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### Topics - Changhao Jiang

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1
##### 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
$y=c_1e^{\frac{t}{3}}+c_2e^{-\frac{4t}{3}}$

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##### 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

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##### 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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##### Quiz-1 / TUT0102 Quiz1
« on: September 27, 2019, 03:05:02 PM »
Question: show it is homogeneous and solve it dy/dx = (x+3y)/(x-y)
Solution： dy/dx = (1+3y/x)/(1-y/x) in this form, which shows it is homogeneous
let u = y/x, y = ux
differentiate on both sides with x, we get
dy/dx = u + xdu/dx
from above, dy/dx = (1+3u)/(1-u)
then (1+3u)/(1-u) =  u + xdu/dx
xdu/dx = (1+3u)/(1-u) - u = (u+1)^2/(1-u)
(1/x) dx = ((1-u) / (u+1)^2) du
integrating on both sides, we get
ln|x| = - 2/u+1 - ln|u+1| + C
-ln|x| - 2/(y/x+1) - ln(y/x+1) + C =0
- ln(y+x) - 2x/y+x + C = 0

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