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### Messages - Yiheng Bian

Pages: 1 [2]
16
##### Quiz-2 / Re: TUT0101
« on: November 18, 2019, 06:16:04 PM »
1

17
##### Term Test 1 / Re: Problem 2 (noon)
« on: October 23, 2019, 09:06:44 AM »
I think in part b when solve y_2 you skip many steps
the equation is
$$xy_2'-y_2=xcosx-sinx$$
$$\int{xy_2'-y_2}=\int{xcos-sinx}$$
So
$$xy_2=xsinx$$
Therefore
$$y_2=sinx$$

18
##### Term Test 1 / Re: Problem 1 (noon)
« on: October 23, 2019, 08:40:10 AM »
Hi, I think you have typo in line4, it should be +but you type=

19
##### Term Test 1 / Re: Problem 2 (noon)
« on: October 23, 2019, 08:25:22 AM »
$$\text{So when you have solve W then through}$$
$$\left\{ \begin{matrix} y_1 & y_2 \\ y_1' & y_2'\\ \end{matrix} \right\} \tag{2} \text{is equal to W solve y_2}$$

20
##### Term Test 1 / Re: Problem 2 (noon)
« on: October 23, 2019, 08:19:56 AM »
$$\text{emm just parts of solution？？}$$

21
##### Term Test 1 / Re: Problem 4 (noon)
« on: October 23, 2019, 07:58:21 AM »
Step 1:from question we get
$$r^2-8r+25=0$$
we solve the equation get
$$r_1=4+3i,r_2=4-3i$$
So we let
$$y_c=c_1e^{4x}cos{3x}+c_2e^{4x}sin{3x}$$
Step2 we solve
$$y''-8y'+25y=18e^{4x}$$
let
$$y_p=Ae^{4x}$$
So we can know get
$$y_p'=4Ae^{4x}, y_p''=16Ae^{4x}$$
We take into original equation and simplify
$$9Ae^{4x}=18e^{4x}$$
So we get
$$A=2$$
$$y_p=2e^{4x}$$
Step3 we solve
$$y''-8y'+25y=104cos{3x}$$
let
$$y_p=Asin{3x}+Bcos{3x}$$
So
$$y_p'=3Acos3x-3Bsin3x$$
$$y_p''=-9Asin3x-9Bcos3x$$
we take into equation and simplify
So
$$6B+4A=0,2B-3A=13$$
Therefore
$$A=-3,B=2$$
Therefore, general solution is
$$y=c_1e^{4x}cos3x+c_2e^{4x}sin3x +2e^{4x}-3sin3x+2cos3x$$

Since
$$y(0)=0 , y'(0)=0}$$
we take into general solution and get
$$c_1=-4,c_2=1/3$$
So
$$y=-4e^{4x}cos3x+1/3e^{4x}sin3x +2e^{4x}-3sin3x+2cos3x$$

22
##### Quiz-4 / TUT0101 QUIZ4
« on: October 18, 2019, 08:41:21 PM »
$$\text{The question is } 9y''+9y'-4y=0$$
$$9r^2+9r-4=0$$
$$\text{So r1}=\frac{1}{3}$$
$$r2=\frac{-4}{3}$$
$$\text{So we can get y}=c_1e^{\frac{t}{3}}+c_2e^{\frac{-4t}{3}}$$

23
##### Quiz-3 / Re: TUT0101 QUIZ3
« on: October 12, 2019, 01:54:57 AM »
$$\text {If the Wronskian W of f and g is } 3e^{4t} \text {, and if f(t)=} e^{2t} \text {, find g(t)}$$
$$\begin{bmatrix} e^{2t} & g(t) \\ 2e^{2t} &g'(t) \end{bmatrix}$$
$$g'(t)*e^{2t}-2e^{2t}=3e^{4t}$$
$$g'(t)-2g(t)=3e^{2t}$$
$$So, p(t)=-2$$
$$Therefore,\mu=e^{-2t}$$
$$e^{-2t}g'(t)-2e^{-2t}g(t)=3e^{2t}*e^{-2t}$$
$$(e^{-2t}g(t))'=3$$
$$e^{-2t}g(t)=3t+c$$
$$g(t)= \frac {3t+c}{e^{-2t}}\$$

24
##### Quiz-3 / TUT0101 QUIZ3
« on: October 12, 2019, 12:57:19 AM »

25
##### Quiz-2 / Re: TUT0101
« on: October 05, 2019, 05:02:20 PM »
$$(x+2)sin(y)+xcos(y)y’=0, \mu(x,y)=xe^x\\$$
$$M=(x+2)cos(y), N=xcos(y)\\$$
$$M_y=(x+2)cos(y), N_x=cos(y)\\$$
$$Since M_y \neq N_x, \text{so the equation is not exact}\\$$
$$\text{Now multiply}\mu=xe^x \text{both sides}\\$$
$$(x+2)xe^xsin(y)+x^2e^xcos(y)y’=0\\$$
$$M’=(x+2)xe^xsin(y), N’=x^2e^xcos(y)\\$$
$$M’_y=cos(y)(x^2e^x+2xe^x), N’_x=cos(y)(2xe^x+x^2e^x)\\$$
$$M’_y=N’_x\\$$
$$Then$$
$$\psi_x=M’=(x+2)xe^xsin(y)=(x^2e^x+2xe^x)sin(y)\\$$
$$\psi=x^2e^xsin(y)+h(y)\\$$
$$\psi=N’=cos(y)x^2e^x+h’(y)=cos(y)x^2e^x\\$$
$$So h’(y)=0\\$$
$$h( y) \text{is constant C}\\$$
$$So \psi=x^2e^xsin(y)+C\\$$
$$So x^2e^xsin(y)=C\\$$

26
##### Quiz-2 / TUT0101
« on: October 05, 2019, 11:39:56 AM »

27
##### Quiz-1 / Re: TUT0101 QUIZ1
« on: October 03, 2019, 03:26:11 PM »
Sure, you can add absolute value sign. But it doesn’t matter actually. Because whatever + or- . At final step. It is a part of constant C.

28
##### Quiz-1 / Re: TUT0101 QUIZ1
« on: September 27, 2019, 03:55:09 PM »
$$\frac{dy}{dx}= \frac{x+3y}{x-y}$$
$$\frac{dy}{dx}=\frac{1+3\frac{y}{x}}{1-\frac{y}{x}}$$
$$Let \frac{y}{x}=u,y=xu$$
$$\frac{dy}{dx}=\frac{d(xu)}{dx}=\frac{1+3u}{1-u}$$
$$u+x\frac{du}{dx}=\frac{1+3u}{1-u}$$
$$u+x\frac{du}{dx}=\frac{1+3u}{1-u}$$
$$x\frac{du}{dx}=\frac{1+3u}{1-u}-u=\frac{(1+u)^2}{1-u}$$
$$\frac{1-u}{(1-u)^2}=\frac{1}{x}dx$$
$$\int\frac{1-u}{(1+u)^2}du=\int\frac{1}{x}dx$$
$$\int\frac{2}{(1+u)^2}-\frac{1+u}{(1+u)^2du}=lnx+c$$
$$-\frac{2}{(1+u)^2}-ln(1+u)=lnx+c$$
$$-\frac{2x}{x+y}-c+ln(x(1+\frac{y}{x}))$$
$$-\frac{2x}{x+y}-c=ln(x+y)$$
$$ln(x+y)+\frac{2x}{x+y}=-c$$
$$ln(x+y)=-c-\frac{2x}{x+y}$$
$$x+y=e^{-c-\frac{2x}{x+y}}=Ce^{-\frac{2x}{x+y}}$$



29
##### Quiz-1 / TUT0101 QUIZ1
« on: September 27, 2019, 03:02:56 PM »