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### Messages - Jingjing Cui

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1
##### Quiz 3 / Quiz3 TUT5101
« on: February 10, 2020, 01:31:56 PM »
$$u_{tt}-c^2u_{xx}=0\\ u|_{t=0}=0\\ u_{t}|_{t=0}=1\\ u_{x}|_{x=0}=0\\$$
general solution: u(x,t)=f(x+ct)+g(x-ct)
when x+ct>0 and x-ct>0:
$$u(x,0)=f(x)+g(x)=0\\ u_{t}(x,0)=cf'(x)-cg'(x)=1\\ f'(x)-g'(x)=\frac{1}{c}\\ f(x)-g(x)=\frac{x}{c}\\$$
Let s>0
$$f(s)=\frac{s}{2c}\\ g(s)=-\frac{s}{2c}\\ u(x,t)=\frac{x+ct}{2c}-\frac{x-ct}{2c}=t\\$$
when x+ct>0 and x-ct<0:
$$u_{x}(0,t)=f'(ct)+g'(-ct)=0\\$$
Let s=-ct<0
$$-f'(-s)=g'(s)\\ f(-s)=g(s)\\ u(x,t)=\frac{x+ct}{2c}+\frac{ct-x}{2c}=t\\$$

2
##### Quiz 2 / Quiz2 TUT5101
« on: January 31, 2020, 03:39:38 PM »
$$2u_{t}+t^2u_{x}=0\\ \frac{dt}{2}=\frac{dx}{t^2}=\frac{du}{0}\\ \int\frac{1}{2}t^2dt=\int1dx\\ \frac{1}{6}t^3+A=x\\ A=x-\frac{1}{6}t^3\\$$
Because c=0, so
$$u(t,x)=g(A)=g(x-\frac{1}{6}t^3)$$

The initial condition given in the question: u(x,0)=f(x)
The characteristics curves ($A=x-\frac{1}{6}t^3$) will always intersect t=0 (x-axis) at a unique point, no matter what value A takes. Thus, the solution always exist.

3
##### Quiz 1 / Quiz1 TUT5101
« on: January 24, 2020, 09:37:59 AM »
Question 1: $u_{xx}+u_{xxyy}+u=0$

This is a 4th order linear homogeneous equation since all the terms in the equation are related to u and the operator of the equation $\frac{d^2u}{dx^2}+\frac{d^2u}{dx^2}\frac{d^2u}{dy^2}+1$ is linear.

Question 2: Find the general solution for $u_{xyz}=xy\\ u_{xy}=xyz+f(x,y)\\ u_{x}=\frac{1}{2}xy^2z+F(x,y)+g(x,z)\\ u=\frac{1}{4}x^2y^2z+\hat{F}(x,y)+G(x,z)+h(y,z)$

4
##### Term Test 2 / Re: Problem 3 (noon)
« on: November 19, 2019, 07:17:44 AM »
b)
$$\phi(t) U'(t)=g(t)\\ \begin{pmatrix} -e^{-t}&2e^{2t}\\ e^{-t}&e^{2t}\\ \end{pmatrix}(\begin{array}{cc} U_1' \\ U_2' \end{array})=(\begin{array}{cc} 0 \\\frac{6e^{3t}}{e^{2t}+1} \end{array})\\ -U_1'e^{-t}+2U_2'e^{2t}=0\\ U_1'e^{-t}+U_2'e^{2t}=\frac{6e^{3t}}{e^{2t}+1}\\ U_1'=\frac{4e^{4t}}{e^{2t}+1}\\ so\;U_1=2e^{2t}-2ln|e^{2t}+1|+C_1\\ U_2'=\frac{2e^{t}}{e^{2t}+1}\\ so\;U_2=2arctan(e^t)+C_2\\ x=\phi(t)U(t)\\ so\;the\;solution\;is\;:\\ x=(2e^{2t}-2ln|e^{2t}+1|+C_1)(\begin{array}{cc} -e^{-t} \\ e^{-t} \end{array})+(2arctan(e^t)+C_2)(\begin{array}{cc} 2e^{2t} \\ e^{2t} \end{array})\\$$

OK, except LaTeX sucks:

1) text should not be a part of math formulae or included like \text{blah blah}
2)  "operators" should be escaped: \cos, \sin, \tan, \ln

5
##### Term Test 2 / Re: Problem 3 (noon)
« on: November 19, 2019, 06:59:22 AM »
$$a)det(A-\lambda I)=0\\ det\begin{vmatrix} 1-\lambda&2\\ 1&-\lambda\\ \end{vmatrix}=0\\ (1-\lambda)(-\lambda)-2=0\\ \lambda_1=-1 \;\; \lambda_2=2\\ (A-\lambda I)x=0\\ \\ when\; \lambda=-1\\ \begin{pmatrix} 2&2\\ 1&1\\ \end{pmatrix}-> \begin{pmatrix} 2&2\\ 0&0\\ \end{pmatrix}\\ so\; 2x_1+2x_2=0\\ let\; x_2=t\;\;\;then\;x_1=-t\\ so\;the\;corresponding\;eigenvector\;is\; (\begin{array}{cc} -1 \\ 1 \end{array})\\ when\; \lambda=2\\ \begin{pmatrix} -1&2\\ 1&-2\\ \end{pmatrix}-> \begin{pmatrix} -1&2\\ 0&0\\ \end{pmatrix}\\ so\; -x_1+2x_2=0\\ let\; x_2=t\;\;\;then\;x_1=2t\\ so\;the\;corresponding\;eigenvector\;is\; (\begin{array}{cc} 2 \\ 1 \end{array})\\ so\;the\;general\;solution\;is\;x=c_1e^{-t}(\begin{array}{cc} -1 \\ 1 \end{array})+c_2e^{2t}(\begin{array}{cc} 2 \\ 1 \end{array})\\$$

6
##### Quiz-5 / TUT0402 Quiz5
« on: November 01, 2019, 01:57:17 PM »
$$(1-t)y''+ty'-y=2(t-1)^2e^{-t} ,\;\; 0<t<1\\ y_1(t)=e^t\\ y_2(t)=t\\ Verify \;\;y_1(t) \;\;and\;\; y_2(t)\;\; satisfy\;\; the\;\; corresponding\;\; homogeneous\;\; equation:\\ y_1'(t)=y_1''(t)=e^t\\ (1-t)e^t+te^t-e^t=0\\ y_2'(t)=t \;\; y_2''(t)=1\\ (1-t)t+t^2-t=0\\ y''+\frac{t}{1-t}y'-\frac{1}{1-t}y=-2(t-1)e^{-t}\\ g(t)=-2(t-1)e^{-t}\\ W=det\begin{vmatrix} e^t&t\\ e^t&1\\ \end{vmatrix} =e^t-te^t=e^t(1-t)\\ W_1=det\begin{vmatrix} 0&t\\ 1&1\\ \end{vmatrix} =-t\\ W_2=det\begin{vmatrix} e^t&0\\ e^t&1\\ \end{vmatrix} =e^t\\ Y(t)=y_1(t)\int\frac{W_1g(t)}{W}dt+y_2(t)\int\frac{W_2g(t)}{W}dt\\ =e^t\int\frac{2t(t-1)e^{-t}}{e^t(1-t)}dt-t\int\frac{2e^t(t-1)e^{-t}}{e^t(1-t)}dt\\ =-e^t\int(2te^{-2t})dt+2t\int(e^{-t})dt\\ =(t+\frac{1}{2})e^{-t}-2te^{-t}\\ =\frac{1}{2}e^{-t}-te^{-t}\\ y(t)=c_1e^t+c_2t+\frac{1}{2}e^{-t}-te^{-t}\\$$

7
##### Term Test 1 / Re: Problem 2 (morning)
« on: October 23, 2019, 07:28:41 AM »
c)
$$y(x)=c_1(e^x)+c_2(\frac{1}{2}x^2e^x)\\ y(1)=c_1(e^1)+c_2(\frac{1}{2}1^2e^1)=c_1e+c_2(\frac{1}{2}e)=0\\ c_1=-\frac{1}{2}c_2\\ y'(x)=c_1(e^x)+c_2xe^x+c_2(\frac{1}{2}x^2e^x)\\ y'(1)=c_1(e^1)+c_2e^1+c_2(\frac{1}{2}1^2e^1)=c_1e+c_2e+\frac{1}{2}c_2e=e\\ c_1+c_2+\frac{1}{2}c_2=1\\ -\frac{1}{2}c_2+c_2+\frac{1}{2}c_2=1\\ c_2=1\\ c_1=-\frac{1}{2}\\ y(x)=-\frac{1}{2}e^x+\frac{1}{2}x^2e^x\\$$

8
##### Term Test 1 / Re: Problem 2 (morning)
« on: October 23, 2019, 07:19:53 AM »
b)
$$y_1(x)=e^x\\ y_1'(x)=y_1''(x)=e^x\\ x(e^x)-(2x+1)e^x+(x+1)e^x=xe^x-2xe^x-e^x+xe^x+e^x=0\\ so \;\;y_1(x)=e^x \;is \;a \;solution\\ \\ take \;c=1, then \;W=xe^{2x}\\ W=det \begin{vmatrix} e^x&y_2(x)\\ e^x&y_2'(x)\\ \end{vmatrix}\\ e^xy_2'(x)-e^xy_2(x)=xe^{2x}\\ y_2'(x)-y_2(x)=xe^{x}\\ p(x)=-1\\ \mu=e^{\int{p(x)}dx}=e^{-x}\\ e^{-x}y_2'(x)-e^{-x}y_2(x)=x\\ e^{-x}y_2(x)=\int{x}dx=\frac{1}{2}x^2+c\\ y_2(x)=\frac{1}{2}x^2e^{x} \;(take\;c=0)\\$$

9
##### Term Test 1 / Re: Problem 2 (morning)
« on: October 23, 2019, 07:06:28 AM »
a)
$$y''-\frac{(2x+1)}{x}y'+\frac{x+1}{x}y=0\\ W=ce^{\int{-p(x)dx}}=ce^{\int{\frac{(2x+1)}{x}dx}}\\ \int{\frac{(2x+1)}{x}dx}=2x+ln|x|\\ W=ce^{2x+ln|x|}=cxe^{2x}\\$$

10
##### Term Test 1 / Re: Problem 3 (main)
« on: October 23, 2019, 06:59:24 AM »
a)
$$y''-2y'-3y=16cosh(x)=8e^{x}+8e^{-x}\\ r^2-2r-3=0\\ (r+1)(r-3)=0\\ r_1=-1\\ r_2=3\\ y_c(x)=c_1e^{-x}+c_2e^{3x}\\ \\ y''-2y'-3y=8e^{x}\\ y_{p1}(x)=ce^x\\ y_{p1}'(x)=y_{p1}''=ce^x\\ ce^x-2ce^x-3ce^x=8e^{x}\\ -4c=8\\ c=-2\\ y_{p1}(x)=-2e^x\\ y''-2y'-3y=8e^{-x}\\ y_{p2}(x)=cxe^{-x}\\ y_{p2}'(x)=ce^{-x}-cxe^{-x}\\ y_{p2}''(x)=-ce^x-ce^{-x}+cxe^{-x}\\ -2ce^{-x}+cxe^{-x}-2ce^{-x}+2cxe^{-x}-3cxe^{-x}=8e^{-x}\\ e^{-x}(-2c-2c)+xe^{-x}(c+2c-3c)=8e^{-x}\\ -4c=8\\ c=-2\\ y_{p2}(x)=-2xe^{-x}\\ y(x)=c_1e^{-x}+c_2e^{3x}-2e^x-2xe^{-x}\\$$

11
##### Quiz-4 / TUT0402 Quiz4
« on: October 18, 2019, 01:59:46 PM »
Given
$$y"(t)+2y'(t)+2=0\\ y(\frac{\pi}{4})=2 \;\;\; y'(\frac{\pi}{4})=-2\\$$
Solution:
$$r^2+2r+2=0\\ r_1=\frac{-b+(b^2-4ac)^{1/2}}{2a}\\ r_2=\frac{-b-(b^2-4ac)^{1/2}}{2a}\\ r_1=\frac{-2+(4-4*2)^{1/2}}{2}=-1+i\\ r_2=\frac{-2-(4-4*2)^{1/2}}{2}=-1-i\\ \lambda=-1 \; \mu=1\\ y(t)=c_1e^{-t}cos(t)+c_2e^{-t}sin(t)\\$$
Substituting the initial conditions:
$$\\ 2=c_1e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})+c_2e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})\\ 2=c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}+c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\ 2\sqrt2=c_1e^{-\frac{\pi}{4}}+c_2e^{-\frac{\pi}{4}} \\ y'(t)=-c_1e^{-t}cos(t)-c_1e^{-t}sin(t)-c_2e^{-t}sin(t)+c_2e^{-t}cos(t)\\ \\ -2=-c_1e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})-c_1e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})-c_2e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})+c_2e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})\\ -2=-c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}-c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}-c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}+c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\ -2=-2c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\ c_1=e^{\frac{\pi}{4}}\sqrt2\\ 2\sqrt2=\sqrt2e^{\frac{\pi}{4}}e^{-\frac{\pi}{4}}+c_2e^{-\frac{\pi}{4}}\\ 2\sqrt2=\sqrt2+c_2e^{-\frac{\pi}{4}}\\ c_2=e^{\frac{\pi}{4}}\sqrt2\\ y(t)=\sqrt2e^{\frac{\pi}{4}}e^{-t}cos(t)+\sqrt2e^{\frac{\pi}{4}}e^{-t}sin(t)\\$$

12
##### Quiz-3 / TUT0402 Quiz3
« on: October 11, 2019, 02:00:24 PM »
Find the Wronskian of the given pair of functions:
$$cos^2(x),\,1+cos(2x)$$
$$W= det\begin{vmatrix} cos^2(x)&1+cos(2x)\\ -2cos(x)sin(x)&-2sin(2x)\\ \end{vmatrix} = det\begin{vmatrix} cos^2(x)&1+cos(2x)\\ -sin(2x)&-2sin(2x)\\ \end{vmatrix}\\ =-2cos^2(x)sin(2x)+sin(2x)+sin(2x)cos(2x)\\ =-sin(2x)[2cos^2(x)-1-cos(2x)]\\ =-sin(2x)[cos(2x)-cos(2x)]\\ =-sin(2x)\times0\\ =0\\ \ Note: cos(2x)=2cos^2(x)-1\\ sin(2x)=2sin(x)cos(x)\\$$

13
##### Quiz-2 / TUT0402 Quiz2
« on: October 04, 2019, 02:00:35 PM »
$$\\ \frac{x}{({x^2+y^2})^{\frac{3}{2}}}+\frac{y}{({x^2+y^2})^{\frac{3}{2}}}y'=0 \\ M=\frac{x}{({x^2+y^2})^{\frac{3}{2}}} \\ N=\frac{y}{({x^2+y^2})^{\frac{3}{2}}} \\ My=\frac{d}{dy}\frac{x}{({x^2+y^2})^{\frac{3}{2}}}=-2xy\frac{3}{2}({x^2+y^2})^{-\frac{5}{2}}=-\frac{3xy}{({x^2+y^2})^{\frac{5}{2}}} \\ Nx=\frac{d}{dx}\frac{y}{({x^2+y^2})^{\frac{3}{2}}}=-2xy\frac{3}{2}({x^2+y^2})^{-\frac{5}{2}}=-\frac{3xy}{({x^2+y^2})^{\frac{5}{2}}} \\ Therefore\; ,\; it's\; exact\; \\ \\ \phi=\int{M}dx=\int\frac{x}{({x^2+y^2})^{\frac{3}{2}}}dx \\ Let\; u=x^2+y^2\; ,\; then\; du=2xdx\; ,\; xdx=\frac{1}{2}du \\ \phi=\frac{1}{2}\int\frac{1}{u^{\frac{3}{2}}}du=-2\frac{1}{2}\frac{1}{u^{\frac{1}{2}}}+h(y)=-\frac{1}{({x^2+y^2})^{\frac{1}{2}}}+h(y) \\ \\ \phi{y}=\frac{1}{2}2y\frac{1}{({x^2+y^2})^{\frac{3}{2}}}+h'(y)=\frac{y}{({x^2+y^2})^{\frac{3}{2}}}+h'(y)=N \\ Therefore\; ,\; h'(y)=0\; h(y)=k \\ \\ \phi=-\frac{1}{({x^2+y^2})^{\frac{1}{2}}}+k \\ k=-\frac{1}{({x^2+y^2})^{\frac{1}{2}}} \\ ({x^2+y^2})^{\frac{1}{2}}=\frac{-1}{k} \\ {x^2+y^2}=\frac{1}{k^2} \\ Thus\; ,\; the\; solution\; is\; {x^2+y^2}=C\; where\; C=\frac{1}{k^2}$$

14
##### Quiz-2 / TUT0402 Quiz2
« on: October 04, 2019, 02:00:03 PM »
$$\\ (3x^2y+2xy+y^3)+(x^2+y^2)y'=0 \\ M=(3x^2y+2xy+y^3) \\ N=(x^2+y^2) \\ My=3x^2+2x+3y^2 \\ Nx=2x \\ R2=\frac{My-Nx}{N}=\frac{3x^2+2x+3y^2-2x}{(x^2+y^2)}=3 \\ \\ \mu=e^{\int{R2}dx}=e^{\int{3}dx}=e^{3x} \\ \\ Multiply\; both\; sides\; by\; \mu\; ,\; we\; get\; \\ e^{3x}(3x^2y+2xy+y^3)+e^{3x}(x^2+y^2)\frac{dy}{dx}=0 \\ M1=e^{3x}(3x^2y+2xy+y^3) \\ N1=e^{3x}(x^2+y^2) \\ \\ \phi=\int{N1}dy=\int{e^{3x}(x^2+y^2) }dy=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3+h(x) \\ \phi{x}=3e^{3x}x^2y+2e^{3x}xy+e^{3x}y^3+h'(x)=e^{3x}(3x^2y+2xy+y^3)+h'(x)=M1 \\ Therefore\; ,\; h'(x)=0\; h(x)=C \\ \phi=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3+C \\ Thus\; ,\; the\; solution\; is\; C=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3$$

15
##### Quiz-1 / TUT0402 Quiz1
« on: September 27, 2019, 02:13:55 PM »
ty’ + 2y = sin(t), t > 0
y' + (2/t)y = sin(t)t^(-1)
p(t) = 2/t
μ = e^( ∫2/t dt) = e^(2ln(t)) = t^2
Multiplying both sides by μ
y’ t^2 + 2ty = sin(t)t
(yt^2)’ = sin(t)t
Let u = t, dv = sin(t)dt
then du = dt, v = -cos(t)
yt^2 = - tcos(t) + ∫cos(t)dt
yt^2 = - tcos(t) + sin(t) + C
y = (- tcos(t) + sin(t) + C) / t^2, t > 0

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