cos(y)+h'(y)$
$h'(y)=10\sin^4(y) \cos(y)$
$h(y)=2\sin^5 (y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+2\sin^5 (y)$

9
##### Final Exam / Re: FE-P6
« on: December 14, 2018, 09:02:51 AM »
part (a)
Find critical points
Let $x'=0, y'=0$
Then $2y(x^2+y^2+4)=0, -2x(x^2+y^2-16)=0$
When $y=0, x=0, x=4, x=-4$
So (0,0) (4,0) (-4,0) are critical points.

Part (b)
$F(x,y)=2y(x^2+y^2+4), G(x,y)=-2x(x^2+y^2-16)$
$F_x=4xy, F_y=2x^2+6y^2+8$
$G_x=-6x^2-2y^2+32, G_y=-4xy$
Then plug in to find J matrix
$$J={ \left[\begin{array}{ccc} 4xy & 2x^2+6y^2+8 \\ -6x^2-2y^2+32 & -4xy \end{array} \right ]},$$
$When (0,0)$
$$J={ \left[\begin{array}{ccc} 0 & 8 \\ 32 & 0 \end{array} \right ]},$$
$When (4,0)$
$$J={ \left[\begin{array}{ccc} 0 & 40 \\ -64 & 0 \end{array} \right ]},$$
This one is a center
$When (-4,0)$
$$J={ \left[\begin{array}{ccc} 0 & 40 \\ -64 & 0 \end{array} \right ]},$$
This one is also a center
Also, the phase portraits are attached in picture 2

Part (c)
$2x(x^2+y^2-16)dx+2y(x^2+y^2+4)dy=0$
Find this equation is exact, then
$H(x,y)=0.5 x^4+x^2y^2-16x^2+h(y)$
$H_y=2x^2 y +h'(y)$
$h'(y)=2y^3+8y$
$h(y)=0.5 y^4+4y^2$
$H(x,y)=0.5 x^4+x^2y^2-16x^2+0.5 y^4+4y^2=c$

Part (d)
Since it is integrable system
Then center is still center in nonlinear system.
See picture 1.

10
##### MAT244--Lectures & Home Assignments / Linear around a point in nonlinear differential equation system
« on: December 11, 2018, 09:26:02 PM »
Do we need to calculate the eigenvalues and eigenvectors when the questions just ask us to linearize the system at the point? Or we just find J matrix at that point is sufficient and no need for eigenvectors?

11
##### Quiz-7 / Re: Q7 TUT 0401
« on: November 30, 2018, 08:33:46 PM »
Here is computer generated picture

12
##### Quiz-7 / Re: Q7 TUT 0201
« on: November 30, 2018, 08:31:23 PM »
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted

13
##### Quiz-7 / Re: Q7 TUT 0301
« on: November 30, 2018, 05:27:47 PM »
c) For the first critical point(s), the eigenvalues are $\pm i$. This means the phaseportrait is centre (CW).
Hi Michael, I think you should mention that phase portrait is counterclockwise for the first matrix since -1<0 rather than imaginary eigenvalues

14
##### Quiz-7 / Re: Q7 TUT 5102
« on: November 30, 2018, 05:17:58 PM »
This is computer generated picture

15
##### Quiz-7 / Re: Q7 TUT 0101
« on: November 30, 2018, 04:35:45 PM »
This is computer generated picture

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