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### Messages - Wei Cui

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1
##### 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*}

2
##### 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 $W(y_1,y_2)(t) = C\cos(t)$.

3
##### 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 $y=0$.

4
##### 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 $2y-4xy'-\ln(y')=0$?

5
##### 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 $t \rightarrow \infty, y \rightarrow 0$

6
##### 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: $y = 2e^{2t}(t-1)+3e^{t}$

7
##### 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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