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### Messages - Ruojing Chen

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1
##### Term Test 2 / Re: Problem 4 (morning)
« on: November 19, 2019, 06:17:30 AM »
a) sketch

2
##### Term Test 2 / Re: Problem 2 (morning)
« on: November 19, 2019, 06:05:06 AM »
$$y'''-2y''+4y'-8y=15cost$$
a) $$p(t)=-2$$
$$w=ce^{-\int p(t)}=ce^{\int2dt}=ce^{2t}$$
b)$$y'''-2y''+4y'-8y=0$$
$$r^3-2r^2+4r-8=0$$
$$(r-2)(r^2+4)=0$$
$$\therefore r=2,\pm2i$$
$$y_c(t)=c_1e^{2t}+c_2cos2t+c_3sin2t$$
$$w(y_1,y_2,y_3)(t)=\left[\begin{matrix} e^{2t} & cos2t & sin2t \\ 2e^{2t} & -2sin2t & 2cos2t \\ 4e^{2t} & -4cos2t & -4sin2t \end{matrix}\right]=16e^{2t}$$
$$\therefore w=ce^{2t}=16e^{2t}, c=16$$
c)Assume $$y_p(t)=Acost+Bsint$$
$$y'=-Asint+Bcost$$$$y''=-Acost-Bsint$$$$y'''=Asint-Bcost$$
By plugging in y,y',y'',y''' into the function
$$Asint-Bcost+2Acost+2Bsint-4Asint+4Bcost-8Acost-8Bsint=15cost$$
$$\left\{ \begin{array}{lr} A+2B-4A-8B=0, & \\ -B+2A+4B-8A=15 \end{array} \right.$$
$$\Rightarrow \left\{ \begin{array}{lr} A+2B=0, & \\ B-2A=5 \end{array} \right.$$
$$\therefore \left\{ \begin{array}{lr} A=-2, & \\ B=1, & \end{array} \right.$$

$$y(t)=c_1e^{2t}+c_2cos2t+c_3sin2t-2cost+sint$$

OK, except LaTeX sucks:

2)  "operators" should be escaped: \cos, \sin, \tan, \ln

$$\boxed{y= -2\cos(t)+\sin(t) + C_1e^{2t} +C_2\cos(2t) +C_3\sin(2t).}$$

3
##### Term Test 2 / Re: Problem 4 (morning)
« on: November 19, 2019, 06:04:30 AM »
$$x'= \left ( \begin{matrix} 2 & -3 \\ 4 & -2 \end{matrix} \right ) x$$

$$det(A-\lambda I)=0$$
$$(2-\lambda)(-2-\lambda)-(-3)*4=0$$
$$-4+\lambda^2+12=0$$
$$\lambda^2=-8$$
$$\Rightarrow \lambda=\pm2\sqrt{2}i$$

$$when \lambda=2\sqrt{2}i$$
$$\left ( \begin{matrix} 2-2\sqrt{2}i & -3 \\ 4 & -2-2\sqrt{2}i \end{matrix} \right ) \left ( \begin{matrix} x_1 \\ x_2 \end{matrix} \right ) =\left ( \begin{matrix} 0\\ 0 \end{matrix} \right )$$
$$(2-2\sqrt{2}i)x_1=3x_2$$
$$\Rightarrow x=t \left ( \begin{matrix} 3 \\ 2-2\sqrt{2}i \end{matrix} \right )$$

$$e^{2\sqrt{2}it}=t \left ( \begin{matrix} 3 \\ 2-2\sqrt{2}i \end{matrix} \right )$$
$$=(cos2\sqrt{2}t+isin2\sqrt{2}t)\left ( \begin{matrix} 3 \\ 2-2\sqrt{2}i \end{matrix} \right )$$

$$\therefore =c_1 \left( \begin{matrix} 3cos2\sqrt{2} \\ 2cos2\sqrt{2}+2\sqrt{2}sin2\sqrt{2} \end{matrix} \right )+ c_2\left(\begin{matrix} 3sin2\sqrt{2} \\ 2sin2\sqrt{2}-2cos2\sqrt{2} \end{matrix} \right )$$

4
##### Term Test 2 / Re: Problem 3 (morning)
« on: November 19, 2019, 06:03:57 AM »
a)$$x'= \left( \begin{matrix} -2 & 1 \\ -1 & 0 \\ \end{matrix} \right )x+\left( \begin{matrix} 0 \\ \frac{e^{-t}}{t^2+1} \\ \end{matrix} \right )$$

$$det(A-\lambda I)=0$$
$$(-2-\lambda)(-\lambda)+1=0$$
$$\lambda^2+2\lambda+1=0$$
$$\lambda_1=\lambda_2=-1$$

$$when \lambda =-1$$
$$(A-\lambda I)x=0$$
$$\left( \begin{matrix} -1 & 1 \\ -1 & 1 \\ \end{matrix} \right )\left( \begin{matrix} x_1 \\ x_2 \\ \end{matrix} \right )=\left( \begin{matrix} 0 \\ 0 \\ \end{matrix} \right )$$
$$x_1=x_2 \Rightarrow x=t\left( \begin{matrix} 1 \\ 1 \\ \end{matrix} \right )$$

$$\left( \begin{matrix} -1 & 1 \\ -1 & 1 \\ \end{matrix} \right )\left( \begin{matrix} x_1 \\ x_2 \\ \end{matrix} \right )=\left( \begin{matrix} 1 \\ 1 \\ \end{matrix} \right )$$
$$x_1+x_2=1 \Rightarrow x=t\left( \begin{matrix} 0 \\ 1 \\ \end{matrix} \right )$$

$$\therefore y=c_1e^{-t}\left( \begin{matrix} 1 \\ 1 \\ \end{matrix} \right )+c_2e^{-t}(\left( \begin{matrix} 1 \\ 1 \\ \end{matrix} \right )t+\left( \begin{matrix} 0 \\ 1 \\ \end{matrix} \right ))$$

B)$$\phi u'=g(t)$$
$$\left( \begin{matrix} e^{-t} & e^{-t}t \\ e^{-t} & e^{-t}t+e^{-t}\\ \end{matrix} \right ) \left( \begin{matrix} u_1' \\ u_2' \\ \end{matrix} \right )=\left( \begin{matrix} 0 \\ \frac{e^{-t}}{t^2+1} \\ \end{matrix} \right )$$
$$\left \{ \begin{array}{lr} u_1'=-\frac{t}{t^2+1} &\\ u_2'=\frac{1}{t^2+1} \end{array} \right.$$
$$\Rightarrow \left \{ \begin{array}{lr} u_1=-\frac{1}{2} \ln(t^2+1)&\\ u_2=\arctan t \end{array} \right.$$
$$x=\phi u$$
$$\therefore x=\left( \begin{matrix} e^{-t} & e^{-t}t \\ e^{-t} & e^{-t}t+e^{-t}\\ \end{matrix} \right )\left( \begin{matrix} -\frac{1}{2}\ln(t^2+1)\\ \arctan t\\ \end{matrix} \right )$$

$$x=(-\frac{1}{2}\ln(t^2+1)+c_1)\left( \begin{matrix} e^{-t} \\ e^{-t} \\ \end{matrix} \right )+(c_2+\arctan t)\left( \begin{matrix} e^{-t}t \\ e^{-t}t+e^{-t} \\ \end{matrix} \right)$$

OK, but LaTeX sucks:
1) \det should be escaped as well
2) Text should not be  included in math formulae, or included  through  \tex{blah blah } to make it upright and properly spaced

3) Directions are opposite

5
##### Quiz-5 / Lec5101 quiz5
« on: November 01, 2019, 02:00:00 PM »
Find the general solution of the given differential equation.
$y''+4y=3csc(2t), o<t<\frac{\pi}{2}$

For homogeneous equation: y''+4y=0
$$r^2+4=0$$
$$r=\pm2i$$
$$y_c(t)=c_1cos2t+c_2sin2t$$

For non-homogeneous equation:y''+4y=3csc(2t)
$$p(t)=0,q(t)=4,g(t)=3csc(2t)$$
$$W=2cos^2t+2sin^2t=2$$
$$u_1=-\int\frac{y_2(t)g(t)}{W}=-\int\frac{sin2t*3csc2t}{2}dt=-\frac{3}{2}t$$
$$u_2=\int\frac{y_1(t)g(t)}{W}=\int\frac{cos2t*3csc2t}{2}dt=\frac{3}{4}ln|2t|$$
$$y_p(t)=u_1y_1(t)+u_2y_2(t) =cos2t*(-\frac{3}{2}t)+sin2t*(\frac{3}{4}ln|2t|)$$

$$\therefore y=c_1cos2t+c_2sin2t+cos2t*(-\frac{3}{2}t)+sin2t*(\frac{3}{4}ln|2t|)$$

6
##### Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 06:45:49 AM »
(a)when $$y''-5y'+6y=0$$
$$r^2-5r+6=0$$
$$(r-2)(r-3)=0$$
$$r_1=-2,r_2=-3$$
$$\therefore y_c(x)=c_1e^{-2x}+c_2e^{-3x}$$

when $$y''-5y'+6y=52Cos(2x)$$
$$y_p(x)=ACos(2x)+BSin(2x)$$
$$y'=-2ASin(2x)+2BCos(2x)$$
$$y''=-4ACos(2x)-4BSin(2x)$$
$$-4ACos(2x)-4BSin(2x)+10ASin(2x)-10BCos(2x)+6ACos(2x)+6BSin(2x)=52Cos(2x)$$
$$Cos(2x)(-4A-10B+6A)=52Cos(2x)$$
$$Sin(2x)(-4B+10A+6B)=0$$
$$\therefore A-5B=26, B+5A=0$$
$$A=1,B=-5$$
$$\therefore y_p(x)=Cos(2x)-5Sin(2x)$$

$$y=y_c(x)+y_p(x)=c_1e^{-2x}+c_2e^{-3x}+Cos(2x)-5Sin(2x)$$

(b) y(0)=0,y'(0)=0
$$y=c_1e^0+c_2e^0+1=0$$
$$c_1+c_2=-1$$
$$y'=-2c_1e^{2x}-3c_2e^{3x}+2Sin(2x)-10Cos(2x)=2c_1e^0+3c_2e^0-10=0$$
$$-2c_1-3c_2=10$$
$$\therefore c_2=-8 c_1=7$$

$$\therefore y=7e^{-2x}-8e^{-3x}+Cos(2x)-5Sin(2x)$$

7
##### Term Test 1 / Re: Problem 1 (morning)
« on: October 23, 2019, 06:37:18 AM »
(a) Let $$M=-ySin(x)+y^3Cos(x)$$
$$N=3Cos(x)+5y^2Sin(x)$$
Then$$M_y=-Sin(x)+3y^2Cos(x)$$
$$N_x=-3Sinx(x)+5y^2Cos(x)$$
$$R=\frac{M_y-N_x}{M}=\frac{2Sin(x)-2Cos(x)}{-ySin(x)+y^3Cos(x)}=\frac{2(Sin(x)-y^2Cos(x))}{-y(Sin(x)-y^2Cos(x))}=-\frac{2}{y}$$
$$\mu=e^{-\int_Rdy}=e^{\int_\frac{2}{y}dy}=e^{2lny}=e^ln(y^2)=y^2$$

Multiple both side by $$\mu=y^2$$
$$y^2(-ySin(x)+y^3Cos(x))+y^2(3Cos(x)+5y^2Sin(x))=0$$
$$M'=-y^3Sin(x)+y^5Cos(x)$$,$$N'=3y^2Cos(x)+5y^4Sin(x)$$
$$\exists\phi_x,y,such that \phi_x=M',\phi_y=N$$
$$\phi=\int_M'dx=\int_-y^3Sin(x)=y^3Cos(x)+y^5Sin(x)+h(y)$$
$$\phi_y=3y^2Cos(x)+5y^4Sin(x)+h'(y)=N'$$
$$h'(y)=0$$
$$h(y)=c$$

$$\therefore \phi=y^3Cos(x)+y^5Sin(x)=c$$
(b)When $$y(\frac{\pi}{4})=\sqrt{2}$$
$$(\sqrt{2})^3Cos(\frac{\pi}{4})+(\sqrt{2}^5)Sin(\frac{\pi}{4})=2\sqrt{2}*\frac{1}{\sqrt{2}}+(4\sqrt{2}*\frac{1}{\sqrt{2}})=2+4=6$$
$$\therefore c=6$$

$$\phi=y^3Cos(x)+y^5Sin(x)=6$$

What is your real life name? I can find it by email, but I am too lazy

8
##### Term Test 1 / Re: Problem 4 (morning)
« on: October 23, 2019, 06:36:58 AM »
(a)When $$y''-6y'+25y=0$$
$$r^2-6r+25=0$$
$$r=\frac{6\pm\sqrt{6^2-4*25}}{2}=\frac{6\pm\sqrt{-64}}{2}=3\pm4i$$
$$\therefore y_c(x)=c_1e^{3x}Cos(3x)+c_2e^{3x}Sin(3x)$$
When$$y''-6y'+25y=16e^{3x}$$
$$y_p(x)=Ae^{3x}$$
$$y'=3Ae^{3x}$$
$$y''=9Ae^{3x}$$
$$9Ae^{3x}-18Ae^{3x}+25Ae^{3x}=16e^{3x}$$
$$16Ae^{3x}=16e^{3x}$$
$$A=1$$
$$\therefore y_p(x)=e^{3x}$$
When $$y''-6y'+25y=102Sin(x)$$
$$y_p(x)=BCos(x)+CSin(x)$$
$$y'=-BSin(x)+CCos(x)$$
$$y''=-BCos(x)-CSin(x)$$
$$-BCos(x)-CSin(x)+6BSin(x)-6CCos(x)+25BCos(x)+25CSin(x)=102Sin(x)$$
$$Cos(x)(-B-6C+25B)=0$$
$$Sin(x)(-C+6B+25C)=102Sin(x)$$
$$4B-C=0$$
$$4C+B=17$$
$$4B=C, 4(4B)+B=17$$
$$B=1$$
$$C=4$$
$$\therefore y_p(x)=Cos(x)+4Sin(x)$$
$$y(x)=y_c(x)+y_p(x)=c_1e^{3x}Cos(4x)+c_2e^{3x}Sin(4x)+e^{3x}+Cos(x)+4Sin(x)$$
(b) y(0)=0, y(0)=0
$$c_1e^0Cos(0)+c_2e^0Sin(0)+e^0+Cos(0)+4Sin(0)=0$$
$$c_1=-2$$
$$y'=3c_1e^{3x}Cos(4x)-4c_1e^{3x}Sin(4x)+3c_2e^{3x}Sin(4x)+4c_2e^{3x}Cos(4x)+3e^{3x}-Sin(x)+4Cos(x)$$
$$3c_1e^0Cos(0)-4c_1e^0Sin(0)+3c_2e^0Sin(0)+4c_2e^0Cos(0)+3e^0-Sin(0)+4Cos(0)=0$$
$$3c_1+4c_2=-7$$
$$plug in c_1=-2$$
$$3(-2)+4c_2=-7$$
$$c_2=-\frac{1}{4}$$
$$y=-2e^{3x}Cos(4x)-\frac{1}{4}e^3xSin(4x)+e^{3x}+Cos(x)+4Sin(x)$$

OK. V.I.

9
##### Term Test 1 / Re: Problem 3 (morning)
« on: October 23, 2019, 06:36:33 AM »
As$$Sinh(2x)=\frac{e^{x}-e^{-x}}{2}$$
$$48Sinh(2x)=48(\frac{e^{2x}-e^{-2x}}{2})=24e^{2x}-24e^{-2x}$$
when $$y''-6y'+8y=0$$ $$r^2-6r+8=0$$
$$(r-4)(r-2)=0$$
$$r_1=4,r_2=2$$
$$\therefore y_c(x)=c_1e^{4x}+c_2e^{2x}$$
when $$y''-6y'+8=24e^{2x}$$
$$y_p(x)=Ae^{2x}x$$
$$y'=2Ae^{2x}x+Ae^{2x}$$
$$y''=4Ae^{2x}x+2Ae^{2x}+2Ae^{2x}=4Ae^{2x}x+4Ae^{2x}$$
$$4Ae^{2x}x+4Ae^{2x}-12Ae^{2x}x-6Ae^{2x}+8Ae^{2x}x=24e^{2x}$$
$$-2Ae^{2x}=24e^(2x)$$
$$A=-12$$
$$\therefore y_p(x)=-12e^{2x}x$$
when$$y''-6y'+8=-24e^{-2x}$$
$$y_p(x)=Be^{-2x}$$
$$y'=-2Be^{-2x}$$
$$y''=4Be^{-2x}$$
$$4Be^{-2x}+12Be^{-2x}+8Be^{-2x}=-24e^{-2x}$$
$$24Be^{-2x}=-24e^{-2x}$$
$$B=-1$$
$$\therefore y_p(x)=-e^{-2x}$$

$$y=y_c(x)+y_p(x)=c_1e^{4x}+c_2e^{2x}-12e^{2x}x-e^{-2x}$$

(b)when y(0)=0 $$c_1e^0+c_2e^0-12e^o*0-e^0=0$$
$$c_1+c_2=1$$
when y'(0)=0
$$y'=4c_1e^{4x}+2c_2e^{2x}-24e^{2x}x-12e^{2x}+2e^{-2x}$$
$$4c_1e^0+2c_2e^0+24e^0*0-12e^0+2e^0=0$$
$$4c_1+2c_2-12+2=0$$
$$2c_1+c_2=5$$
$$\therefore c_1=4,c_2=-3$$

$$y=4e^{4x}-3e^{2x}+12e^{2x}x-e^{-2x}$$ Some errors. V.I.

10
##### Quiz-4 / tut 0202 quiz4
« on: October 18, 2019, 02:06:36 PM »
Find the general solution for equation y''-2y'+6y=0

$$set \ r^2-2r+6=0$$

$$r=\frac{2\pm\sqrt{(-2)^2-4*6}}{2}$$

$$r=\frac{2\pm2i\sqrt{5}}{2}$$

$$r=1\pm\sqrt{5}i$$

$$y_c(t)=c_1e^tCos(\sqrt{5}t)+c_2e^tSin(\sqrt{5}t)$$

11
##### Quiz-4 / tut 0202 quiz4
« on: October 18, 2019, 02:02:34 PM »
Find the general solution for equation y''-2y'+6y=0

$$set \ r^2-2r+6=0$$

$$r=\frac{2\pm\sqrt{(-2)^2-4*6}}{2}$$

$$r=\frac{2\pm2i\sqrt{5}}{2}$$

$$r=1\pm\sqrt{5}i$$

$$y_c(t)=c_1e^tCos(\sqrt{5}t)+c_2e^tSin(\sqrt{5}t)$$

12
##### Quiz-3 / TUT 0202 Quiz3
« on: October 11, 2019, 02:00:02 PM »
0202 solution

13
##### Quiz-2 / TUT 0202 Quiz2
« on: October 04, 2019, 03:35:10 PM »

14
##### Quiz-1 / TUT 0202 Quiz
« on: September 27, 2019, 03:00:21 PM »
solution

15
##### Quiz-1 / TUT 0202 Quiz
« on: September 27, 2019, 02:11:52 PM »
dy/dx = ((x^2)+(3y^2))/(2xy)

dy/dt = (1+3y^2/x^2)/(2y/x)

let u = y/x, then y = ux

dy/dx = d(ux)/dx = (du/dx)*x + u = (1+3u^2) / (2u) = 1/(2u) + 3u/2
(du/dx)*x= 1/(2u) + u/2 = (1+u^2)/(2u)
∫ ( (2u) / (1+u^2) ) du = ∫ (1/x) dx
ln|1+(y/x)^2| = ln|x| + c

1+u = ax, a = e^c
1 + (y^2)/(x^2) - ax = 0
y^2 + x^2 - ax^3 = 0

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