### 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 - Xinqiao Li

Pages: [1]
1
##### Chapter 3 / Re: Confused about the notation in maximum modulus principle
« on: December 16, 2020, 04:16:44 PM »
woww thank you, that makes sense now.

2
##### Chapter 3 / Confused about the notation in maximum modulus principle
« on: December 15, 2020, 10:20:52 AM »
Hello, this is the problem 5 from the sample exam 2020F, I am kind of confused by the notation here. What is meant by find instead $max_D Re(f(z))$ and $\in_D Re(f(z))$? Are we suppose to find the maximum of the real part of the function f on the bounded domain D?

3
##### Quiz 2 / Quiz2 LEC0101 C
« on: October 03, 2020, 05:47:44 AM »
Questions: Find all points of continuity of the given function: $f(z)=(Imz-Rez)^{-1}$

Solutions:

Let $z=x+iy$ where x,y are real numbers.

Then $f(z)=f(x,y)=(Im(x+iy)-Re(x+iy))^{-1}=(y-x)^{-1}=\frac{1}{y-x}$

The function is not valid when the denominator equals 0, that is, $y=x$.

Therefore, the function is discontinuous only at all points of the line $y=x$.

4
##### Quiz 1 / LEC0101 Quiz1
« on: September 25, 2020, 11:17:28 AM »
Problem:
Describe the locus of points z satisfying the given equation.
$|z+1|^2+2|z|^2=|z−1|^2.$

Solution:
Let $z=x+yi$ where x,y are real numbers
$|z+1|^2 = |(x+yi)+1|^2=|(x+1)+yi|^2=(x+1)^2+y^2=x^2+2x+1+y^2$
$2|z|^2=2|x+yi|^2=2(x^2+y^2)=2x^2+2y^2$
$|z-1|^2 = |(x+yi)-1|^2=|(x-1)+yi|^2=(x-1)^2+y^2=x^2-2x+1+y^2$

Hence,
$|z+1|^2+2|z|^2=|z−1|^2$
$x^2+2x+1+y^2+2x^2+2y^2=x^2-2x+1+y^2$
$2x^2+4x+2y^2=0$
$2(x^2+2x+y^2)=0$
$x^2+2x+y^2=0$
$x^2+2x+1+y^2=0$
$(x+1)^2+y^2=1$

The locus of point z is a circle centered at (-1,0) with radius 1.

5
##### Term Test 1 / Re: Problem 3 (main)
« on: October 23, 2019, 04:29:59 PM »
a) Find the general solution for equation $y'' - 2y' -3y = 16coshx$
b) Find solution, satisfying $y(0) = 0, y'(0) = 0$

a)
Consider $y'' - 2y' -3y = 0$

Assume $y = e^{rx}$, then the characteristic polynomial of the equation is given by $r^2 -2r -3 = 0$

This simplifies to $(r - 3)(r +1) = 0$ which gives us two roots $r_1 = 3$ and $r_2 = -1$

Then the complementary solution is $y_c = c_1e^{3x} + c_2e^{-x}$

Since $y'' - 2y' -3y = 16cosh = 8e^x + 8e^{-x}$

We guess the particular solution is of the form

$y = Ae^x + Bxe^{-x}$
$y' = Ae^x + B(e^{-x} - xe^{-x}) = Ae^x + Be^{-x} - Bxe^{-x}$
$y'' = Ae^x - Be^{-x} - B(e^{-x} - xe^{-x}) = Ae^x - 2Be^{-x} + Bxe^{-x}$

Substituting back to the original equation:

$y'' - 2y' -3y = Ae^x - 2Be^{-x} + Bxe^{-x} - 2Ae^x - 2Be^{-x} + 2Bxe^{-x} - 3Ae^x - 3Bxe^{-x} = -4Ae^x - 4Be^{-x} = 8e^x + 8e^{-x}$

Therefore, $A = -2$ and $B = -2$
The particular solution is $y_p = - 2e^x - 2xe^{-x}$
The general solution is :
$$y = c_1e^{3x} + c_2e^{-x} - 2e^x - 2xe^{-x}$$

b)
$$y = c_1e^{3x} + c_2e^{-x} - 2e^x - 2xe^{-x}$$
$$y' = 3c_1e^{3x} - c_2e^{-x} - 2e^x + 2xe^{-x} - 2e^{-x}$$

Plug in the initial conditions $y(0) = 0, y'(0) = 0$, we have,
$0 = c_1 + c_2 - 2$
$0 = 3c_1 - c_2 - 2 - 2$

so, $c_1 = \frac{3}{2}$ and $c_2 = \frac{1}{2}$
The particular solution is
$$y = \frac{3}{2}e^{3x} + \frac{1}{2}e^{-x} - 2e^x - 2xe^{-x}$$

6
##### Term Test 1 / Re: Problem 1 (main sitting)
« on: October 23, 2019, 04:18:11 PM »
Here is another method to find $\varphi$. (By integrating N with respect to y)

Find integrating factor and then a general solution of the ODE
$$(y+3y^2e^{2x}) + (1+2ye^{2x})y' = 0, y(0) = 1$$

Let $M = y+3y^2e^{2x}$ and $N = 1+2ye^{2x}$

We can see that $M_y = 1 + 6e^{2x}y$ and $N_x = 4e^{2x}y$

They are not equal, so not exact. Our goal is to find an integrating factor and make $M_y$ equals $N_x$

$$R_2 = \frac{M_y - N_x}{N} = \frac{1 + 2e^{2x}y}{1+2e^{3x}y} = 1$$

So $\mu = e^{\int R_2dx} = e^{\int1dx} = e^x$

Multiply $\mu$ on both side of the orignial equation and we got
$$(e^xy+3y^2e^{3x}) + (e^x+2ye^{3x})y' = 0$$
Now $M_y = e^x + 6e^{3x}y$ and $N_x = e^x + 6e^{3x}y$

Therefore $\exists\varphi_{(x,y)}$ satisfy $\varphi_x = M$ and $\varphi_y = N$

$$\varphi = \int Ndy = \int (e^x+2ye^{3x})dy = e^xy+y^2e^{3x} + h(x)$$

Then $\varphi_x =e^xy +3y^2e^{3x} + h'(x)$

Since $\varphi_x = M = e^xy+3y^2e^{3x}$

So $h'(x) =0$ and $h(x)=c$
$$\varphi = e^xy+y^2e^{3x} = c$$
Given initial condition $y(0) = 1$

We have $1\times 1 + 1^2 \times 1 = c$ and $c = 2$

Therefore, the particular solution is
$$e^xy+y^2e^{3x} = 2$$

7
##### Term Test 1 / Re: Problem 2 (main)
« on: October 23, 2019, 04:09:34 PM »
a) Find Wronskian $W(y_1, y_2)(x)$ of a fundamental set of solutions $y_1(x)$, $y_2(x)$ for ODE $x^2y'' -2xy' + (x^2+2)y = 0$
b) Check that $y_1(x) = xcosx$ is a solution and find another linearly independent solution.
c) Write the general solution, and find solution that $y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$

a)
$$y'' - \frac{2}{x}y' + (1 + \frac{2}{x^2}) = 0$$
We see that $p(x) = -\frac{2}{x}$ is continuous everywhere except at $x=0$, $q(x) = (1 + \frac{2}{x^2})$ is continuous everywhere except at $x=0$.

Then by Abel's Theorem,
$$W(y_1, y_2)(x) = ce^{-\int p(x)dx} = ce^{\int(\frac{2}{x})dx} = ce^{2lnx} = cx^2$$
b)
Let's verify $y_1(x) = xcosx$ is a valid solution.

Since $y_1(x) = xcosx$, $y'_1(x) = cosx-xsinx$, $y''_1(x) = -sinx-xcosx-sinx = -xcosx-2sinx$

Plug in: $x^2y'' -2xy' + (x^2+2)y = 0 = -x^3cosx-2x^2sinx - 2xcosx + 2x^2sinx + x^3cosx + 2xcosx = 0$

So $y_1(x) = xcosx$ is a solution.

Take $c = 1, W(y_1, y_2)(x) = x^2$.

By Reduction of Order, we have:
$$y_2 = y_1\int(\frac{ce^{-\int p(x)dx}}{(y_1)^2})dx = xcosx\int(\frac{x^2}{(x^2cos^2x})dx = xcosx\int sec^2xdx = xcosx tanx = xsinx$$

c)
Then by part b, the general solution is $y = c_1xcosx +c_2xsinx$

The derivative is given by $y' = c_1(cosx -xsinx) + c_2(xcosx + sinx)$

Plug in the initial conditions $y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$, we have

$1 = \frac{\pi}{2}c_2$ and $0 = (-\frac{\pi}{2})c_1 + c_2$

So $c_2 = \frac{2}{\pi}$ and $c_1 = \frac{4}{(\pi)^2}$

The particular solution is,
$$y = \frac{4}{(\pi)^2}xcosx +\frac{2}{\pi}xsinx$$

8
##### Term Test 1 / Re: Problem 2 (main)
« on: October 23, 2019, 08:23:07 AM »
b) Forgot to check $y_1(x) = xcosx$ is a valid solution.

Since $y_1(x) = xcosx$, $y'_1(x) = cosx-xsinx$, $y''_1(x) = -sinx-xcosx-sinx = -xcosx-2sinx$

Plug in: $x^2y'' -2xy' + (x^2+2)y = 0 = -x^3cosx-2x^2sinx - 2xcosx + 2x^2sinx + x^3cosx + 2xcosx = 0$

So $y_1(x) = xcosx$ is a solution.

9
##### Term Test 1 / Re: Problem 2 (main)
« on: October 23, 2019, 08:12:27 AM »
Could also use Reduction of Order to solve part b).

Take $c = 1, W(y_1, y_2)(x) = x^2$.

By Reduction of Order, we have:
$$y_2 = y_1\int(\frac{ce^{-\int p(x)dx}}{(y_1)^2})dx = xcosx\int(\frac{x^2}{(x^2cos^2x})dx = xcosx\int sec^2xdx = xcosx tanx = xsinx$$

10
##### Quiz-4 / TUT0502 Quiz4
« on: October 18, 2019, 09:59:31 PM »
Find the general solution of the given differential equation:
$$9y'' + 6y' + y = 0$$
Solve:

The characteristic polynomial is given by $9r^2 + 6r + 1 = 0$

Factor it we obtain $(3r + 1)(3r + 1)$

Then $r_1 = -\frac{1}{3}$ and $r_2 = -\frac{1}{3}$

We observe repeated roots here.

Then the general solution of the given differential equation is $y(t) = c_1e^{-\frac{1}{3} t} + c_2te^{-\frac{1}{3}t}$

11
##### Quiz-3 / TUT0502 Quiz 3
« on: October 11, 2019, 02:00:35 PM »
Find the solution of the given initial problem:
$$y''+4y'+3y = 0, y(0) = 2, y'(0) = -1$$
Assume $y = e^{rt}$, then it must follow that r is the root of the characteristic polynomial
$$r^2+4r+3=0\\ (r+1)(r+3)=0$$
We have $r_1 = -1$ or $r_2 = -3$.

The general solution of the second order differential equation has the form of
$$y = c_1e^{r_1t} + c_2e^{r_2t}$$

Thus, we have
$$y = c_1e^{-t} + c_2e^{-3t}$$

The derivative of this general solution is
$$y' = -c_1e^{-t} - 3c_2e^{-3t}$$

To satisfy both initial conditions $y(0) = 2$ and $y'(0) = -1$,

We have $2 = c_1 + c_2$ and $-1 = -c_1 -3c_2$

This gives us $c_1 = \frac{5}{2}$ and $c_2 = -\frac{1}{2}$

Therefore, the solution of the initial value problem is
$$y = \frac{5}{2}e^{-t} -\frac{1}{2}e^{-3t}$$

12
##### Chapter 3 / Re: Initial conditions evaluated at different $t_0$'s?
« on: October 05, 2019, 10:42:21 PM »
From what I understand, when you are solving the second order initial value problems, you first get a general solution of that equation with two arbitrary constant of the form y(t)=C1ert+C2ert. Then you find the derivative y' of this general solution, and plug in the initial conditions to solve what the two constants are. So you might not necessary need both initial conditions to be defined at t0.

13
##### Quiz-2 / TUT0502 Quiz2
« on: October 04, 2019, 02:00:01 PM »
Determine whether the equation given below is exact. If it is exact, find the solution
$$(e^xsin(y)-2ysin(x))-(3x-e^xsin(y))y'=0\\ (e^xsin(y)-2ysin(x))dx-(3x-e^xsin(y))dy=0$$
Let $M(x,y) = e^xsin(y)-2ysin(x)$

Let $N(x,y) = -(3x-e^xsin(y)) = e^xsin(y)-3x$
$$M_y(x,y) = \frac{\partial}{\partial y} M(x,y)= \frac{\partial}{\partial y} (e^xsin(y)-2ysin(x)) = e^xcos(y)-2sin(x)\\ N_x(x,y) = \frac{\partial}{\partial x} N(x,y) = \frac{\partial}{\partial x} (e^xsin(y)-3x) = e^xsin(y)-3$$
Clearly, we see that $e^xcos(y)-2sin(x) \neq e^xsin(y)-3$

Therefore, $M_y(x,y) \neq N_x(x,y)$

By definition of exact, we can conclude the equation is not exact.

14
##### Quiz-1 / TUT0502
« on: September 27, 2019, 02:08:27 PM »

15
##### Quiz-1 / Re: TUT5103 quiz1
« on: September 27, 2019, 10:45:42 AM »
a very small typo: the final solution should be just y, not y^2.

Pages: [1]