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### Messages - Siyan Chen

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1
##### 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, $$\int_{|z|=2} \frac{e^z \ dz}{z(z-3)} = 2 \pi i f(z_0) =2 \pi i \frac{e^0}{0-3}\ = -\frac{2 \pi i}{3}$$

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##### 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

5
##### Quiz-1 / Tut0301-Quiz1
« on: September 27, 2019, 02:13:45 PM »
$$\frac{dy}{dx} = \frac {x^2-3y^2}{2xy}$$
It’s not homogeneous, since: $$\frac{dy}{dx}*2xy - x^2 + 3y^2 = 0$$

Using substitution to solve the equation:
Let $$u = \frac {y}{x} \\then, y = ux$$
$$\frac {dy}{dx} = \frac {du}{dx} * x + u$$
and,
$$\frac {dy}{dx} = \frac {x^2-3y^2}{2xy} \\ = \frac {1-3(y/x)^2}{2(y/x)} \\ = \frac {1-3(u^2)}{2u}$$
then,
$$\frac {1-3(u^2)}{2u} = \frac {du}{dx} * x + u \\ \frac {1-3(u^2)}{2u} - u = \frac {du}{dx} * x \\ \frac {1-5(u^2)}{2u} = \frac {du}{dx} * x \\ \frac{2u}{1-5(u^2)} du = \frac {1}{x} dx$$
integrating both sides:
$$\int\frac{2u}{1-5(u^2)} du = \int\frac{1}{x} dx$$
Using substitution again: $$let \ a = 1-5(u^2),$$
$$\frac{da}{du} = -10u => du = \frac{da}{-10u}$$
$$\int\frac{2u}{1-5(u^2)} du = \int\frac{1}{-5a} da \\= -\frac{1}{5}\ln{|a|}+C$$
$$=> \int\frac{2u}{1-5(u^2)} du = \int\frac{1}{x} dx \\ -\frac{1}{5}\ln|1-5(u^2)| = \ln{x}+C \\ -\frac{1}{5}\ln|1-5(\frac{y}{x})^2| = \ln|x|+C$$

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