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### Messages - Wei Cui

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1
##### MAT244--Misc / How do we calculate the WB bonus mark?
« on: December 21, 2018, 01:36:12 PM »
Happy holidays everyone!

Can someone please tell me how do we calculate the WB bonus mark ?  Thanks in advance!

2
##### MAT244--Misc / Re: Not getting the marks for quiz 6 and 7
« on: December 11, 2018, 01:26:11 AM »
Yes, I have the same issue with you. I always go to my original tutorial section and I still haven't receive my marks for quiz 6 and quiz 7 now. I am wondering when they will be released.

3
##### Quiz-4 / Re: Q4 TUT 0501
« on: October 27, 2018, 11:58:12 AM »
Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given inhomogeneous equation.
\begin{gather*}
x^2y'' + xy' + (x^2 - 0.25)y = 3x^{3/2} \sin (x),\qquad x > 0;\\
y_1(x) = x^{-1/2} \sin (x),\quad y_2(x) = x^{-1/2} \cos (x).
\end{gather*}

Hence,
\begin{gather*}
\begin{cases}
y_1(x) = x^{-\frac{1}{2}}sin(x)\\
y_1'(x) = -\frac{1}{2}x^{-\frac{3}{2}}sin(x) + x^{-\frac{1}{2}}cos(x)\\
y_1''(x) = \frac{3}{4}x^{-\frac{5}{2}}sin(x)-x^{-\frac{3}{2}}cos(x)-x^{-\frac{1}{2}}sin(x)
\end{cases}
\end{gather*}

\begin{gather*}
\begin{cases}
y_2(x) = x^{-\frac{1}{2}}cos(x)\\
y_2'(x) = -\frac{1}{2}x^{-\frac{3}{2}}cos(x) - x^{-\frac{1}{2}}sin(x)\\
y_2''(x) = \frac{3}{4}x^{-\frac{5}{2}}cos(x) + x^{-\frac{3}{2}}sin(x)-x^{-\frac{1}{2}}cos(x)
\end{cases}
\end{gather*}

Substitute back into the homogeneous equation:
\begin{gather*}
x^2y'' + xy' + (x^2 - 0.25)y = 0
\end{gather*}

Verified that $y_1(x)$ and $y_2(x)$ both satisfy the corresponding homogeneous equation.
And the complementary solution $y_c(x) = c_1x^{-\frac{1}{2}}sin(x) + c_2x^{-\frac{1}{2}}cos(x)$
Now divide both sides of the original equation by $x^2$:
\begin{gather*}
y'' + \frac{1}{x}y' + \frac{x^2-0.25}{x^2}y = 3x^{-\frac{1}{2}}sin(x)
\end{gather*}

Then
\begin{gather*}
p(t) = \frac{1}{x}, q(t) = \frac{x^2 - 0.25}{x^2}, g(t) = 3x^{-\frac{1}{2}}sin(x)\\
\ \\
W[y_1,y_2](x) =
\begin{vmatrix}
y_1(x) & y_2(x)\\
y'_1(x) & y'_2(x)
\end{vmatrix}
= -x^{-1} \neq 0
\end{gather*}

Since the particular solution has the form:
\begin{gather*}
Y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)
\end{gather*}

and
\begin{align*}
u_1(x) &= - \int \frac{y_2(x) g(x)}{W[y_1,y_2](x)}dx\\
&= -\int \frac{x^{-\frac{1}{2}}cos(x)\cdot 3x^{-\frac{1}{2}}sin(x)}{-x^{-1}} dx\\
&= \frac{3}{2}\int 2sin(x)cos(x) dx\\
&= \frac{3}{2}\int sin(2x)dx\\
&= -\frac{3}{4}cos(2x)
\end{align*}

\begin{align*}
u_2(x) &= \int \frac{y_1(x) g(x)}{W[y_1,y_2](x)}dx\\
&= \int \frac{x^{-\frac{1}{2}}sin(x)\cdot 3x^{-\frac{1}{2}}sin(x)}{-x^{-1}} dx\\
&= -3\int sin^2(x) dx\\
&= -3\int \frac{1-cos(2x)}{2}dx\\
&= -\frac{3}{2}x + \frac{3}{4}sin(2x)
\end{align*}

Therefore,

\begin{align*}
Y_p(x) &= -\frac{3}{4}cos(2x)\cdot x^{-\frac{1}{2}} sin(x) + (-\frac{3}{2}x + \frac{3}{4}sin(2x))\cdot x^{-\frac{1}{2}}cos(x)\\
&= \frac{3}{4}x^{-\frac{1}{2}}(sin(2x)cos(x)-cos(2x)sin(x))- \frac{3}{2}x^{-\frac{1}{2}}cos(x)\\
&= \frac{3}{4}x^{-\frac{1}{2}}(2sin(x)cos^2(x)-(2cos^2(x)- cos(x))sin(x))-\frac{3}{2}x^{\frac{1}{2}}cos(x)\\
&= \frac{3}{4}x^{-\frac{1}{2}}sin(x) - \frac{3}{2}x^{\frac{1}{2}}cos(x)
\end{align*}

Hence, the general solution:
\begin{align*}
y(x) &= y_c(x) + Y_p(x)\\
&= c_1x^{-\frac{1}{2}}sin(x) + c_2x^{-\frac{1}{2}}cos(x) + \frac{3}{4}x^{-\frac{1}{2}}sin(x) - \frac{3}{2}x^{\frac{1}{2}}cos(x)\\
&= (c_1 + \frac{3}{4})x^{-\frac{1}{2}}sin(x) + c_2x^{-\frac{1}{2}}cos(x) - \frac{3}{2}x^{\frac{1}{2}}cos(x)\\
&= c_3x^{-\frac{1}{2}}sin(x) + c_2x^{-\frac{1}{2}}cos(x)-\frac{3}{2}x^{\frac{1}{2}}cos(x)
\end{align*}

where $c_3 = c_1 + \frac{3}{4}$
Therefore, the particular solution of the given non homogenous equation is
\begin{gather*}
Y_p(x) = -\frac{3}{2}x^{\frac{1}{2}}cos(x)
\end{gather*}

4
##### Quiz-3 / Re: Q3 TUT 0701
« on: October 13, 2018, 10:49:13 AM »
First, we divide both sides of the equation by $\cos(t)$ and we get:
$y''+\tan(t)y'-\frac{t}{\cos(t)}y = 0$

Then the equation in the form of $L(y) = y'' +p(t)y' +q(t)y=0$, then in this case $p(t) = tan(t)$
According to Abel's Theorem, then the Wronskian                         $W(y_1,y_2)(t) = Ce^{-\int \tan(t)dt}$
$=Ce^{\int \frac{1}{\cos(t)}d\cos(t)}$
$= Ce^{\ln(\cos(t))}$
$= C\cos(t)$

Therefore, the solution is $W(y_1,y_2)(t) = C\cos(t)$.

5
##### MAT244--Lectures & Home Assignments / Re: non-homogenous equation
« on: October 10, 2018, 10:45:38 AM »
If the equation is a non-homogeneous and the right-hand side is a constant, I think you can assume that the particular solution $y(t) = at + b$ and try to solve the equation.

6
##### Thanksgiving Bonus / Re: Thanksgiving bonus 5
« on: October 07, 2018, 06:35:43 PM »
Find general and singular solutions to:
$2y-4xy'-\ln(y')=0$

1. We transform the equation into the form:
$y=x\cdot 2y'+\frac{1}{2}\ln(y')$

We plug $p=y'$ and differentiate the equation, we get:

$pdx = 2pdx+2x\cdot dp+\frac{1}{2}\frac{1}{p}\cdot dp$

Then:
$p\frac{dx}{dp}=2p\frac{dx}{dp}+2x+\frac{1}{2p}$

$-px'-2x=\frac{1}{2p}$

$x'+\frac{2}{p}x=-\frac{1}{2p^2}$

Then $p(p) = \frac{2}{p}$, $u(p)=e^{2\int \frac{1}{p}dp} = e^{2lnp}=p^2$,
Multiply both sides with integrating factor $u(p) = p^2$ and we get:

$p^2x'+2px=-\frac{1}{2}$

$(p^2x)'=-\frac{1}{2}$

$p^2x=-\int \frac{1}{2}dp$

$p^2x=-\frac{1}{2}p+C$

$x=-\frac{1}{2p}+\frac{C}{p^2}$

Then the solution in the parametric form is: $\begin{cases}x=-\frac{1}{2p} + \frac{C}{p^2}\\ y=x\cdot 2p + \frac{1}{2}\ln p\end{cases}$

2. To find the singular solution, we solve the equation $\varphi(p) -p=0 \implies 2p-p=0 \implies p=0$

It follows from this that $y=C$, we can make direct substitution to make sure that the constant $C$ is equal to 0.

Therefore, the differential equation has the singular solution $y=0$.

7
##### Thanksgiving Bonus / Re: Thanksgiving bonus 5
« on: October 07, 2018, 02:37:38 PM »
Find general and singular solutions to:
$2y-4xy'-\ln(y')=0$

1. We transform the equation into the form:
$y=x\cdot 2y'+\frac{1}{2}\ln(y')$

We plug $p=y'$ and differentiate the equation, we get:

$pdx = 2pdx+2x\cdot dp+\frac{1}{2}\frac{1}{p}\cdot dp$

Then:
$p\frac{dx}{dp}=2p\frac{dx}{dp}+2x+\frac{1}{2p}$

$-px'-2x=\frac{1}{2p}$

$x'+\frac{2}{p}x=-\frac{1}{2p^2}$

Then $p(p) = \frac{2}{p}$, $u(p)=e^{2\int \frac{1}{p}dp} = e^{2lnp}=p^2$,
Multiply both sides with integrating factor $u(p) = p^2$ and we get:

$p^2x'+2px=-\frac{1}{2}$

$(p^2x)'=-\frac{1}{2}$

$p^2x=-\int \frac{1}{2}dp$

$p^2x=-\frac{1}{2}p+C$

$x=-\frac{1}{2p}+\frac{C}{p^2}$

Then the solution in the parametric form is: $\begin{cases}x=-\frac{1}{2p} + \frac{C}{p^2}\\ y=x\cdot 2p + \frac{1}{2}\ln p\end{cases}$

2. I am confused about the singular solution to this question. Since $\varphi(c) = 2c$ and $\varphi(c) -c=0 \implies 2c-c=0\implies c=0$.

However, for $\psi(c)$, since the domain of $\ln x$ is $x>0$. Then $\ln c=\ln 0$ will have no definition. Therefore, how am I supposed to solve the singular solution to the equation $2y-4xy'-\ln(y')=0$?

8
##### Quiz-1 / Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« on: September 29, 2018, 09:51:28 PM »
Question: $ty^{'}+2y=sin(t)$,   $t>0$

standard equation form: $y^{'}+\frac{2}{t}y=\frac{sin(t)}{t}$

$p(t) = \frac{2}{t}$,  $g(t) = \frac{sin(t)}{t}$

$u = e^{\int p(t)}dt = e^{2\int \frac{1}{t}dt} = t^2$, then we multiply both sides with $u$, and we get:

$t^2y^{'} + 2ty = tsin(t)$

$(t^2y)^{'}=tsin(t)$

$d(t^2y) = tsin(t)dt$

$t^2y = \int tsin(t) dt$

(integrating by parts, $u=t \implies du =dt$ and $dv=sin(t)$ and $v=-cos(t)$

$\int tsin(t)dt = uv - \int vdu$

$=-tcos(t)-\int (-cos(t))dt$

$=-tcos(t) +\int cos(t)dt$

$=-tcos(t) +sin(t) + C$)

Therefore, $t^2y=-tcos(t)+sin(t) + C$

$y=\frac{-tcos(t)+sin(t)+C}{t^2}$.

Since $t>0$, and when $t \rightarrow \infty, y \rightarrow 0$

9
##### Quiz-1 / Re: Q1: TUT 0801
« on: September 29, 2018, 09:35:40 PM »
Question: $y^{'} - y = 2te^{2t}$,   $y(0) = 1$

$p(t) = -1$,   $g(t) = 2te^{2t}$

$u(t) = e^{\int -1dt} = e^{-t}$

multiply both sides with $u$, then we get:

$e^{-t}y^{'}-e^{-t}y=2te^{t}$

$(e^{-t}y)^{'} = 2te^{t}$

$d(e^{-t}y)= 2te^{t}dt$

$e^{-t}y=\int 2te^{t}dt$

$e^{-t}y = 2e^{t}(t-1)+C$

$y = 2e^{2t}(t-1)+Ce^{t}$

Since $y(0) = 1 \implies 1= 2\times e^{0}(0-1)+Ce^{0}$, then we get $C =3$

Therefore, general solution is: $y = 2e^{2t}(t-1)+3e^{t}$

10
##### Quiz-1 / Re: Q1: TUT0401
« on: September 29, 2018, 09:16:33 PM »
tut0401   $y^{'} + \frac{2}{3}y=1-\frac{1}{2}t$, $y(0) = y_0$

$p(t) = \frac{2}{3}$, $g(t) = 1-\frac{1}{2}t$, then $u = e^{\int \frac{2}{3}dt}$,

multiply both sides with $u$, then we get:

$e^{\frac{2}{3}t}y^{'} + \frac{2}{3}e^{\frac{2}{3}t}y = e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t}$

$(e^{\frac{2}{3}t}y)^{'}=e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t}$

$d(e^{\frac{2}{3}t}y) = (e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t})dt$

$e^{\frac{2}{3}t}y=\int (e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t})dt$

$e^{\frac{2}{3}t}y=\frac{3}{2}e^{\frac{2}{3}t}-\frac{3}{4}te^{\frac{2}{3}t}+\frac{9}{8}e^{\frac{2}{3}t}+C$    (integrate by parts)

$y= \frac{3}{2}-\frac{3}{4}t+\frac{9}{8}+Ce^{-\frac{2}{3}t}$

$y=\frac{21}{8}-\frac{3}{4}t+Ce^{-\frac{2}{3}t}$. Consider $y(0) = y_0 \implies y_0=\frac{21}{8}-0+C \implies C=y_0-\frac{21}{8}$

Therefore, $y=\frac{21}{8}-\frac{3}{4}t+(y_0-\frac{21}{8})e^{-\frac{2}{3}t}$

If $y(t)$ touches, but does not cross the $t$-axis at some point $t_0$ s.t. $y(t_0) = 0$ and $y^{'}(t_0) = 0$

Thus, $y^{'} +\frac{2}{3}y = 1- \frac{1}{2}t$ when $t=t_0$

$0+0 = 1-\frac{1}{2}t_0$

$\frac{1}{2}t_0 = 1 \implies t_0 = 2$

substitute $t_0 = 2$, then $y(2) = 0 \implies 0 = \frac{21}{8} - \frac{3}{4}\times 2 + (y_0-\frac{21}{8})e^{-\frac{4}{3}}$

Therefore, we get $y_0 = \frac{21}{8}e^{-\frac{4}{3}}-\frac{9}{8}$

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##### MAT244--Lectures & Home Assignments / Re: Turning inexact solutions into exact solutions
« on: September 27, 2018, 01:43:07 AM »
If you ask how to turn the inexact equation into exact equation, then there are three cases:

Try 1: check $\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} =$ $f(x)$ function of $x$ only,

then let $\frac{u^{'}}{u} = f(x)$

$\frac{du}{u} = f(x)dx$

$u = e^{\int f(x)dx}$

Try 2: check $\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M} = g(y)$ function of $y$ only, and then same as the first one you let $\frac{u^{'}}{u} = g(y)$,

$u = e^{\int g(y)dy}$

Try 3: if $\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{Mx-Ny} = z(x, y)$, then

$u = u(x, y)$ and $u = e^{\int z(x,y)}$

In each of these cases,  $u$ is the integrating factor, when you solve $u$ and you multiply the equation both sides with $u$ then you will turn the inexact equation into an exact equation.

12
##### MAT244--Lectures & Home Assignments / Re: problem 6 in 1.3
« on: September 23, 2018, 05:56:09 PM »
A linear differential equation is in the form of $a_0(t)y^{(n)} + a_1(t)y^{(n-1)}+ ... + a_n(t)y = g(t)$, and also $a_i(t)$ can be a constant function which is: $a_i(t) = C$.

In this case, $a_1(t) = 0$, so the second derivative $y^{''}$ disappears. However, it's still a linear differential equation.

13
##### MAT244--Lectures & Home Assignments / Different methods
« on: September 20, 2018, 04:31:45 PM »
I find that there are different methods to solve for one differential question, even different profs have different methods (I guess they are the same in essence). Let's say when we solve the problems on our test or quiz, can we use a method which may differ from the one in the sample solution? Is there any formal format we should follow or as long as we solve the question, then we are fine?

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##### MAT244--Lectures & Home Assignments / Re: linear differential equations
« on: September 18, 2018, 09:48:30 PM »
This equation is in the form of $a_0(x)y^{(n)} + a_1(x)y^{(n-1)} + ... + a_n(x)y = g(x)$. Therefore, $xy'=1$ is a linear equation.

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##### MAT244--Lectures & Home Assignments / Re: last question from todays lecture
« on: September 13, 2018, 10:57:41 PM »
First of all, I wrote $a-b=0$ and $a+b-2=0$ in my notes (not $a-b-2=0$). And, I also wrote $x=t+a$, $y=w+b$, $x=t+1$, $y=w+1$, which may be the initial condition. In this case, $a=1$, and $b=1$, then the equation will make sense.

Well, actually I am confused about it too...

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