$$\begin{cases} \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} = 0, &amp;x\in\mathbb{R},\ t&gt;0\\ u(x,0) = u_0(x), &amp;x\in\mathbb{R}. \end{cases}$$  Which choices of the following functions for $u_0$ yield a $C^1$ Solution $u(x,t)$ for all  $x\in\mathbb{R}$  and $t&gt;0$  $1)$  $u_0 (x) = \frac{1}{1+ x^2}$  $2)$  $u_0(x) =  x$  $3)$  $u_0(x) = 1+x^2$  $4)$ $u_0(x) = 1+2x$  My answer key says the answer is $2$ and $4$. My  question is why we should reject option $1$ and $3$?  $$\frac{dt}{1}  =\frac{dx}{u}= \frac{du}{0}$$    gives me  $u=c_1$  and  $ut-x=c_2$ where $c_1$ and $c_2$ are constants.  Now  by condition given in Option $1$ I get a relation between constants  I.e    $$c_1= \frac{1}{1+(c_2)^2}.$$  Then I have the solution    $$u = \frac{1}{1+ (ut-x)^2}.$$   Then why  we reject option $1$?
123562
Consider the Cauchy Problem of finding $u(x,t)$ such that    $$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=0,x\in\mathbb{R},t&gt;0$$ $$u(x,0) = u_0(x), x\in\mathbb{R}$$ Which choices of the following functions for   $u_{0}$ yield a $C^{1}$ solution $u(x,t)$ for all   $x\in\mathbb{R},t&gt;0$         $u_{0}(x)=\frac{1}{1+x^{2}}$   $u_{0}(x)=x$   $u_{0}(x)=1+x^{2}$   $u_{0}(x)=1+2x$.      If I use the existence and uniqueness theorem for Cauchy problem i get the corresponding determinant is non zero so all are true according to me. But in answer key only option 2nd and 4th is given. Please help me to solve the problem. Thanks a lot.
89608
Really stuck with this, \begin{cases}  2tx'(t)-1=x^2t \\  x(t=1)=1 \\  \end{cases} I don't understand the variables. What does it mean that $x$ differentiated have $t$ as a function value, like this: $x'(t)$? Don't get it :(
69627
Let $U, V \sim U(0,1)$ be independent. What is $P(U \leq V)$?  My attempt:  We are looking for $P(U-V \leq 0)$. For a given $t \in [0,1]$ this is equivalent to $$P(U \leq t, V \gt t)= P(U \leq t)P(V \gt t)=P(U \leq t)(1-P(V\leq t))=t(1-t)=t-t^2$$  Then summing over $t$ on $[0,1]$, I get:  $$\int_0^1 t-t^2 \, dt = \frac{1}{6}$$  Which seems like a reasonable value, but I'm not exactly sure whether my approach with doing this in terms of t was correct, can anybody please validate my solution?
16164
Is the following statement $$\frac{\partial^2 f}{\partial x \, \partial y}=\frac{\partial^2 f}{\partial y \, \partial x}$$ always true? If not what are the conditions for this to be true?
33120
The task is as follows:     Given: $$F(x,y,z) = 0$$    Goal:  Show    $\frac{\partial  z}{\partial y}|_x \frac{\partial  y}{\partial x}|_z \frac{\partial x}{\partial z} |_y = -1$   Here is my work so far:  (1) Differentiate with respect to y, I get:  $0 + F_2 + F_3  \frac{\partial z}{\partial y} = 0$  So $ F_3  \frac{\partial z}{\partial y} = - F_2$  (2) Differentiate with respect to x, I get:  $F_1 + F_2 \frac{\partial y}{\partial x}  + 0 = 0$  So $ F_2  \frac{\partial y}{\partial x} = - F_1$  (3) Differentiate with respect to z, I get:  $F_1  \frac{\partial x}{\partial z}    + 0  + F_3 = 0$  So $ F_1  \frac{\partial x}{\partial z} = - F_3$  (4) After some manipulations with the $F_i$, I get to the conclusion that $\frac{\partial z}{\partial y}* \frac{\partial y}{\partial x} * \frac{\partial x}{\partial z}  = -1$, so when evaluated with x, z, y respectively, conclusion is still true  My questions are:  (1) Is my proof correct?  (2) For example, when I differentiate with respect to y, I "let" $F_1$ be 0 and find partials for other coordinates.    I had a hard time trying to explain to my friend on the reason(s) why I can do such "let be 0" thing.  Although I think if I can't do that, then there is no way that I can reach the conclusion, but I somehow feel confused about the fact too.  Since my book is doing it that way, my understanding is that I can do such "let be 0" thing based on the independece of x with respect to y, when I differentiate with respect to y.  But is my thought ok?  Would someone please help me on this question? Thank you very much ^_^
116269
$$\int_0^1\sqrt\frac x{1-x}\,dx$$ I saw in my book that the solution is $x=\cos^2u$ and $dx=-2\cos u\sin u\ du$. I would like to see different approaches, can you provide them?
112982
Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{(2x)}  \, dx = \frac{\ln|2x|}{2}+ C$$   Check: By taking derivative we get $\frac{1}{x}$  Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{2x}  \, dx = \frac{1}{2}\int  \frac{1}{x}  \, dx =\frac{\ln|x|}{2}+ C$$  Check: By taking derivative we get $\frac{1}{x}$  Question: Is  $\int  \frac{1}{2x}  \, dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$?
65966
Regarding: Let $f:(0, \infty) \to \mathbb{R}$ be a differentiable function for which $f'(x)f'(\frac{1}{x})&lt;0$ for every $0&lt;x \ne 1$. Then $f'(1)=0$. -- So I am trying to prove this, and I said: We know that $f$ is differentiable at $(0, \infty)$, so there exists a $0&lt; \delta &lt; 1$ such that $f$ is differentiable at $[1- \delta, 1+\delta]$ . So there exists a $x_{0} \in [1- \delta, 1+\delta]$ such that $f'(x_{0})f'(\frac{1}{x_{0}})&lt;0$ So by Darboux Theorem, there exists a $x_{0}&lt;c&lt; \frac{1}{x_{0}}$ such that $f'(c)=0$. But now, I do not know how to link $c$ with $1$... Any hints? Thanks! :)
64872
The full question is "Define the operator $\frac{d}{d \overline{z}} = \frac{1}{2}(\frac{d}{dx}+i\frac{d}{dy})$. Show that a  function $f(z)=u(x,y)+iv(x,y)$ satisfies the Cauchy-Riemann equation if and only if $\frac{df}{d\overline{z}}=0$." I have already figured out the $\rightarrow$ direction and only need help with the $\leftarrow$ direction.   So far I have: Let $\frac{df}{d\overline{z}}=0$ and $f(z)=u(x,y)+iv(x,y)$. Then $$0=\frac{df}{d\overline{z}}$$ $$=\frac{1}{2}\frac{d}{dx}(u+iv)+\frac{i}{2}\frac{d}{dy}(u+iv)$$ $$=\frac{1}{2}(u_x-v_y)+\frac{i}{2}(u_y+v_x)$$ And this is where I get stuck. I am not sure where to go from there so any help would be much appreciated. (Also feel free to fix my formatting I know it is atrocious."
116863
I am trying to solve the wave equation:  $u_{tt}$  = $u_{xx}$   With initial values:  $u(x,0) =\begin{cases} x^3 - x, &amp;\text{for }|x|\le 1,\ \\0, &amp;\text{for }|x|\ge1\end{cases}$  $u_t(x,0) =\begin{cases} 1 - x^2, \text{for } |x|\le 1, \\ 0, \text{for } |x|\ge1    \end{cases}$  In the domain $D= \{(x,t) | - \infty &lt; x&lt;\infty, \space\space t&gt;0\}  $  I am using D'Alembert's Formula and after inputting the values in I have:  $u(x,t) = \frac{1}{2}[\frac{2}{3}(x+t)^3 + \frac{4}{3}(x-t)^3  - 2(x-t)] $  I am not sure where to go from here.  I think because the initial conditions are not differentiable at some points  then there will be points in the solution that are not differentiable   So I'd need to consider the different cases of $|x+t|\le 1$?  I could be going down the complete wrong path so any help is appreciated.  Thanks
90668
Take the function  $$ \begin{cases}f(x,y)=(x^2+y^2)/\sin(\sqrt{x^2+y^2}) &amp; \text{when } 0&lt;\lvert (x,y)\rvert&lt;\pi\\   0 &amp; \text{when } (x,y)=(0,0) \end{cases}$$  I got that this function is differentiable at $(0,0)$. I calculated the partial derivatives at $(0,0)$ of which both were $0$ and trying to show the definition of differentiability is not satisfied but I keep getting that it is. Plz help.
20398
The solution of the wave equation $$u_{tt}-c^2u_{xx}=0$$ is given as $$u(x,t)=f(x+ct)+g(x-ct)$$ where $f,g$ are arbitrary functions of one variable.  One way to prove this is the following:  $$(\partial_{tt}-c^2 \partial_{xx})u=0$$ $$(\partial_t-c \partial_x)(\partial_t+c \partial_x)u=0$$  So solutions must satisfy  $$ (\partial_t-c \partial_x)u=0 \;\;\cup\;\;  (\partial_t+c \partial_x)u=0 $$  $\displaystyle{(\partial_t-c \partial_x)u=0 \Rightarrow u=f(x+ct)}$  $\displaystyle{(\partial_t+c \partial_x)u=0 \Rightarrow u=g(x-ct)}$    Could you explain me why the last two equations stand??
88696
Suppose the function $f:[0,\delta) \to \mathbb{R}$ is continuous, differentiable in $(0,\delta)$ and $f(0)=0$.  If the limit $\displaystyle \lim_{x \to 0+}\frac{f(x)}{f'(x)}= L$ exists, then is it always the case that $L = 0$.  This seems to be true both for functions with well-behaved derivatives such as $f(x) = x,$  $$\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x}{1}=0,$$  as well as functions with "bad" derivatives such as   $$f(x) = \begin{cases}x \ln x &amp;\mbox{if }x&gt;0 ,\\0  &amp;\mbox{if } x=0,  \end{cases},$$  where  $$\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x \ln x }{1+ \ln x}=0.$$  Is there a simple proof or counterexample?
122611
Finding all differentiable $f(z) = u(x) + iv(y)$ in $\mathbb{C}$ where $u(x),v(y)$ are real valued functions.   I’m not sure what to do. Would $f$ be differentiable simply if and only if both $u$ and $v$ were differentiable?   My thinking is that if the limit $$ \lim_{h \to 0}\frac{f(z+h)-f(z)}{h} $$  exists then it would be equal to  $$ \lim_{h \to 0}\frac{u(x+h)-u(x)}{h} + i\lim_{h \to 0}\frac{v(y+h)-v(y)}{h} $$  and therefore would exist if they existed. But that doesn’t seem quite right.  I also thought about using Cauchy-Riemann equations, but seeing that I’m trying to find ones that are differentiable rather than those that aren’t, I thought they wouldn’t be of much help.
20082
I have the example pde $u_t(x,t)=u_{xx}(x,t)-b(t)u(x,t)+q_0$, where $b(t)$ is a function of only $t$ and $q_0$ is a constant, $0&lt;x&lt;\pi$, $t&gt;0$. The subscripts denote derivatives. I also have some boundary conditions: $u(x,0)=0$ and $u(0,t)=0=u(\pi,t)$. The problem solution using variation of parameters is $$u(x,t)=\frac{4q_0}{\pi a(t)}\sum\limits_{n=1}^\infty \frac{\sin((2n-1)x)}{2n-1}\int\limits_0^t e^{-(2n-1)^2(t-\tau)}a(\tau)d\tau$$ where $$a(t)=e^{\int\limits_0^t b(\sigma)d\sigma}.$$  The problem I'm working on is slightly different. It gives the pde $u_t(x,t)=u_{xx}(x,t)-bu(x,t)+q(t)$ where $b$ is a constant and $q(t)$ is a function of only $t$, $0&lt;x&lt;\pi$, $t&gt;0$. The solution should be very similar to that of the example.  I don't know where to start. I'm using the textbook "Fourier Series and Boundary Value Problems" by Churchill and Brown, 7th edition, if anyone's interested. The "example" is problem 10 page 117 of that book.  Any help/hints would be greatly appreciated. Thanks!
114165
$$\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$   $\bf{My\; Try::}$ Let $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$  Put $\displaystyle x = \frac{1}{u},$ Then $\displaystyle dx = -\frac{1}{u^2}du$ and changing limits, We get  $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{1}{u(1+u^2)}du$$  So $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{t}{1+t^2}dt = \frac{1}{2}\int_{\frac{1}{e}}^{e}\frac{2t}{1+t^2}dt$$  So $$f(x) = \frac{1}{2}\ln (1+t^2)|_{\frac{1}{e}}^{e} = \ln(e) = 1$$  My question is how can we solve it in some shorter way, Help required, Thanks
114648
What is ${\partial\over \partial x_i}(x_i !)$ where $x_i$ is a discrete variable?  Do you consider $(x_i!)=(x_i)(x_i-1)...1$ and do product rule on each term, or something else? Thanks.