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### Topics - Jingjing Cui

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##### 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
$$-f'(-s)=g'(s)\\ f(-s)=g(s)\\ u(x,t)=\frac{x+ct}{2c}+\frac{ct-x}{2c}=t\\$$

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##### 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.

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##### 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 $u_{xyz}=xy\\ u_{xy}=xyz+f(x,y)\\ u_{x}=\frac{1}{2}xy^2z+F(x,y)+g(x,z)\\ u=\frac{1}{4}x^2y^2z+\hat{F}(x,y)+G(x,z)+h(y,z)$

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

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

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

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

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##### 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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##### 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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##### Quiz-1 / TUT0402 Quiz1
« on: September 27, 2019, 02:12:57 PM »
xy' = (1-y^2)^(1/2)
solution: x(dy/dx) = (1-y^2)^(1/2)
∫ 1/ (1-y^2)^(1/2) dy =  ∫1/x dx
arcsin(y) = ln|x| + C
y = sin(ln|x| + C)

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