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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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I think (-4,0) and (4,0) are maximum and (0,0) is minimum

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Since system is integrable, so just saddle and center as in linear system

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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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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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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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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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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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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?

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Here is computer generated picture

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I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted

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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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This is computer generated picture

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This is computer generated picture

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