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### Messages - Changhao Jiang

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##### Term Test 2 / Re: Problem 4 (noon)
« on: November 19, 2019, 05:50:48 AM »
To find eigenvalues, $det(A-\lambda x)=(1-\lambda)(-3-\lambda)-(-2)(3)=0$,we can get $\lambda = -1-\sqrt{2}$ or $\lambda = -1+\sqrt{2}$
To find eigenvectors, when $\lambda=-1-\sqrt{2}$,
$\begin{pmatrix} 2+\sqrt{2} & 3 \\ -2 & -2+\sqrt{2} \end{pmatrix} ~ \begin{pmatrix} 2 & 2-\sqrt{2} \\ 0 & 0 \end{pmatrix}$
the eigenvector is $\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix}$
so $e^{(-1-\sqrt{2})t}\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix} = e^{-t}(\cos\sqrt{2}t-i\sin\sqrt{2}t)\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix} = e^{-t} (\begin{bmatrix} 2\cos\sqrt{2}t-\sqrt{2}cos\sqrt{2}t \\ -2cos\sqrt{2}t \end{bmatrix} + i\begin{bmatrix} -2\sin\sqrt{2}t+\sqrt{2}sin\sqrt{2}t \\ 2sin\sqrt{2}t \end{bmatrix})$
therefore, the general solution is $x(t)=c_1 e^{-t} \begin{bmatrix} 2\cos\sqrt{2}t-\sqrt{2}cos\sqrt{2}t \\ -2cos\sqrt{2}t \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} -2\sin\sqrt{2}t+\sqrt{2}sin\sqrt{2}t \\ 2sin\sqrt{2}t \end{bmatrix}$

2
##### Term Test 2 / Re: Problem 3 (noon)
« on: November 19, 2019, 05:15:16 AM »
(a)
To find eigenvalues, $(1-\lambda x)(-\lambda)-2=0$,we can get $\lambda = 2$ or $\lambda = -1$
To find eigenvectors, when $\lambda=2$,
$\begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix} ~ \begin{pmatrix} 1 & -2 \\ 0 & 0 \end{pmatrix}$
let $x_2=t, x_1=2t$, so the eigenvector is \begin{bmatrix}1 \\ 2\end{bmatrix}
when $\lambda = -1$
$\begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix}$
~
$\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$
let $x_2=-t, x_1=t$, so the eigenvector is \begin{bmatrix}1 \\ -1\end{bmatrix}
Therefore, the general solution is y=$c_1 \begin{bmatrix}1 \\ 2\end{bmatrix} e^{2t}$ + $c_2 \begin{bmatrix}1 \\ -1\end{bmatrix} e^{-t}$

(b)
from (a), we can know $\phi(t)= \begin{pmatrix} e^{2t} & e^{-t} \\ 2e^{2t} & -e^{-t} \end{pmatrix}$
then
$\begin{pmatrix} e^{2t} & e^{-t} \\ 2e^{2t} & -e^{-t} \end{pmatrix} \begin{bmatrix}u_1'\\ u_2'\end{bmatrix} = \begin{bmatrix} 0 \\ \frac{6e^{3t}}{e^{2t}+1}\end{bmatrix}$
By ref form, we can know $u_1'= \frac{6e^t}{e^2t+1}, u_2' = 0$, then by integrating, we can get $u_1=6arctan(e^t)+c_1, u_2=c_2$
then the  general solution is $x(t)=(6arctan(e^t)+c_1)\begin{bmatrix}1 \\ 2\end{bmatrix} e^{2t} + c_2\begin{bmatrix}1 \\ -1\end{bmatrix} e^{-t}$

3
##### Term Test 1 / Re: Problem 2 (afternoon)
« on: October 23, 2019, 07:39:00 AM »
(a)
Rewrite the equation:$y''+\frac{2x+2}{x(2x+1)}y'+\frac{2}{x(2x+1)}y=0$
$p(x)=\frac{2x+2}{x(2x+1)}, W=e^{-\int p(x) dx}$
then, by Abel's theorem, $W(y_1,y_2)(x)=ce^{- \int \frac{2x+2}{x(2x+1)}}dx$
We can first calculate the integral $\int \frac{2x+2}{x(2x+1)} dx$
let $\frac{A}{x}+\frac{B}{2x+1}=\frac{2x+2}{x(2x+1)}, then \frac{(2x+1)A+Bx}{x(2x+1)}=\frac{2x+2}{x(2x+1)}$
then we get A=2, B=-2, so the integral $\int \frac{2x+2}{x(2x+1)} dx = \int 2(\frac{1}{x}-\frac{1}{2x+1}) dx = 2(lnx-\ln(2x+1))=\ln \frac{x^2}{2x+1}$
therefore, $W=ce^{-\ln \frac{x^2}{2x+1}}=c\frac{2x+1}{x^2}$
let c=1, $W=\frac{2x+1}{x^2}$
(b)
$y_1=x+1, y_1'=1, y_1''=0$, put $y_1, y_1'$ and $y_1''$ into the equation, we can get $(2x+1)x \cdot 0+(2x+2) \cdot 1+2(x+1)=0$; therefore; $y_1$ is the solution for ODE
Let another linearly independent solution be $y_2$
Since we know $W=y_1y_2'-y_2y_1'=(x+1)y_2'-y_2$
then we need to solve the equation $(x+1)y_2'-y_2=\frac{2x+1}{x^2}$
Rewrite the equation: $y_2'-\frac{1}{x+1}y_2=\frac{2x+1}{x^2(x+1)}$
$\mu(x)=e^{\int -\frac{1}{x+1} dx}=\frac{1}{x+1}$, multiply on both sides with $\mu(x)$
We can get $\frac{1}{x+1}y_2'-\frac{1}{(x+1)^2}y_2=\frac{2x+1}{x^2(x+1)^2}$
Integrate on both sides, we can get $\frac{1}{1+x}y_2 = -\frac{1}{x(x+1)}+c$
Then $y_2 = -\frac{1}{x}+c(x+1)$, let c=1, $y_2=-\frac{1}{x}+x+1$
(c)
the general solution $y=c_1(x+1)+c_2(-\frac{1}{x}+x+1)$
since y(-1)=1, we can get $c_2=1$
$y'=c_1+c_2+\frac{c_2}{x^2}$, and y'(-1)=0, so $c_1=-2$
Therefore, $y=-(\frac{1}{x}+x+1)$

4
##### Quiz-4 / TUT0102 Quiz4
« on: October 18, 2019, 10:35:54 PM »
Find the general solution of the given differential equation 9y''+9y'-4y=0
Solution: We can write as $9r^2+9r-4=0$. Then by factorization, we can get
(3r-1)(3r+4)=0, $r_1=\frac{1}{3}$ or $r_2=-\frac{4}{3}$, and these are two
distinct real roots, so the general solution is
$y=c_1e^{\frac{t}{3}}+c_2e^{-\frac{4t}{3}}$

5
##### Quiz-3 / TUT0102 Quiz3
« on: October 11, 2019, 02:04:50 PM »
\documentclass{article}
\usepackage[utf8]{inputenc}

\usepackage{natbib}
\usepackage{graphicx}
\usepackage{amsmath}
\begin{document}

If the Wronskian W of f and g is $3e^{4t}$, and if $f(t)=e^{2t}$, find g(t)

Solution: We calculate the Wronskian of f and g:

$W = \begin{vmatrix} f(t) & g(t) \\ f'(t) & g'(t) \\ \end{vmatrix} = \begin{vmatrix} e^{2t} & g(t) \\ 2e^{2t} & g'(t) \\ \end{vmatrix}$
= $e^{2t}g'(t)-2e^{2t}g(t)=3e^{4t}$ \newline
$e^{2t}g'(t)-2e^{2t}g(t)=3e^{4t}$ \newline
Divided by $e^{2t}$ on both sides, we get $g'(t)-2g(t)=3e^{3t}$ \newline
$\mu(t)$ = $e^{\int-2dt}$ = $e^{-2t}$ \newline
multiply $e^{-2t}$ on both sides, $e^{-2t}g'(t)-2e^{-2t}g(t)=3$ \newline
$(e^{-2t}g(t))'=3$ \newline
Integral on both sides, $e^{-2t}g(t)=3t+C$ \newline
g(t) = $3te^2t+Ce^{2t}$

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##### Quiz-3 / TUT0102 Quiz3
« on: October 11, 2019, 02:03:03 PM »
\documentclass{article}
\usepackage[utf8]{inputenc}

\usepackage{natbib}
\usepackage{graphicx}
\usepackage{amsmath}
\begin{document}

If the Wronskian W of f and g is $3e^{4t}$, and if $f(t)=e^{2t}$, find g(t)

Solution: We calculate the Wronskian of f and g:

$W = \begin{vmatrix} f(t) & g(t) \\ f'(t) & g'(t) \\ \end{vmatrix} = \begin{vmatrix} e^{2t} & g(t) \\ 2e^{2t} & g'(t) \\ \end{vmatrix}$
= $e^{2t}g'(t)-2e^{2t}g(t)=3e^{4t}$ \newline
$e^{2t}g'(t)-2e^{2t}g(t)=3e^{4t}$ \newline

7
##### Quiz-2 / TUT0102 Quiz2
« on: October 04, 2019, 02:06:57 PM »
Question: e^x+（e^x  cot⁡(y)+2y csc⁡(y) ） y^'=0, find an integrating factor and solve the given equation.
Solution:
My=0.
Nx=∂/∂x [e^x  cot⁡(y)+2y csc⁡(y) ]=cot⁡(y) e^x.
We know the given equation is not exact, so we need to find μ(t).
R=(My-Nx)/M=(0-cot⁡(y) e^x)/e^x =-cot⁡(y).
μ(t)=e^(-∫Rdy)=e^(-∫cot⁡y dy )=e^ln⁡|sin⁡y | =sin⁡y.
Multiply sin⁡y on both sides.
We get sin⁡y e^x+(e^x  cos⁡y+2y) y^'=0.
There exists ψ(x,y) such that ψ_x (x,y)=sin⁡y e^x.
Integrating on both sides with x, ψ(x,y)=sin⁡y e^x+g(y).
Differentiate on both sides with y, ψ_y (x,y)=cos⁡y e^x+g^' (y).
Since ψ_y (x,y)=N=e^x  cos⁡y+2y, then g^' (y)=2y and so g(y)=y^2+c.
Combine all results above, we get cos⁡y e^x+y^2=c.

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