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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
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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 
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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.
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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.
We already know that Wolfram Alpha provides rather crappy pictures here. V.I.
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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
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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 $\sin^{3}(y)$
$M'=2x\sin^4(y)+\sin^3(y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+h(y)$
$\psi_y=4x^2\sin^3(y)\cos(y) + x \sin^2(y)\cos(y)+h'(y)$
$h'(y)=10\sin^4(y) \cos(y)$
$h(y)=2\sin^5 (y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+2\sin^5 (y)$
$M_y=2x\cos(y), N_x=8x\cos(y)+3\cot(y)$
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 $\sin^{3}(y)$
$M'=2x\sin^4(y)+\sin^3(y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+h(y)$
$\psi_y=4x^2\sin^3(y)\cos(y) + x \sin^2(y)\cos(y)+h'(y)$
$h'(y)=10\sin^4(y) \cos(y)$
$h(y)=2\sin^5 (y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+2\sin^5 (y)$
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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
\begin{equation} J={ \left[\begin{array}{ccc} 4xy & 2x^2+6y^2+8 \\ -6x^2-2y^2+32 & -4xy \end{array} \right ]}, \end{equation}
$When (0,0)$
\begin{equation} J={ \left[\begin{array}{ccc} 0 & 8 \\ 32 & 0 \end{array} \right ]}, \end{equation}
This is a saddle
$When (4,0)$
\begin{equation} J={ \left[\begin{array}{ccc} 0 & 40 \\ -64 & 0 \end{array} \right ]}, \end{equation}
This one is a center
$When (-4,0)$
\begin{equation} J={ \left[\begin{array}{ccc} 0 & 40 \\ -64 & 0 \end{array} \right ]}, \end{equation}
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.
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
\begin{equation} J={ \left[\begin{array}{ccc} 4xy & 2x^2+6y^2+8 \\ -6x^2-2y^2+32 & -4xy \end{array} \right ]}, \end{equation}
$When (0,0)$
\begin{equation} J={ \left[\begin{array}{ccc} 0 & 8 \\ 32 & 0 \end{array} \right ]}, \end{equation}
This is a saddle
$When (4,0)$
\begin{equation} J={ \left[\begin{array}{ccc} 0 & 40 \\ -64 & 0 \end{array} \right ]}, \end{equation}
This one is a center
$When (-4,0)$
\begin{equation} J={ \left[\begin{array}{ccc} 0 & 40 \\ -64 & 0 \end{array} \right ]}, \end{equation}
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.
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Quiz-7 / Re: Q7 TUT 0401
« on: November 30, 2018, 08:33:46 PM »
Here is computer generated picture
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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
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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
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Quiz-7 / Re: Q7 TUT 5102
« on: November 30, 2018, 05:17:58 PM »
This is computer generated picture
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Quiz-7 / Re: Q7 TUT 0101
« on: November 30, 2018, 04:35:45 PM »
This is computer generated picture
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Quiz-7 / Re: Q7 TUT 0501
« on: November 30, 2018, 04:29:02 PM »
To find critical points, let $x'=0, y'=0$
Thus, $-2x-y-x(x^2+y^2)=0, x-y+y(x^2+y^2)=0$
$ (0,0)$ is one crtical point
Also, $(0.330757,1.09242), (0.330757,-1.09242)$ are also critical points
Let $f(x)=-2x-y-x(x^2+y^2), g(x)=x-y+y(x^2+y^2)$
$f_x=-2-3x^2-y^2, f_y=-1-2xy$
Also, $ g_x=1+2xy, g_y=-1+x^2+3y^2$
$J=\begin{bmatrix}
-2-3x^2-y^2&-1-2xy\\
1+2xy&-1+x^2+3y^2\\
\end{bmatrix}$
Plug in the critical points to find eigenvalues of each linear system
For $(0,0)$,
We get $\begin{bmatrix}
-2&--1\\
1&-1\\
\end{bmatrix}$,
$r=\frac{-3\pm\sqrt{-3}i}{2}$
Therefore, it is stable at $(0,0)$
Similarly, for others critical points,
Plug in and eigenvalues are $-3.5092, 2.6771$
Thus, it is a saddle point therefore unstable
Thus, $-2x-y-x(x^2+y^2)=0, x-y+y(x^2+y^2)=0$
$ (0,0)$ is one crtical point
Also, $(0.330757,1.09242), (0.330757,-1.09242)$ are also critical points
Let $f(x)=-2x-y-x(x^2+y^2), g(x)=x-y+y(x^2+y^2)$
$f_x=-2-3x^2-y^2, f_y=-1-2xy$
Also, $ g_x=1+2xy, g_y=-1+x^2+3y^2$
$J=\begin{bmatrix}
-2-3x^2-y^2&-1-2xy\\
1+2xy&-1+x^2+3y^2\\
\end{bmatrix}$
Plug in the critical points to find eigenvalues of each linear system
For $(0,0)$,
We get $\begin{bmatrix}
-2&--1\\
1&-1\\
\end{bmatrix}$,
$r=\frac{-3\pm\sqrt{-3}i}{2}$
Therefore, it is stable at $(0,0)$
Similarly, for others critical points,
Plug in and eigenvalues are $-3.5092, 2.6771$
Thus, it is a saddle point therefore unstable
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