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Messages - Siyan Chen

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Quiz 4 / TUT0401 Quiz4
« on: February 14, 2020, 10:59:04 AM »
Evaluate the given integral using Cauchy’s Formula or Theorem.

$$\int_{|z|=2} \frac{e^z \ dz}{z(z-3)}$$


First, we can find that $\frac{e^z \ dz}{z(z-3)}$ is not analytic when $z=0$ and $z=3$,

also, $z=3$ is outside the circle $|z|=2$ and $z=0$ is inside the circle $|z|=2$.

Hence, $$\int_{|z|=2} \frac{e^z \ dz}{z(z-3)} = \int_{|z|=2} \frac{ \frac{e^z }{z-3}}{z}dz$$

By Cauchy Formula,  we can get $$f(z)= \frac{e^z}{z-3} , \ and \ z_{0} = 0$$

Therefore, $$\int_{|z|=2} \frac{e^z \ dz}{z(z-3)} = 2 \pi i f(z_0) =2 \pi i \frac{e^0}{0-3}\ = -\frac{2 \pi i}{3}$$

2
Term Test 1 / Re: Problem 4 (afternoon)
« on: October 23, 2019, 02:29:18 PM »
a) Find the general solution of  $y’’+2y’+17y=40e^x + 130 \sin(4x)$

First, for the complimentary solution, consider the homogeneous equation:
$y’’+2y’+17y=0$

and then, the characteristic equation is: $r^2+2r+17=0$,
$r \ = \ \frac{-2 \pm \sqrt{4-4 \times 17}}{2} \ = \ \frac{-2 \pm 8i}{2} \ = \ -1 \pm 4i$

These roots are a pair of complex conjugates in the form of $ \lambda \pm i \mu$,
so the differential equation has a general solution in the form of $y(x)=C_{1} e^{\lambda x} \cos(\mu x)+C_{2} e^{\lambda x} \sin(\mu x)$.

In this case, we have: $y_{c}(x)= e^{-x} (C_{1} \cdot \cos(4x) +C_{2} \cdot \sin(4x))$

Then, for the particular solution, by the method of undetermined coefficients,
suppose $y_{p}=A \cdot e^x$ satisfies the equation: $y’’+2y’+17y=40e^x$

Since, $y_{p}=A \cdot e^x$
$y’_{p}=A \cdot e^x$
$y’’_{p}=A \cdot e^x$

plug into the equation: $A \cdot e^x + 2A \cdot e^x + 17 A \cdot e^x = 20A \cdot e^x = 40e^x$
so, $A=2$
$\therefore y_{p}=2e^x$

For another particular solution, suppose $y_{p}=B \cdot \cos(4x) + C \cdot \sin(4x)$ satisfies the equation: $y’’+2y’+17y=130 \cdot \sin(4x)$

Since, $y_{p}=B \cdot \cos(4x) + C \cdot \sin(4x)$
$y’_{p}=-4B \cdot \sin(4x) + 4C \cdot \cos(4x)$
$y’’_{p}=-16B \cdot \cos(4x) -16 C \cdot \sin(4x)$

plug into the equation:  $-16B \cdot \cos(4x) -16 C \cdot \sin(4x)-8B \cdot \sin(4x) + 8C \cdot \cos(4x) + 17B \cdot \cos(4x) + 17C \cdot \sin(4x) \\ =(B+8C) \cdot \cos(4x) + (C-8B) \cdot \sin(4x) \\ =130 \sin(4x)$

Then, $
\begin{cases}
C-8B=130 \\
B+8C=0
\end{cases}$

=> $\begin{cases}
C=2 \\
B=-16
\end{cases}$

$\therefore y_{p}=-16 \cdot \cos(4x) + 2 \cdot \sin(4x)$

So, the general solution is: $y(x)=e^{-x} (C_{1} \cdot \cos(4x) +C_{2} \cdot \sin(4x))+2e^x-16 \cdot \cos(4x) + 2 \cdot \sin(4x)$


b) when $y(0)=0, y’(0)=0$, and we have
$y’(x)=-e^{-x} (C_{1} \cdot \cos(4x) +C_{2} \cdot \sin(4x))+e^{-x} (-4C_{1} \cdot \sin(4x) +4C_{2} \cdot \cos(4x))+2e^x+64 \cdot \sin(4x) + 8 \cdot \cos(4x)$,

plug $y(0)=0$ into $y(x)$ equation, we get: $0=C_{1}+2-16$ => $C_{1}=14$,

then, plug $y’(0)=0$ into $y’(x)$ equation, we have:
$-C_{1}+4C_{2}+2+8=0 \\
-14+4C_{2}+10=0 \\
C_{2} = 1$

$\therefore y(x)=e^{-x} (-14\cdot \cos(4x) + \cdot \sin(4x))+2e^x-16 \cdot \cos(4x) + 2 \cdot \sin(4x)$

3
Quiz-3 / TUT0301 Quiz3
« on: October 11, 2019, 02:22:45 PM »
quiz 2 (tut 0301)
$$ y’’ + 4y = 0,\ y_{1}(t) = \cos(2t),\ y_{2}=\sin(2t)$$
The given solution of differential equation is: $y_{1}(t) = \cos(2t)$
So, $y_{1}’(t) = -2 \sin(2t)$
$y_{1}’’(t) = -4 \cos(2t)$

Then, plug into the given differential equation: $y’’ + 4y = 0$,
i.e. $-4 \cos(2t)\ + 4(\cos(2t)) = 0$ => 0=0
So, $y_{1}(t) = \cos(2t)$ is a solution of this equation

Similarly, we have $y_{2}(t)=\sin(2t)$
So, $y_{2}’(t) = 2 \cos(2t)$
$y_{2}’’(t) = -4 \sin(2t)$

Then, plug into the given differential equation: $y’’ + 4y = 0$,
i.e. $-4 \sin(2t)\ + 4(\sin(2t)) = 0$ => 0=0
So, $y_{2}(t)=\sin(2t)$ is also a solution of this equation

To check whether $y_{1}$ and $y_{2}$ constitute a fundamental set of solutions, we will find the Wronskian $W(y_{1}(t), y_{2})(t)\ = \begin{vmatrix}
y_{1}(t) & y_{2}(t)\\
y_{1}’(t) & y_{2}’(t)\\
\end{vmatrix} $
= $ \begin{vmatrix}
\cos(2t) & \sin(2t)\\
-2\sin(2t) & 2\cos(2t)\\
\end{vmatrix} $
= $2\cos^2(2t)+2\sin^2(2t)$
= $2(\cos^2(2t)+\sin^2(2t))$
= $2 (1)$
= $2$
Since, $W(y_{1}(t), y_{2})(t)\neq0$, then we say that $y_{1}$ and $y_{2}$ constitute a fundamental set of solutions.

4
Quiz-2 / TUT0301 Quiz2
« on: October 04, 2019, 02:01:10 PM »
$$1 + (\frac{x}{y}-\sin(y))y’ = 0$$
Let $M=1, N=\frac{x}{y}-\sin(y)$
Then, we can get: $M_{y}=0, N_{x}=\frac{1}{y}$
Define $R=\frac{M_{y}-N_{x}}{M}=\frac{-\frac{1}{y}}{1}=-\frac{1}{y}$,
=> The integrating factor should be: $\mu (x,y) = e^{-\int R dy} = e^{ln|y|} = y$

Multiple both sides by y, we can define the new $M = y, N = x-\sin(y)y$,

Since it’s exact now, there exists a function $\phi (x,y)$, such that: $\phi_{x}(x,y) = M(x,y)\ and\ \phi_{y}(x,y) = N(x,y)$
Integrating with respect to x:
$\phi(x,y) = \int M(x,y) \ dx \\ i.e. \phi(x,y) = \int y\ dx \\ \phi(x,y) = xy + h(y)$
Then, we have: $\phi_{y} =  x + h’(y) = N = x-y\sin(y)$, i.e. $h’(y)=-y\sin(y)$
=> $h(y) = -\int y \sin(y) \ dy$

Using integrating by parts,  $u = y, \ v = -\cos(y),\\du = 1, \ dv = \sin(y) $

=> $h(y) = -(-y \cos(y)- \int(-\cos(y))\ dy) = y \cos(y) - \sin(y)$

So, the solution of the given equation is: $\phi(x,y) = xy + y \cos(y) - \sin(y) = C$, where C is the constant

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