text: 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]
metadata: {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}