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### Messages - Qi Cui

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1
##### Final Exam / Re: FE-P2
« on: December 14, 2018, 10:27:15 AM »
I think Cui calculated the homogeneous solution wrong:
$(r-1)(r-1)(r-1) = r^3 -3r^2+3r+1$

Actually:
$r^3 -3r^2+4r-2= (r-1)(r^2-2r-2)$ = 0
$r=1$ or $r=1-i, 1+i$

So, Homogeneous solution is
$y= c_1e^t + c_2e^t\cos(t) + c_3e^t\sin(t)$

Non homogenous part for $10e^t$
We should assume $Y= Ate^t$
$Y’=Ate^t+Ae^t$
$Y’’=Ate^t+2Ae^t$
$Y’’’=Ate^t+3Ae^t$

$Y’’’-3Y’’+4Y’-2Y =(A-3A+4A-2A)te^t+(3A-6A+4A-2A)e^t =-Ae^t =10e^t$
$A =-10, Y=-10te^t$
fixed,thx!

2
##### Final Exam / Re: FE-P2
« on: December 14, 2018, 10:10:14 AM »
Non homo part should be 10tet-e-t+2𝑐𝑜𝑠𝑡+6𝑠𝑖𝑛𝑡
How come?

3
##### Final Exam / Re: FE-P3
« on: December 14, 2018, 09:23:35 AM »
$$Homo: r^3 -2r^2-r+2=0$$
$$(r+1)(r-1)(r-2)=0$$
$$r_1=1, r_2=-1, r_3=2$$
$$\therefore y_c(t)=c_1e^t+c_2e^{-t}+c_3e^{2t}$$
$$W = \left| \begin {array}{ccc} {e^t}&e^{-t}&e^{2t}\\ e^t& -e^{-t}&2e^{2t}\\ e^t&e^{-t}& 4e^{2t} \end {array} \right| = -6e^{2t}$$
$$W_1 = \left| \begin {array}{ccc} {0}&e^{-t}&e^{2t}\\ 0& -e^{-t}&2e^{2t}\\ 1&e^{-t}& 4e^{2t} \end {array} \right| = 3e^{t}$$
$$W_2 = \left| \begin {array}{ccc} {e^t}&0&e^{2t}\\ e^t&0&2e^{2t}\\ e^t&1& 4e^{2t} \end {array} \right| = e^{3t}$$
$$W_3 = \left| \begin {array}{ccc} {e^t}&e^{-t}&0\\ e^t& -e^{-t}&0\\ e^t&e^{-t}& 1 \end {array} \right| = -2$$
substitute above into the formula:
$$y_p(t)=y_1{\int}{{w_1(s)}{g(s)}\over{W(s)}} ds+y_2{\int}{{w_2(s)}{g(s)}\over{W(s)}}ds+y_3{\int}{{w_2(s)}{g(s)}\over{W(s)}}ds$$
$$=-e^t{\int}{{6e^s}\over{e^s+1}} ds-e^{-t} {\int}{{2e^{3s}}\over{e^s+1}}ds+e^{2t}{\int}{{4}\over{e^s+1}}ds$$
$$=-6e^tln|e^t+1|-e^{-t}ln|e^t+1|-e^{-t}(e^t-1)^2-e^{2t}4ln|e^{-x}+1|$$
$$y(t)=y_c(t)+y_p(t)$$

4
##### Final Exam / Re: FE-P2
« on: December 14, 2018, 08:52:58 AM »
$$Homo: r^3 -3r^2+4r-2=0$$
$$r_1=1r_2=1+ir_3=1-i$$
$$\therefore y_c(t)=c_1e^t+c_2e^tcost+c_3e^tsint$$
Non-Homo:
$$y^{'''}-3y{''}+4y{'}-2y=10e^t+10e^{-t} +20cost$$
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
$$y_{p1}(t)=Ate^t$$
$$y^{'}=Ate^t+ Ae^t$$
$$y^{''} = Ate^t+ 2Ae^t$$
$$y^{'}= Ate^t+ 3Ae^t$$
substitute above into the function:
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
$$we have: A=-10$$
$$\therefore y_{p1}(t)=-10e^t$$
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
$$y^{'''}-3y{''}+4y{'}-2y=10e^{-t}$$
$$y_{p2}(t)=Ae^{-t}$$
$$y^{'}= -Ae^{-t}$$
$$y^{''}= Ae^{-t}$$
$$y^{'''}= -Ae^{-t}$$
substitute above into the function:
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
we have A=1
$$\therefore y_{p2}(t)=e^{-t}$$
$$y^{'''}-3y{''}+4y{'}-2y=20cost$$
$$y_{p3}(t)=Acost+Bsint$$
$$y^{'}= -Asint+Bcost$$
$$y^{''}= -Acost-Bsint$$
$$y^{'''}= Asint-Bcost$$
substitute above into the function:
we have A=2, B=6
$$\therefore y_{p3}(t)=2cost+6sint$$
$$\therefore y(t)= c_1e^t+c_2e^tcost+c_3e^tsint-10e^t+ e^{-t} + 2cost+6sint$$

5
##### Quiz-6 / Re: Q6 TUT 0701
« on: November 17, 2018, 07:10:29 PM »
$$det(A-\lambda I) = \left| \begin {array}{ccc} {3- \lambda}&2&4\\ 2& {- \lambda}&2\\ 4&2&{3- \lambda} \end {array} \right| = 0$$
$${-\lambda}^3 + 6 {\lambda}^2+ 15{\lambda}+8 = 0$$
$$By\ long\ devision\ method, we\ get\ -({\lambda+1})^{2}(\lambda-8) = 0$$
$$\quad\therefore \lambda = -1,-1,8$$
$when\ \lambda = 8:$
$$(A-\lambda I)x = 0$$
$$\left[ \begin {array}{ccc} -5&2&4\\ 2&-8&2\\ 4&2&-5 \end {array} \right]x = 0$$
$$By\ row\ operation, we\ get: \left[ \begin {array}{ccc} 1&0&-1\\ 0&2&-1\\ 0&0&0 \end {array} \right]\left[ \begin {array}{c} x_1\\ x_2\\ x_3 \end {array} \right]= 0$$
$let x_3 = t:$
$$x_1= t$$
$$2x_2= t$$
$we\ have$:
$$\left[ \begin {array}{c} 2\\ 1\\ 2 \end {array} \right]$$
$When \lambda = -1:$
$$\left[ \begin {array}{ccc} 4&2&4\\ 2&1&2\\ 4&2&4 \end {array} \right] x=0$$
$$By\ row\ operation, we\ get: \left[ \begin {array}{ccc} 2&1&2\\ 0&0&0\\ 0&0&0 \end {array} \right]\left[ \begin {array}{c} x_1\\ x_2\\ x_3 \end {array} \right]=0$$
$let\ x_3=t,x_2=s:$
$$2x_1=-s-2t$$
$$x_1={{-1}\over {2}}s-t$$
$$x_2=s$$
$$x_3=t$$
$we\ have:$
$$\left[ \begin {array}{c} {{-1}\over {2}}s-t\\ s\\ t \end {array} \right] = t\left[ \begin {array}{c} -1\\ 0\\ 1 \end {array} \right] +s \left[ \begin {array}{c} {{-1}\over {2}}\\ 1\\ 0 \end {array} \right]$$
$\quad\therefore we\ have\ eigenvector: \left[ \begin {array}{c} {{-1}\over {2}}\\ 1\\ 0 \end {array} \right] , \left[ \begin {array}{c} -1\\ 0\\ 1 \end {array} \right]$
$$\quad\therefore x(t) = c_1e^{8t}\left( \begin {array}{c} 2\\ 1\\ 2 \end {array} \right)+c_2e^{-t} \left( \begin {array}{c} {{-1}\over2}\\ 1\\ 0 \end {array} \right)+c_3e^{-t}\left( \begin {array}{c} -1\\ 0\\ 1 \end {array} \right)$$

6
##### Quiz-6 / Re: Q6 TUT 0102
« on: November 17, 2018, 06:22:50 PM »
$$let\ h(z) = z, g(z) = {sin(z)}^2$$
$As \ z _0 = 0:$
$$h(0) = 0, h'(0)= 1 \ne1\quad\therefore order = 1$$
$$g(z) = {sin(z)}^2 = 0$$
$$g'(z) = 2sin(z)cos(z), g'(0) = 0$$
$$g''(z) = −2(sin^2(z)-cos^2(z)), g''(0) = 2\ne0, \quad\therefore order = 2$$
$$\quad\therefore 2-1= 1 \quad\therefore \ order\ of \ pole = 1$$
$$\quad\therefore{{z}\over sin^{2}(z)} = a_{-1}z^{-1} + a_{0} + a_{1}z^{1} +...$$
$$z= (z-{z^{3}\over 3!} + {z^{5}\over 5!} -... )^{2}(a_{-1}z^{-1} + a_{0} + a_{1}z^{1} + ... )$$
$After\ equating\ coefficients\ of\ equal\ powers\ of\ z, we\ can\ get:$
$$a_{-1}=1,a_{0}=0, a_{1}={{1}\over{3}}, a_{2}= 0, a_{3}={{1}\over{15}},a_{4}=0,a_{5}= {{2}\over{189}}$$

$$\quad\therefore {{1}\over {z}} + {{z}\over3} + {{1z^3}\over15} +{{2z^5}\over{189}} +..., Res(f;0)=1$$

7
##### Quiz-6 / Re: Q6 TUT 0101
« on: November 17, 2018, 05:51:54 PM »
$Since\ f(z)\ has\ a\ zero\ of\ order\ m\ at \ z_{0}$,
$$\quad\therefore f(z) = (z-z_{0})^mg(z),where g'(z_{0})\ne0$$
$$\quad\therefore f'(z) = (z-z_{0})^mg'(z)+m(z-z_{0})^{m-1}g(z)$$
$$f'(z) = (z-z_{0})^m(g'(z)+m(z-z_{0})^{-1}g(z))$$
$$\quad\therefore {{f'(z)}\over {f(z)}} ={{(z-z_{0})^m(g'(z)+m(z-z_{0})^{-1}g(z))}\over {(z-z_{0})^mg(z)}}$$
$$= {{g'(z)}\over {g(z)}}+m(z-z_{0}^{-1})$$
$$\quad\therefore Res({{f'}\over f}, z_{0})=m$$

8
##### Quiz-6 / Re: Q6 TUT 0501
« on: November 17, 2018, 05:38:40 PM »
$$det(A-\lambda I) = \left| \begin {array}{ccc} {-3- \lambda}&0&2\\ 1& {-1- \lambda}&0\\ -2&-1&{- \lambda} \end {array} \right| = 0$$
$${\lambda}^3 + 4 {\lambda}^2+ 7{\lambda}+6 = 0$$
$$By\ long\ devision\ method, we\ get\ (\lambda+2)({\lambda}^2+ 2{\lambda} +3) = 0$$
$$({\lambda}^2+ 2{\lambda} +3) : \lambda = {-2 \pm \sqrt{-8} \over 2} = -1\pm \sqrt{2}i$$
$$\quad\therefore \lambda = -2; {-1\pm \sqrt{2}i}$$
$when\ \lambda = -2:$
$$(A-\lambda I)x = 0$$
$$\left[ \begin {array}{ccc} -1&0&2\\ 1&1&0\\ -2&-1&2 \end {array} \right]x = 0$$
$$By\ row\ operation, we\ get: \left[ \begin {array}{ccc} -2&-1&2\\ -1&0&2\\ -0&0&0 \end {array} \right]\left[ \begin {array}{c} x_1\\ x_2\\ x_3 \end {array} \right]= 0$$
$let x_3 = t:$
$$-2x_1-x_2+2t= 0$$
$$-x_1+2t= 0$$
$we\ have$:
$$span(\left[ \begin {array}{c} 2\\ -2\\ 1 \end {array} \right])$$
$By\ similar\ procedure\ as\ above\ shown,we\ can\ get\ other\ two\ eigenvector\ for \lambda = -1\pm \sqrt{2}i : span(\left[ \begin {array}{c} 2- \sqrt{2}i\\ -1- \sqrt{2}i\\ 1 \end {array} \right], \left[ \begin {array}{c} 2+ \sqrt{2}i\\ -1+ \sqrt{2}i\\ 1 \end {array} \right] )$
$$\quad\therefore x(t) = c_1e^{-2t}\left( \begin {array}{c} 2\\ -2\\ 1 \end {array} \right)+c_2e^{(-1+ \sqrt{2}i)t }\left( \begin {array}{c} 2- \sqrt{2}i\\ -1- \sqrt{2}i\\ 1 \end {array} \right)+c_3e^{(-1- \sqrt{2}i)t}\left( \begin {array}{c} 2+ \sqrt{2}i\\ -1+ \sqrt{2}i\\ 1 \end {array} \right)$$

9
##### Quiz-6 / Re: Q6 TUT 5201
« on: November 17, 2018, 04:47:08 PM »
$$let\ h(z) = 0, g(z) = 1- cos(z)$$
$$As \ z _0 = 0, h(0) = 1\ne0; order = 0$$
$$g(z) = 1- cos(z)$$
$$g'(z) = sin(z), g'(0) = 0$$
$$g''(z) = cos(z), g''(0) = 1\ne0, \quad\therefore order = 2$$
$$\quad\therefore 2-0= 2 \quad\therefore \ order\ of \ pole = 2$$
$$\quad\therefore{{1}\over1-cos(z)} = a_{-2}z^{-2} + a_{-1}z^{-1} + a_{0} + a_{1}z^{1} + a_{2}z^{2} +...$$
$${{1}\over 1-(1-{{z^{2}\over {2 !}}+{z^{4}\over 4!}+...})} = a_{-2}z^{-2} + a_{-1}z^{-1} + a_{0} + a_{1}z^{1} + a_{2}z^{2} +...$$
$$1= ({z^{2}\over 2!} - {z^{4}\over 4!} + {z^{6}\over {6!}}... )(a_{-2}z^{-2} + a_{-1}z^{-1} + a_{0} + a_{1}z^{1} + a_{2}z^{2} +... )$$
$Corresponding\ coefficients \ equal\ with\ each\ other:$
$${{a_{-2}}\over {2!}} = 1 \quad\therefore a_{-2} = 2$$
$${{a_{-1}}\over {2!}} = 0 \quad\therefore a_{-1} = 0$$
$${{a_{0}}\over {2!}} - {{1}\over4!}a_{-2} = 0 \quad\therefore a_{0} = {{1}\over6}$$
$${{a_{0}}\over {2!}} - {{a_{-1}}\over {4!}} = 0 \quad\therefore a_{1} = 0$$
$$similar\ like \ the\ above\ procedure\ we\ can\ get \ a_{2} = {{1}\over120}$$
$$\quad\therefore {{2}\over {z^2}} + {{1}\over6} + {{1}\over120} {z^2} +...,\ Res(f;0) = 0$$

10
##### Term Test 1 / Re: TT1 Problem 4 (morning)
« on: October 16, 2018, 05:06:00 PM »
The solution attached below.

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