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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 »
Answer

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 »
ANSWER

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 »
answer

Pages: 1 [2]