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

5
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:
$$
\\
2=c_1e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})+c_2e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})\\
2=c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}+c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\
2\sqrt2=c_1e^{-\frac{\pi}{4}}+c_2e^{-\frac{\pi}{4}}
\\
y'(t)=-c_1e^{-t}cos(t)-c_1e^{-t}sin(t)-c_2e^{-t}sin(t)+c_2e^{-t}cos(t)\\
\\
-2=-c_1e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})-c_1e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})-c_2e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})+c_2e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})\\
-2=-c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}-c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}-c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}+c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\
-2=-2c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\
c_1=e^{\frac{\pi}{4}}\sqrt2\\
2\sqrt2=\sqrt2e^{\frac{\pi}{4}}e^{-\frac{\pi}{4}}+c_2e^{-\frac{\pi}{4}}\\
2\sqrt2=\sqrt2+c_2e^{-\frac{\pi}{4}}\\
c_2=e^{\frac{\pi}{4}}\sqrt2\\
y(t)=\sqrt2e^{\frac{\pi}{4}}e^{-t}cos(t)+\sqrt2e^{\frac{\pi}{4}}e^{-t}sin(t)\\
$$

6
Quiz-3 / TUT0402 Quiz3
« on: October 11, 2019, 02:00:24 PM »
Find the Wronskian of the given pair of functions:
$$
cos^2(x),\,1+cos(2x)
$$
$$
W=
det\begin{vmatrix}
cos^2(x)&1+cos(2x)\\
-2cos(x)sin(x)&-2sin(2x)\\
\end{vmatrix}
=
det\begin{vmatrix}
cos^2(x)&1+cos(2x)\\
-sin(2x)&-2sin(2x)\\
\end{vmatrix}\\
=-2cos^2(x)sin(2x)+sin(2x)+sin(2x)cos(2x)\\
=-sin(2x)[2cos^2(x)-1-cos(2x)]\\
=-sin(2x)[cos(2x)-cos(2x)]\\
=-sin(2x)\times0\\
=0\\
\
Note:
cos(2x)=2cos^2(x)-1\\
sin(2x)=2sin(x)cos(x)\\
$$


7
Quiz-2 / TUT0402 Quiz2
« on: October 04, 2019, 02:00:35 PM »
$$
\\ \frac{x}{({x^2+y^2})^{\frac{3}{2}}}+\frac{y}{({x^2+y^2})^{\frac{3}{2}}}y'=0
\\ M=\frac{x}{({x^2+y^2})^{\frac{3}{2}}}
\\ N=\frac{y}{({x^2+y^2})^{\frac{3}{2}}}
\\ My=\frac{d}{dy}\frac{x}{({x^2+y^2})^{\frac{3}{2}}}=-2xy\frac{3}{2}({x^2+y^2})^{-\frac{5}{2}}=-\frac{3xy}{({x^2+y^2})^{\frac{5}{2}}}
\\ Nx=\frac{d}{dx}\frac{y}{({x^2+y^2})^{\frac{3}{2}}}=-2xy\frac{3}{2}({x^2+y^2})^{-\frac{5}{2}}=-\frac{3xy}{({x^2+y^2})^{\frac{5}{2}}}
\\ Therefore\; ,\; it's\; exact\;
\\
\\ \phi=\int{M}dx=\int\frac{x}{({x^2+y^2})^{\frac{3}{2}}}dx
\\ Let\; u=x^2+y^2\; ,\; then\; du=2xdx\; ,\; xdx=\frac{1}{2}du
\\ \phi=\frac{1}{2}\int\frac{1}{u^{\frac{3}{2}}}du=-2\frac{1}{2}\frac{1}{u^{\frac{1}{2}}}+h(y)=-\frac{1}{({x^2+y^2})^{\frac{1}{2}}}+h(y)
\\
\\ \phi{y}=\frac{1}{2}2y\frac{1}{({x^2+y^2})^{\frac{3}{2}}}+h'(y)=\frac{y}{({x^2+y^2})^{\frac{3}{2}}}+h'(y)=N
\\ Therefore\; ,\; h'(y)=0\; h(y)=k
\\
\\ \phi=-\frac{1}{({x^2+y^2})^{\frac{1}{2}}}+k
\\ k=-\frac{1}{({x^2+y^2})^{\frac{1}{2}}}
\\ ({x^2+y^2})^{\frac{1}{2}}=\frac{-1}{k}
\\ {x^2+y^2}=\frac{1}{k^2}
\\ Thus\; ,\; the\; solution\; is\; {x^2+y^2}=C\; where\; C=\frac{1}{k^2}
$$

8
Quiz-2 / TUT0402 Quiz2
« on: October 04, 2019, 02:00:03 PM »
$$
\\ (3x^2y+2xy+y^3)+(x^2+y^2)y'=0
\\ M=(3x^2y+2xy+y^3)
\\ N=(x^2+y^2)
\\ My=3x^2+2x+3y^2
\\ Nx=2x
\\ R2=\frac{My-Nx}{N}=\frac{3x^2+2x+3y^2-2x}{(x^2+y^2)}=3
\\
\\ \mu=e^{\int{R2}dx}=e^{\int{3}dx}=e^{3x}
\\
\\ Multiply\; both\; sides\; by\; \mu\; ,\; we\; get\;
\\ e^{3x}(3x^2y+2xy+y^3)+e^{3x}(x^2+y^2)\frac{dy}{dx}=0
\\ M1=e^{3x}(3x^2y+2xy+y^3)
\\ N1=e^{3x}(x^2+y^2)
\\
\\ \phi=\int{N1}dy=\int{e^{3x}(x^2+y^2) }dy=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3+h(x)
\\ \phi{x}=3e^{3x}x^2y+2e^{3x}xy+e^{3x}y^3+h'(x)=e^{3x}(3x^2y+2xy+y^3)+h'(x)=M1
\\ Therefore\; ,\; h'(x)=0\; h(x)=C
\\ \phi=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3+C
\\ Thus\; ,\; the\; solution\; is\; C=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3
$$

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