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Topics - Wanying Zhang

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Home Assignment 2 / 2.4 problem4
« on: February 27, 2019, 10:30:59 PM »
The problem is given:
$$u_{tt} - u_{xx} = (x^2 -1)e^{-\frac{x^2}{2}}$$
$$u(x,0) = -e^{-\frac{x^2}{2}}, u_t(x,0) = 0$$

I have already got the general solution as followed, but I have trouble solving the integral,
$$\int_{0}^{t} \int_{x-t+s}^{x+t-s} (y^2-1)e^{-\frac{x^2}{2}}dyds$$
and I tried $\Delta$ method, but it seems to make the equation more complex. Professor, could you please give a hint of solving this problem?

Home Assignment 3 / Problem2(17) even or odd?
« on: February 01, 2019, 10:40:32 AM »
Given the conditions:
$$g(x) = 0,   
h(x) =
       \text{1} & {|x| < 1}\\
       \text{0} & {|x| \geq 1} \\
I know since both $g, h$ are even functions, as a result, $u$ is even with respect to $x$. But I don't quite understand how to get the conclusion that $u$ is odd with respect to $t$?

I have difficulties obtaining the general solution of the equation $u_{tt} - c^2u_{xx} = 0$. From the online textbook Section2.3, it mentions $v = u_t +cu_x$, and it gets the result $v_t - cv_x = 0$ by chain rule. But when I expand $v_t - cv_x = 0$, I get an extra term $x'(t)$ when applies chain rule to $v_t$ because I think $u_{t}$ in $v$ can be two separated into to parts which are $u_t(t)$ and $u_t(x(t))$. Then applying chain rule, it becomes $v_t = u_{tt} + x'(t) +c(u_{xt} + x'(t))$ where the term $x'(t)$ can not be cancelled. I wonder where is the problem of my thought.

Home Assignment 2 / Assignment 2.2 Problem3 (16)
« on: January 24, 2019, 12:25:14 AM »
Given the question $u_t + 3u_x -2u_y = xyu$. My steps are:
$$\frac{dt}{1} = \frac{dx}{3} = \frac{dy}{-2} = \frac{du}{xyu} \Rightarrow x=C_1+3t, y=C_2-2t$$
$$xydt = \frac{du}{u} \Rightarrow (C_1+3t)(C_2-2t)dt = \frac{du}{u}$$
$$\ln u = C_1C_2 t + \frac{t^2}{2} (3C_2 - 2C_1) - 2t^3 +\phi(C_1, C_2)$$
I know then I need to insert base $e$ into both sides to get $u$, but it would be complex. I wonder is there any simple way to solve this? Or do I have something wrong in above equations so that leads to the complex solution?

Home Assignment 2 / problem 5 (23)
« on: January 20, 2019, 12:53:01 AM »
I have trouble solving this problem: $yu_x - xu_y = x^2$. I know the characteristic equation is $\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x^2}$ and then have $C = \frac{x^2}{2} + \frac{y^2}{2}$. Then the following should be the integration relative to $du$, but either $\frac{du}{dx}$ or $\frac{du}{dy}$ will contain not only one variable, like $\frac{du}{dx} = \frac{x^2}{y}$ contain both $x$ and $y$. I wonder if $x$ and $y$ are independent here. If not, should I rewrite the expression $C = \frac{x^2}{2} + \frac{y^2}{2}$ in order to get the expression of y in terms of x , and then applies it into the integration relative to $du$? Any reply would be appreciated.

Home Assignment 2 / Secondary Textbook Chp1.2 Exercise 10
« on: January 18, 2019, 06:16:16 PM »
I have difficulties solving this problem: $u_x + u_y + u = e^{x+2y}$. I get $\frac{dy}{dx} = 1$ from the characteristic equation and have no ideas what should be followed. Any reply would be appreciated.

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