I have been attempting to solve the above differential equation for some time now, and I remain stuck on one step. After substituting $u=x+y$, separating the variables, and integrating both sides, I am left with $$\frac{\sin(u)-1}{\cos(u)}=x+c$$ I have to solve for $y$, and thus, for $u$, but I cannot think of any identity which helps me here. Any help would be appreciated.  I followed the step in the solution, and set $$\frac{\tan(\frac{u}{2})-1}{\tan(\frac{u}{2})+1}=x+c$$ Then, I used the substitution $z=\frac{1}{1+\tan(\frac{u}{2})}$, which led to $1-2z=c+x$, which in turn, simplified to $z=\frac{1}{2}-\frac{c}{2}-\frac{x}{2}$. Then, substituting back: $$u=-2\arctan\left(1-\frac{1}{\frac{1}{2}-\frac{c}{2}-\frac{x}{2}} \right)$$ Thus, for all $n \in \mathbb Z$, the solution of the differential equation is $$y=-x-2\arctan\left(\frac{c+x+1}{c+x-1}\right)+2\pi n$$ Is my process correct, and do I require the $2\pi n$ generalization in my final expression?  Edit: The solution can also be expressed in indefinite form as $$\tan(x+y)-\sec(x+y)=x+c$$
One whose solution never makes me happy is the following: $$y'(t)=\sin(y+t)\text{.}$$ I would start by substituting $z(t)=y(t)+t$ to get an ODE in $z(t)$, but then I'm not sure about how to substitute back my solution to check if it's correct or not$\dots$
Until now, I have been compiling my LaTeX documents with pdfLaTeX. My standard document preamble includes the "usual suspects" for non-English speakers (, , ) and a bunch of other packages. Assuming I were to convert to XeLaTeX, what modifications of my preamble are advisable? I'm aware that the expansion=true option of the  package is not (yet) available for XeLaTeX, and that  is sort of a default package for OpenType fonts. What other packages (and package options) should be removed and, vice versa, incorporated when switching from pdfLaTeX to XeLaTeX? Proving an identity for Bernoulli polynomials Consider the Bernoulli polynomials $B_n(x)$ given by the expansion  $$\frac{te^{xt}}{e^t-1} = \sum\limits_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}.$$  I want to prove the identity $$B_n(1-x)=(-1)^nB_n(x).$$  This formula is given in a lot of books, but unfortunately without any proof. Therefore I'm wondering how to prove it. Do you have any idea?  I tried for instance:  $\sum\limits_{n=0}^{\infty}B_n(1-x)\frac{t^n}{n!}=\frac{te^{-xt}}{e^t-1}e^t=\left(\sum\limits_{n=0}^{\infty}B_n(-x)\frac{t^n}{n!}\right)\sum\limits_{n=0}^{\infty}\frac{t^n}{n!}=\sum\limits_{n=0}^{\infty}\left(\sum\limits_{l=0}^n\binom{n}{l}B_{l}(-x)  \right)\frac{t^n}{n!}$.  Hence  $B_n(1-x)=\sum\limits_{l=0}^n\binom{n}{l}B_{l}(-x)$.  But I do not know how to proceed.  On the other hand I know that $B_n(x+1)-B_n(x)=n x^{n-1}$. Hence  $B_{n}(1-x)=B_{n}(-x)-n(-x)^{n-1}$. But again, I do not know how I can go further.   Hopefully someone can help me.  Best regards Is my phone liable to explode if I charge it with a charger that came with a different phone? Lost space after removing bootcamp Schengen visa application rejected because my US visa stamp expires in December. What are my options? I went to the Italian consulate in Washington, DC today and they would not accept my visa application because they said my US visa stamp cannot expire less than 3 months after the end of my trip to Italy. Here's my relevant visa and trip info:   US visa stamp expires on 16 Dec 2015 I-797 valid until 3 Mar 2017 Trip to Italy is from 15 Oct to 25 Oct 2015   Is it true that my visa stamp needs to be valid for 3 months after my trip, even though I can legally stay in the US until 2017? Does anyone have any suggestions for getting a Schengen visa in my situation? I am firmly convinced he is innocent and his accusation is the result of the police conspiracy.     I am firmly convinced he is innocent and his accusation is the result of the police's conspiracy.    Which sentence do you think is a better choice? I'm preparing a document using the book class. In the table of contents, the space between the subsection number and title is too wide. How to reduce it? Equality holds in triangle inequality iff both numbers are positive, both are negative or one is zero How do we show that equality holds in the triangle inequality $|a+b|=|a|+|b|$ iff both numbers are positive, both are negative or one is zero? I already showed that equality holds when one of the three conditions happens. I need to customize my references in the following style using biblatexs Alpha style having the columns Label, Author, and Title and Other Information.    Till now with your help I could already change the square brackets to slashes.   My code so far is:   \documentclass[12pt]{scrartcl}  .  .   \usepackage[% natbib        = true, backend       = bibtex8, %bibtex, biber style         = alphabetic, %citestyle     = alphabetic, maxcitenames  = 2, mincitenames  = 1, %bibstyle      = alphabetic, %authoryear, %draft, %, sorting       = nyvt, maxbibnames   = 6, minbibnames   = 6, language      = ngerman, date          = long, backref       = false, backrefstyle  = none,  % none, three, two, two+, three+, all+ firstinits=true  ]{biblatex}    \usepackage[babel,german=quotes]{csquotes}    \bibliography{input/literatur}      \newrobustcmd{\mkbibslashes}[1]{/#1/}      \DeclareFieldFormat{labelalphawidth}{\mkbibslashes{#1}}    \DeclareFieldFormat{shorthandwidth}{\mkbibslashes{#1}}     \DeclareCiteCommand{\cite}[\mkbibslashes]    {\usebibmacro{prenote}}    {\usebibmacro{citeindex}% \usebibmacro{cite}}    {\multicitedelim}    {\usebibmacro{postnote}}    \renewcaptionname{ngerman}{\refname}{Literaturverzeichnis}   .   .   .   Thanks again How to add USB 3.0 port to your laptop Bevel not...beveling properly I'm taking a course and the instructor has beveled an edge and I cannot get the same result. My bevel only bevels on segment on one side then the rest, about 4 or 5 segments bevel on the other side. Other words the bevel is not beveling evenly on the top of a cylinder. Pictures below Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$ It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$  Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}{4})$  I was thinking of making polar transformations $X=r \cos \theta, Y=r \sin \theta$  Then I am getting stuck at ranges of $\theta$ I have noticed that while asking questions and scrolling to the bottom of the page, the right sidebar containing recent questions covers the footer partially. Also, a part of it goes beyond the browser's boundary. It may not be a big problem but it still doesn't look good.    This also happens on the  page In a bash script, how can I redirect all standard outputs to a log file and tee the output on the screen using exec ?  log_file="$HOME/logs/install.txt-`date +'%Y-%m-%d_%H-%M-%S'`" [ -f "$log_file" ] || touch "$log_file" exec 1&gt;&gt; $log_file 2&gt;&amp;1   This code redirect all the log to the log file but not to the screen . Why is the minimum size of a generating set for a finite group at most $\log_2 n$? It seems to be known that the minimum size of a generating set for a finite group of order $n$ is at most $\log_2 n$. Can someone explain why this is true?  Edit: noted that the logarithm is base 2, and $n$ represents the order of the group. When I was younger (11-12 years old, so the early 2000s) I once read a story where a young boy, who may have been either a young prince or a squire of some sort, is recruited by a wizard and taken to a city to train as a mage.  I seem to remember him living in dorms with the other children at this school, until he gets into difficulty with another of the boys in the dorm. He is kicked out of the dorms and ends up living above a blacksmith shop (I think).  I remember it being the first book in a series of books of which there were about 3 out at the time. I think one of the stories was about the boy going to a tower to do something or other.  Does anyone have any idea what this book series is? I haven't been able to find it for years now but it was a story I really loved. “While” loop without “do”; “if” statement without “then” I have a similar question as this :     How can I get rid of the do in a for loop and the then in an if   statement when typesetting algorithms using the algorithm and   algorithmicx packages?   but I want the same thing for algorithm2e package.  A minimal compilable example:  \documentclass[10pt]{article} \usepackage[linesnumbered,ruled]{algorithm2e} \SetKwRepeat{Do}{do}{while} \SetKw{KwGoTo}{go to}  \begin{document}  \begin{algorithm} \DontPrintSemicolon \KwIn{$A,B$} \KwOut{$C$} \While{$B &gt; 0$} {      $C = C+1$;      $B = B-1$;     \uIf{$C &gt; A$}        {            \textbf{break}\;        } } \caption{ABC} \end{algorithm}  \end{document} text-overflow: ellipsis in table-cell without width How does damage dealt increase as you level up How can I recover files from partitioned disk?