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Messages - Lan Cheng

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1
Term Test 2 / Re: Problem 4 (morning)
« on: November 19, 2019, 05:59:51 AM »
a) First we should solve for eigenvalues:

let $det\begin{bmatrix}2-\lambda & -3\\
4 & -2-\lambda
\end{bmatrix}=0.\lambda_{1}=2\sqrt{2}i,\lambda_{2}=-2\sqrt{2}i.$

Second we need to solve for eigenvectors:

Let $(A-\lambda I)v=0.$ When $\lambda=2\sqrt{2}i,v=\begin{bmatrix}\sqrt{2}i+1\\
2
\end{bmatrix}.$

Therefore, $e^{(2\sqrt{2}i)t}\begin{bmatrix}\sqrt{2}i+1\\
2
\end{bmatrix}=(cos(2\sqrt{2}t)+isin(2\sqrt{2}t))\begin{bmatrix}\sqrt{2}i+1\\
2
\end{bmatrix}.$

$x(t)=C_{1}\begin{bmatrix}-\sqrt{2}sin(2\sqrt{2}t)+cos(2\sqrt{2}t)\\
2cos(2\sqrt{2}t)
\end{bmatrix}+C_{2}\begin{bmatrix}\sqrt{2}cos(2\sqrt{2}t)+sin(2\sqrt{2}t)\\
2sin(2\sqrt{2}t)
\end{bmatrix}.$

b) See photo below.

2
Term Test 2 / Re: Problem 2 (morning)
« on: November 19, 2019, 05:58:31 AM »
a) We can see the coefficient of y” is -2. Therefore, $W=Ce^{2t}.$

b) First, we should solve for $y'''-2y"+4y'-8y=0.$

Let $r^{3}-2r^{2}+4r-8=0.$

$r_{1}=2,r_{2}=2i,r_{3}=-2i.$

thus, $y_{c}(t)=C_{1}e^{2t}+cos(2t)+sin(2t).$

$W=det\begin{bmatrix}e^{2t} & cos(2t) & sin(2t)\\
2e^{2t} & -2sin(2t) & 2cos(2t)\\
4e^{2t} & -4cos(2t) & -4sin(2t)
\end{bmatrix}=16e^{2t}.$

Therefore, the answer in b) respond that the differential equation in a) is correct.

c) Let $y_{p}(t)=Acos(t)+Bsin(t).$

$y'=-Asin(t)+Bcos(t).y"=-Acos(t)-Bsin(t).y'''=Asin(t)-Bcos(t).$

Plug in, we get $\begin{cases}
-3A-6B=0 & -6A+3B=15.\end{cases}\begin{cases}
A=-2 & B=1.\end{cases}$

Therefore, $y(t)=C_{1}e^{2t}+cos(2t)+sin(2t)-2cos(t)+sin(t).$ Two constants missing

3
Term Test 2 / Re: Problem 1 (morning)
« on: November 19, 2019, 05:57:54 AM »
a) First, we should solve for $y"-y=0.$

Let $r^{2}-1=0,r_{1}=1,r_{2}=-1.$

thus, $y_{c}(t)=C_{1}e^{t}+C_{2}e^{-t}.$

Second, we should calculate W.

$W=\begin{bmatrix}e^{t} & e^{-t}\\
e^{t} & -e^{-t}
\end{bmatrix}=-2,W_{1}=\begin{bmatrix}0 & e^{-t}\\
1 & -e^{-t}
\end{bmatrix}=-e^{-t},W_{2}=\begin{bmatrix}e^{t} & 0\\
e^{t} & 1
\end{bmatrix}=e^{t}.$

therefore, $y_{p}(t)=e^{t}\int\frac{W_{1}(t)g(t)}{W(t)}dt-e^{-t}\int\frac{W_{2}(t)g(t)}{W(t)}dt.$

$y_{p}(t)=6e^{t}\int\frac{e^{-t}}{e^{t}+1}dt+6e^{-t}\int\frac{e^{t}}{e^{t}+1}dt.$

$y_{p}(t)=6e^{t}(-e^{-t}-t+ln(e^{t}+1))-6e^{-t}ln(e^{t}+1).$

Thus, $y(t)=C_{1}e^{t}+C_{2}e^{-t}+6e^{t}(-e^{-t}-t+ln(e^{t}+1))-6e^{-t}ln(e^{t}+1).$

b) $y'(t)=C_{1}e^{t}-C_{2}e^{-t}+6e^{t}ln\left|e^{t}+1\right|+6e^{t}\frac{e^{t}}{e^{t}+1}-6te^{t}-6e^{t}-6e^{-t}ln\left|e^{t}+1\right|+6e^{-t}\frac{e^{t}}{e^{t}+1}.$

plug in, we can get $C_{1}=6-6ln(2),C_{2}=6ln(2).$

Therefore, $y(t)=(6-6ln(2))e^{t}+6ln(2)e^{-t}+6e^{t}(-e^{-t}-t+ln(e^{t}+1))-6e^{-t}ln(e^{t}+1).$


OK, except LaTeX sucks:

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


$$
\boxed{  y= 6\Bigl(-e^{-t}+\ln (1+e^{-t})+1-\ln(2) \Bigr)e^{t} + 6\Bigl(\ln (e^t+1)-\ln(2)\Bigr)e^{-t}. }
$$




4
Quiz-5 / LEC 5101 QUIZ 5
« on: November 01, 2019, 02:00:17 PM »
Find the general solution of the given differential equation:

$y"+4y'+4y=t^{-2}e^{-2t},t>0.$

Let $y"+4y'+4y=0.$

$r^{2}+4r+4=0,r_{1}=-2=r_{2}.$

Thus,

$y_{c}(t)=C_{1}e^{-2t}+C_{2}te^{-2t}.$

$W=\begin{vmatrix}e^{-2t} & te^{-2t}\\
-2e^{-2t} & -2te^{-2t}+e^{-2t}
\end{vmatrix}=-2te^{4t}+e^{4t}+2te^{4t}=e^{4t}.$

$W_{1}=\begin{vmatrix}0 & te^{-2t}\\
1 & -2te^{-2t}+e^{-2t}
\end{vmatrix}=-te^{-2t}.$

$W_{2}=\begin{vmatrix}e^{-2t} & 0\\
-2e^{-2t} & 1
\end{vmatrix}=e^{-2t}.$

Let $\mu_{1}=\int\frac{W_{1}(t)g(t)}{W(t)}dt=\int\frac{-te^{-2t}\times t^{-2}e^{-2t}}{e^{4t}}dt=-ln(t). $

$\mu_{2}=\int\frac{W_{2}(t)g(t)}{W(t)}dt=\int\frac{e^{-2t}\times t^{-2}e^{-2t}}{e^{4t}}dt=-\frac{1}{t}.$

Therefore,

$y(t)=y_{1}(t)\mu_{1}+y_{2}(t)\mu_{2}+y_{c}(t)=-ln(t)e^{-2t}-e^{-2t}+C_{1}e^{-2t}+C_{2}te^{-2t}.$

5
Quiz-5 / Re: LEC5101 Quiz5
« on: October 31, 2019, 09:54:55 PM »
Hi, I think your general solution should be
$y(t)=Y(t)+y_{c}(t)=-ln(t)e^{-2t}-e^{-2t}+C_{1}e^{-2t}+C_{2}te^{-2t}.$


6
Term Test 1 / Re: Problem 2 (afternoon)
« on: October 23, 2019, 07:17:28 AM »
Quote from: Hongling Liu on October 23, 2019, 06:43:41 AM
2:
(2x+1)xy’’ + (2x+2)y’ -2y = 0
Solution:
a):
y’’ + [(2x+2)/(2x+1)x]y’ -[2/(2x+1)x]y= 0
W = C•e^∫-[(2x+2)/(2x+1)x]dx
W = C•(2x+1/x^2) let C = 0
W  = (2x+1/x^2)


7
Term Test 1 / Re: Problem 2 (afternoon)
« on: October 23, 2019, 07:04:14 AM »
a) divide each side by $x(2x+1)$:

$y”+\frac{2x+2}{x(2x+1)}y’-\frac{2}{x(2x+1)}y=0.$

$p(x)=\frac{2x+2}{x(2x+1)},W=ce^{-\int p(x)dx}.$

let $p(x)=\frac{A}{x}+\frac{B}{2x+1}.$

$p(x)=\frac{(2A+B)x+A}{x(2x+1)}, A=2,B=-2.$

$p(x)=\frac{2}{x}-\frac{2}{2x+1}.$

$\int-p(x)dx=\int\frac{2}{2x+1}-\frac{2}{x}dx=ln(2x+1)-2ln(x).$

$W=ce^{ln(2x+1)-2ln(x)}=ce^{ln(2x+1)}e^{ln(x^{-2})}=c(2x+1)(\frac{1}{x^{2}})=c(\frac{2}{x}+\frac{1}{x^{2}})$

b) let $c=1,W=\frac{2}{x}+\frac{1}{x^{2}}=\begin{array}{cc}
x+1 & y_{2}\\
1 & y_{2}'
\end{array}=(x+1)y_{2}'-y_{2}.$

divide each side by $(x+1): y_{2}'-\frac{1}{x+1}y_{2}=\frac{2}{x(x+1)}+\frac{1}{x^{2}(x+1)}.$

$\mu=e^{\int p_{2}(x)dx}=e^{\int-\frac{1}{x+1}dx}=e^{-ln(x+1)}=\frac{1}{x+1}.$

multiply each side by $\mu:\frac{1}{x+1}y_{2}'-\frac{1}{(x+1)^{2}}y_{2}=\frac{2}{x(x+1)^{2}}+\frac{1}{x^{2}(x+1)^{2}}.$

$\frac{1}{x+1}y_{2}=-\frac{1}{x(x+1)},y_{2}=-\frac{1}{x}.$

Therefore, $y_{1}=x+1,y_{2}=-\frac{1}{x}.$

c)$y(-1)=1,y'(-1)=0.$

$y(x)=C_{1}y_{1}+C_{2}y_{2}=C_{1}(x+1)+C_{2}(-\frac{1}{x}).$

$y'(x)=C_{1}+C_{2}x^{-2}.$

$\begin{cases}
C_{1}=-1 & C_{2}=1\end{cases}.$

Therefore, the general solution is $y(x)=-x-1-\frac{1}{x}.$

8
Term Test 1 / Re: Problem 1 (afternoon)
« on: October 23, 2019, 06:53:28 AM »
a) Let $M=-y^{2}sin(xy),N=-xysin(xy)+2cos(xy)+3y,$

$My=-2ysin(xy)-xy^{2}cos(xy), Nx=-ysin(xy)-xy^{2}cos(xy)-2ysin(xy).$ Should be $M_y$ and so on

Let $R_{1}=\frac{My-Nx}{M}=\frac{ysin(xy)}{-y^{2}sin(xy)}=-\frac{1}{y},$

$\mu=e^{-\int R_{1}dy}=e^{\int\frac{1}{y}dy}=e^{ln(y)}=y.$

multiply each side by y,

$-y^{3}sin(xy)+(-xy^{2}sin(xy)+2ycos(xy)+3y^{2})y'=0.$

$My=-3y^{2}sin(xy)-xy^{3}cos(xy),Nx=-3y^{2}sin(xy)-xy^{3}cos(xy)=My.$

Thus, there exist $\varphi(x,y) such that \varphi x=M,\varphi y=N.$

$\varphi=y^{2}cos(xy)+h(y),\varphi y=2ycos(xy)-xy^{2}sin(xy)+h'(y)=-xy^{2}sin(xy)+2ycos(xy)+3y^{2}.$

$h'(y)=3y^{2},h(y)=y^{3}+C.$

Thus, $y^{2}cos(xy)+y^{3}=C.$

b) $y(\frac{\pi}{3})=1,cos(\frac{\pi}{3})+1=C.$

$C=\frac{3}{2},y^{2}cos(xy)+y^{3}=\frac{3}{2}.$

9
Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 06:37:47 AM »
a) 1. let $y"-5y'+6y=0.$

$r^{2}-5r+6=0.$

$\begin{cases}
r_{1}=2 & r_{2}=3.\end{cases}$

Thus, $y_{c}(x)=C_{1}e^{2x}+C_{2}e^{3x}.$

2.let $y"-5y'+6y=52cos(2x).$

let $y_{p}(x)=Acos(2x)+Bsin(2x). y'=-2Asin(2x)+2Bcos(2x),y"=-4Acos(2x)-4Bsin(2x).$

$-4Acos(2x)-4Bsin(2x)+10Asin(2x)-10Bcos(2x)+6Acos(2x)+6Bsin(2x)=52cos(2x).$

$\begin{cases}
-4A-10B+6A=52 & -4B+10A+6B=0\end{cases}.$

$A=1,B=-5. $

$y_{p}(x)=cos(2x)-5sin(2x).$

Therefore, $y(x)=C_{1}e^{2x}+C_{2}e^{3x}+cos(2x)-5sin(2x).$

b) $y’(x)=+2C_{1}e^{2x}+3C_{2}e^{3x}-2sin(2x)-10cos(2x).$

$\begin{cases}
C_{1}+C_{2}+1=0 & 2C_{1}+3C_{2}-10=0\end{cases}.$

$C_{1}=-13,C_{2}=12.$

Therefore, $y(x)=12e^{3x}-13e^{2x}+cos(2x)-5sin(2x).$

OK. V.I.

10
Quiz-2 / tut 5103 quiz 2
« on: October 04, 2019, 02:00:01 PM »
show the given equation is not exact but becomes exact when multiplied by the given integrating factor, then solve the equation.

x^(2) * y^(3) + x * (1+y^2) * y' = 0, u(x, y) = 1/(xy^3).

First, let's show the given DE x^(2) * y^(3) + x * (1+y^2)y' = 0 is not exact.

Define M = x^(2) * y^(3), N = x * (1 + y^2).

M_y = d/(dy) [x^(2) * y^(3)] = 3x^(2) * y^(2)

N_x = d/(dx) [x * (1 + y^2)] = 1 + y^2

Since 3x^(2)y^(2) != 1 + y^2, the given DE is not exact.

multiply each side of given DE by integrating factor u, we get

1/(xy^3) * x^(2) * y^(3) + 1/(xy^3) * x * (1 + y^2)y' = x + (y^(-3) + y^(-1)) * y' = 0

Let the new M = 1/(xy^3) x^(2) * y^(3), new N =  1/(xy^3) * x * (1 + y^2)

M_y = 0, N_x = 0, M_y = N_x and the new DE is exact.

there exist \phy (x, y) such that

\phy x = M, \phy y = N.

\phy x = M = 1/(x * y^3) * x^(2) * y^(3)

Integrating both side by x, we have

\phy = 1/2 * x^2 + h(y)

\phy y = h'(y) = N =  1/(xy^3) *  x * (1 + y^2)

h'(y) =  1/(xy^3) *  x * (1 + y^2)

Integrating both side by y, we have

h(y) = -0.5 * y^(-2) + ln|y| + C

Altogether, we have

\phy (x,y) = 0.5 * x^2 - 0.5 * y^(-2) + ln|y| + C

So our general solution is

0.5 * x^2 - 0.5 * y^(-2) + ln|y| = C.

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