### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Tzu-Ching Yen

Pages: 1 [2]
16
##### Quiz-2 / Re: Q2 TUT 0701
« on: October 05, 2018, 05:42:24 PM »
\begin{gather*}
N_x = \frac{2x}{y} - \frac{3y}{x^2}, M_y = -\frac{6}{y^2}, xM = 3x^2 + \frac{6x}{y}, yN = x^2 + 3\frac{y^2}{x}, \\\frac{N_x - M_y}{xM - yN} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{2x^2 - \frac{3y^2}{x} + \frac{6x}{y}} = R = \frac{1}{xy}
\end{gather*}

From assignment
\begin{gather*}
\mu (z) = \exp(\int \frac{1}{z} d(z)) = z,\\
\mu M = 3x^2y + 6x, \mu N = x^3 + 3y^2,\\
\int \mu M dx = x^3y + 3x^2 + f(y) + c_0,\\
\int \mu N dy = x^3y + y^3 + g(x) + c_1.
\end{gather*}
Combine the previous two result gives
$$\phi(x, y) = x^3y + 3x^2 + y^3 = c,$$
where
$\frac{\partial\phi}{\partial x} = uM, \frac{\partial\phi}{\partial x} = uN$
We do not need this

17
##### Quiz-1 / Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« on: September 29, 2018, 09:23:56 PM »
Rephrase equation
$y' + \frac{2}{t}y = \frac{sin(t)}{t}$
Find integrating factor
$u(t) = e^{\int \frac{2}{t}} = t^2$
the constant from integration is chosen to be zero. Now
$y = \frac{1}{u(t)}\int u(t)\frac{sin(t)}{t}$
Use int by parts,
$y = -\frac{cos(t)}{t} + \frac{sin(t)}{t^2} + \frac{c_1}{t^2}$
Since $t$ is in denominator and the nominators are bounded, as $t \to \infty$, $y \to 0$

18
##### Quiz-1 / Re: Q1: TUT 0701
« on: September 28, 2018, 06:00:02 PM »
First divide by $t^3$ on both side of the equation, we get
$$y' + \frac{4}{t}y = \frac{e^{-t}}{t^3}$$
Using the method of integrating factor we have equation for $u(t)$
$$u(t) = e^{\int \frac{4}{t}dt} = e^{4\ln(t) + c} = t^4$$
where constant $c$ is arbitrary, it's chosen to be 0 here. Then
$$\bigl(y u(t)\bigr)' = u(t)\frac{e^{-t}}{t^3}$$
rearranging gives equation
$$y = \frac{1}{u(t)}\int u(t)\frac{e^{-t}}{t^3}$$
substitute in $u(t) = t^4$
$$y = \frac{1}{t^4}\int te^{-t}$$
use integration by parts
$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} + \frac{c_1}{t^4}$$
to check $c_1$, plug in condition $y(-1) = 0$
$$y(-1) = e - e + c_1 = c_1= 0$$
Plug in $c_1 = 0$ gets
$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4}$$

Pages: 1 [2]