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Messages - Yiting Zhang

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1
Quiz-3 / Re: Q3 TUT 5102
« on: October 12, 2018, 06:36:23 PM »
$$y'' + \frac{x}{x^2}y' + \frac{x^2-v^2}{x^2}y = 0$$
$$W = ce^{-\int p(t)dt}$$
$$p(t) = \frac{1}{x}$$
$$-\int p(t)dt = -\ln{x}$$
$$W = ce^{-\ln{x}} = ce^{\ln{\frac{1}{x}}} = \frac{c}{x}$$

2
Quiz-3 / Re: Q3 TUT 5101
« on: October 12, 2018, 06:26:58 PM »
$$y′′−\frac{2x}{1−x^2}y′+\frac{\alpha(\alpha+1)}{1−x^2}y=0$$
$$W=ce^{\int −p(x) dx}, p(x) = −\frac{2x}{1−x^2}$$
$$W = ce^{\int \frac{2x}{1−x^2} dx}$$
$$u=1−x^2, du=−2xdx$$
$$ce^{-\int \frac{1}{u}du} = ce^{-ln(u)+C}=ce^{−ln(1−x^2)+C}=\frac{ce^C}{1-x^2}$$
Since $ce^C$ is constant. Let $ce^C = c$
$$W=\frac{c}{1−x^2}$$

3
Quiz-3 / Re: Q3 TUT 0101
« on: October 12, 2018, 06:11:17 PM »
$$y′′−2y′−2y=0$$
$$r^2−2r−2=0$$
$$r = \frac{−b \pm \sqrt{b^2−4ac}}{2a}$$
$$r_1=1+\sqrt{3}, r_2=1−\sqrt{3}$$

$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$

4
Quiz-1 / Re: Q1: TUT 0101
« on: September 29, 2018, 06:56:03 PM »
$$\begin{split} y' &= \frac{x^2+3y^2}{2xy} \\ &= \frac{x}{2y} + \frac{3y}{2x} \\ &= \frac{1}{2} (\frac{y}{x})^{-1} + \frac{3}{2}(\frac{y}{x}) \end{split}$$

$$v = \frac{y}{x} \implies y = vx$$

$$\begin{split} y' &= v'x+ v \\ &= \frac{1}{2} v^{-1} + \frac{3}{2} v \\ \end{split}$$

$$\begin{split} v'x &= y' - v \\ &= \frac{1}{2} v^{-1} + \frac{1}{2} v \\ &= \frac{1+v^2}{2v} \end{split}$$

$$\begin{split} \frac{2v}{1+v^2} dv &= \frac{1}{x} dx \\ \ln |1+v^2| &= \ln |x| + c \\ 1+v^2 &= Cx, C = e^c \\ \end{split}$$

$$\begin{split} 1 + \frac{y^2}{x^2} - Cx &= 0\\ y^2 + x^2 - Cx^3 &= 0 \end{split}$$

5
Quiz-1 / Re: Q1: TUT 0501
« on: September 28, 2018, 08:18:05 PM »
$p(t) = -1, q(t) = 1+3 \sin(t)$

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

$\mu y' - \mu y = \mu (1+3\sin(t))$

$e^{-t} y' - e^{-t}y = e^{-t} (1+3\sin(t))$

$\frac{d}{dt}(e^{-t} y) = e^{-t} +3e^{-t}\sin(t)$

$\int \frac{d}{dt}(e^{-t} y) dt= \int e^{-t} +3e^{-t}\sin(t) dt$

$e^{-t} y = -e^{-t} - \frac{3}{2}\sin te^{-t} - \frac{3}{2}\cos te^{-t} + c$

$y(t) = -1 - \frac{3}{2}\sin t - \frac{3}{2}\cos t + ce^t$

$y(0) = y_0 = -1 - \frac{3}{2}\sin(0) - \frac{3}{2}\cos(0) + ce^0 = -1 - \frac{3}{2} + c$

$c = y_0 + \frac{5}{2}$

$y(t) = -1 - \frac{3}{2}\sin t - \frac{3}{2}\cos t + (y_0 + \frac{5}{2}) e^t$

If $y_0 < - \frac{5}{2}, y_0 + \frac{5}{2} < 0$,

$$\lim_{t\to\infty} y(t) = \lim_{t\to\infty} (y_0 + \frac{5}{2}) e^t = -\infty$$
If $y_0 > - \frac{5}{2}, y_0 + \frac{5}{2} > 0$,
$$\lim_{t\to\infty} y(t) = \lim_{t\to\infty} (y_0 + \frac{5}{2}) e^t = +\infty$$
If $y_0 = - \frac{5}{2}, y_0 + \frac{5}{2} = 0$,
$$\lim_{t\to\infty} y(t) = \lim_{t\to\infty} (-1 - \frac{3}{2}\sin t - \frac{3}{2}\cos t)$$
is the only value of $y_0$ that makes the solution finite
$y_0 = - \frac{5}{2}$

6
Quiz-1 / Re: Q1: TUT 0601
« on: September 28, 2018, 06:30:48 PM »
\begin{gather*}
\frac{dy}{dx} = \frac{x + 3y}{x - y}\implies
\frac{dy}{dx} = \frac{x+3y}{x-y} = \frac{1+3(\frac{y}{x})}{1 - (\frac{y}{x})}
\end{gather*}
Let $v = \frac{y}{x}, y=xv$
\begin{gather*}\frac{dy}{dx} = v+x \frac{dv}{dx}\implies\\
x \frac{dv}{dx} = \frac{dy}{dx} -v = \frac{1+3v}{1-v} - v = \frac{(1+v)^2}{1-v}\implies\\
\frac{1-v}{(1+v)^2}dv = \frac{1}{x}dx\implies
\int \frac{1-v}{(1+v)^2}dv = \int \frac{1}{x}dx\implies\\
-\frac{2}{1+v} - \ln (1+v) + c = \ln x\implies\\
-\frac{2}{1+(\frac{y}{x})} - \ln (1+(\frac{y}{x})) + c = \ln x\implies\\
c-\frac{2x}{x+y} = \ln (x (1+\frac{y}{x})) \implies\\
c-\frac{2x}{x+y} = \ln (x+y)\implies\\
\frac{2x}{x+y} + \ln (x+y) = c\implies
x + y = Ce^{-\frac{2x}{x+y}}
\end{gather*}
I improved formatting. V.I.

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