Messages

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 »

$$
\\ (3x^2y+2xy+y^3)+(x^2+y^2)y'=0
\\ M=(3x^2y+2xy+y^3)
\\ N=(x^2+y^2)
\\ My=3x^2+2x+3y^2
\\ Nx=2x
\\ R2=\frac{My-Nx}{N}=\frac{3x^2+2x+3y^2-2x}{(x^2+y^2)}=3
\\
\\ \mu=e^{\int{R2}dx}=e^{\int{3}dx}=e^{3x}
\\
\\ Multiply\; both\; sides\; by\; \mu\; ,\; we\; get\;
\\ e^{3x}(3x^2y+2xy+y^3)+e^{3x}(x^2+y^2)\frac{dy}{dx}=0
\\ M1=e^{3x}(3x^2y+2xy+y^3)
\\ N1=e^{3x}(x^2+y^2)
\\
\\ \phi=\int{N1}dy=\int{e^{3x}(x^2+y^2) }dy=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3+h(x)
\\ \phi{x}=3e^{3x}x^2y+2e^{3x}xy+e^{3x}y^3+h'(x)=e^{3x}(3x^2y+2xy+y^3)+h'(x)=M1
\\ Therefore\; ,\; h'(x)=0\; h(x)=C
\\ \phi=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3+C
\\ Thus\; ,\; the\; solution\; is\; C=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3
$$