Initial value problem, show y(x) = 1-x is a solution I'm not really sure on how to go about solving this problem...  Consider the initial value problem.       $$y' (x) = \frac{1}{2}(-x + \sqrt{x^2 + 4y})$$   $$y(2) = -1$$   $a)$ Show that $y(x) = 1-x$ and $y(x) = -\frac{x^2}{4}$ are two solutions to the above IVP.
$y'=$ ${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$ Prove that the solutions $y_1$ and $y_2$ for the initial value problem   $y'=$${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$  are:  $y_1=1-x$  $y_2={{-x^2}\over 4}$  And explain why the two solutions not contradiction with the theory of existence and uniqueness.  ( sorry I don't speak English well )  My answer:  For the initial condition :  $y_1=1-x$ Satisfy the initial condition  $y(2)=-1$  $y_1=1-2=-1$   And Satisfy the differential equation  ${y_1}^{\prime}=-1$  ${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4(1-x)}}\over 2}$  $={{-x+\sqrt{(x-2)^2}}\over 2}$  $={{-x+x-2}\over 2}=-1$  L.h.s.=R.h.s.  $y^{\prime}={{-x+\sqrt{x^2+4y}}\over 2}$  Also $y_2$ Satisfy the initial condition  $y_2={{-x^2}\over4}={-4\over4}=-1$   And satisfy the differential equation.   $y_2^{\prime}={-x\over 2}$  ${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4({-x^2\over 4})}}\over 2}={-x\over 2}$  And for second part of exercise:  ${\partial f\over \partial y}={1\over \sqrt{x^2+4y}}$  and this not defined when $x=2, \ y=-1$  So $f(x,y)$ no satisfy lipshtize condition in any rectangular has (2,-1), Lipshtize condition including uniqueness of solution, so no contradict with the existence and uniqueness.  True ?  If this also wrong I well be delet it , I answered by the same way of my teacher :(
Complete residue system proof I was given a proof of this in class, but it does not make sense to me, can someone provide a better proof, or explain the one I have provided, thanks.  Suppose $\gcd(a,n) = 1$, where $a$ and $n$ are natural numbers. Prove that $0$, $a$, $2a$, $3a,\ldots, (n-1)a$, is complete residue system modulo $n$.  Proof given in class:  \begin{align*} ka &amp;\equiv ra \pmod{n}\\ &amp;\Rightarrow  k \equiv r \pmod{n}\\ &amp;\Rightarrow  \text{$k-r$ divides $n$}\\ &amp;\Rightarrow \text{$k=r$ because $0\leq k\lt n$, $0\leq r\lt n$} \end{align*} , Laser Interferometer Gravitational-Wave Observatory, is a large-scale physics experiment aiming to directly detect gravitational waves. The device measures the phase shift laser beams.  If I understand this well, the measurement assumes that a suitable gravitational wave reduces and prolongs one of the tubes and the extension will take effect in the phase-shifting laser beams.  But I think if there's a gravitational wave that passes through it, it will affect everything, even the laser beam. This means that when the wave lengthens or shortens the tube, it also lengthens or shortens the wavelength of the laser beam.  In my opinion, therefore, this device does not work. Or am I wrong? PCIe Bus error severity=Corrected I have a new HP Pavilion Gaming Notebook and a new installation of Ubuntu 16.04. When I press Ctrl + Alt + F1 I go start seeing the errors shown in the following image and it doesn't allow me to interact with the console:    I also see these errors for a while everytime I boot. I need to do Ctrl + Alt + F1 to access a non graphical terminal to install some Nvidia drivers. What's going on?  What's causing the problem seems to be: 00:1c.5 PCI bridge [0604]: Intel Corporation Sunrise Point-H PCI Express Root Port #6 [8086:a115] (rev f1)  jpiabrantes@joao:~$ lspci -nn 00:00.0 Host bridge [0600]: Intel Corporation Sky Lake Host Bridge/DRAM Registers [8086:1910] (rev 07) 00:01.0 PCI bridge [0604]: Intel Corporation Sky Lake PCIe Controller (x16) [8086:1901] (rev 07) 00:02.0 VGA compatible controller [0300]: Intel Corporation Skylake Integrated Graphics [8086:191b] (rev 06) 00:04.0 Signal processing controller [1180]: Intel Corporation Skylake Processor Thermal Subsystem [8086:1903] (rev 07) 00:14.0 USB controller [0c03]: Intel Corporation Sunrise Point-H USB 3.0 xHCI Controller [8086:a12f] (rev 31) 00:14.2 Signal processing controller [1180]: Intel Corporation Sunrise Point-H Thermal subsystem [8086:a131] (rev 31) 00:16.0 Communication controller [0780]: Intel Corporation Sunrise Point-H CSME HECI #1 [8086:a13a] (rev 31) 00:17.0 SATA controller [0106]: Intel Corporation Sunrise Point-H SATA Controller [AHCI mode] [8086:a103] (rev 31) 00:1c.0 PCI bridge [0604]: Intel Corporation Sunrise Point-H PCI Express Root Port #5 [8086:a114] (rev f1) 00:1c.5 PCI bridge [0604]: Intel Corporation Sunrise Point-H PCI Express Root Port #6 [8086:a115] (rev f1) 00:1c.6 PCI bridge [0604]: Intel Corporation Sunrise Point-H PCI Express Root Port #7 [8086:a116] (rev f1) 00:1f.0 ISA bridge [0601]: Intel Corporation Sunrise Point-H LPC Controller [8086:a14e] (rev 31) 00:1f.2 Memory controller [0580]: Intel Corporation Sunrise Point-H PMC [8086:a121] (rev 31) 00:1f.3 Audio device [0403]: Intel Corporation Sunrise Point-H HD Audio [8086:a170] (rev 31) 00:1f.4 SMBus [0c05]: Intel Corporation Sunrise Point-H SMBus [8086:a123] (rev 31) 01:00.0 3D controller [0302]: NVIDIA Corporation GM107M [GeForce GTX 950M] [10de:139a] (rev a2) 07:00.0 Unassigned class [ff00]: Realtek Semiconductor Co., Ltd. RTS522A PCI Express Card Reader [10ec:522a] (rev 01) 08:00.0 Network controller [0280]: Realtek Semiconductor Co., Ltd. RTL8723BE PCIe Wireless Network Adapter [10ec:b723] 09:00.0 Ethernet controller [0200]: Realtek Semiconductor Co., Ltd. RTL8101/2/6E PCI Express Fast/Gigabit Ethernet controller [10ec:8136] (rev 0a) How do I go about proving this? I know one method is:  $\eqalign{    \cos (90^\circ + \theta ) &amp;\equiv \cos90^\circ \cos\theta  - \sin90^\circ \sin\theta   \cr    &amp;  \equiv (0)(\cos\theta ) - (1)(\sin \theta )  \cr    &amp;  \equiv  - \sin \theta  \cr} $  I'd appreciate others, particularly ones that will allow me to visualize this identity on the unit circle.  Thanks alot. Smallest positive element of $ \{ax + by: x,y \in \mathbb{Z}\}$ is $\gcd(a,b)$ Let $S = \{ax + by: x,y \in \mathbb{Z}\}$  and let $e &gt; 0$ be the smallest element in $S$. Prove that $e \mid a$, and hence prove that $e = \gcd(a,b)$  I'm afraid I can't provide much of my initial working here because I'm at a loss of how to start.  I suppose that to show $e\mid a$ I would have to show that $eg = a$ for some $g \in \mathbb{Z}$. But I have no idea how I would do this. After the upgrade I can't figure out how the new shield system works. I have the option to discard the shield, and I notice I have a new potion in the inventory.  More importantly, I got attacked (and lost) twice in a short amount of time and don't have a shield. When do you now get shield? I would like to prove the following equality: $$k^{\phi(l)} + l^{\phi(k)} \equiv1\pmod{lk}$$if $\gcd(l,k)=1$. What methods can I use? Thank you for your help. Adding two numbers concatenates them instead of calculating the sum Testing whether two regression coefficients are significantly different (in R ideally) If this is a duplicate question, please point to the right way, but the similar questions I've found here haven't been sufficiently similar. Suppose I estimate the model $$Y=\alpha + \beta X + u$$  and find that $\beta&gt;0$. However, it turns out that $X=X_1+X_2$, and I suspect $\partial Y/\partial X_1 \ne \partial Y / \partial X_2$, and in particular, that $\partial Y/\partial X_1 &gt; \partial Y / \partial X_2$. So I estimate the model $$Y=\alpha + \beta_1 X_1 + \beta_2 X_2 +u$$and find significant evidence for $\beta_1,\beta_2&gt;0$. How can I then test whether $\beta_1&gt; \beta_2$? I considered running another regression $$Y=\alpha +\gamma(X_1 - X_2) + u$$ And testing whether $\gamma&gt;0$. Is this the best way?  Also, I need to generalize the answer to many variables, i.e. suppose we have $$Y=\alpha + \beta^1 X^1 + \beta ^2 X^2 + \dots + \beta^nX^n + u$$ where for each $j=1,\dots,n$, $X^j=X_1^j+X_2^j$, and I would like to test for each $j$ whether $\partial Y/\partial X_1^j \ne \partial Y / \partial X_2^j$.  By the way, I am primarily working in R. There has been some discussion about getting notified on comments to questions or answers that don't belong to you, and the long-standing wish to be able to keep track of new answers or developments in questions asked by other people.  Previous discussion in regards to subscribing to questions:                  (very similar to this one)      In regards to subscribing to comments:                    The general feeling - at least here on meta - seems to be against broadening the notification rules, out of fears of useless notifications to flood users' inboxes, turning SO into more of a discussion forum than it should be, etc.   Here's a suggestion that would keep the user 100% in control what they get notified about, not generate any notifications by default, and require only minimal modifications to the User Interface.  Each question and each comment area would get an envelope symbol.    it is inactive (=white) by default.  When those envelopes are hovered over, the tooltip says click to subscribe or something.  When clicked on, the envelope lights up in green or some other colour (but different from the actual notification envelope's orange to avoid confusion.) You are then subscribed to any updates to that question or comment thread.    For each subscription, you receive notifications in the "responses" tab, as if you were the owner of the comment or question.     To unsubscribe from a question or comment thread, you open the page, and click the green envelope. It turns back white, and the notifications disappear from your responses tab (if technically possible - otherwise, there just won't be any new ones added.)    No further functionality, like managing subscriptions or anything, is necessary IMO. If the flood of notifications from a subscribed question starts to annoy you, you just open it and unclick the envelope. Done.   It would make it possible to leave a comment to a newbie's question and make sure their answer reaches you even if they don't know they need to address you using @username You could subscribe to those "oh, I'd like to know the answer to that" questions and receive active notification, instead of them getting forgotten in your favourites list You could subscribe to discussions you are interested in the outcome of, but don't want to participate in. It would (in my opinion) sort out a number of woes aired by many users in a good way, without adding annoyances to those who like the system just fine the way it is. It would not require a change in the visible layout. As far as I can see, the space the envelope would need to be is empty everywhere.   I don't know SO under the hood of course, but technically, the system would be doing what it's doing already, notify question owners of updates - just that the circle of internal "owners" would grow. Is it true that $A^A\simeq P(A)$? Let $A$ be an infinite set and $$A^A=\{f | f:A\to A\,\,\, \text{is a function}\}.$$ Is there any natural bijection between $A^A$ and $P(A)$? I'm creating side skirts for a truck and I am duplicating the frames I want. The problem is it's making the duplicated object into another object. How do I put it into the same one? Problem  I have various footnotes, which are hyperlinked from the text to the bottom of their respective pages. When I click on the hyperlinks (in the main text), they always take me to the first page of my document. I also have footnotebackref which puts a hyperlink at the footnote to take the reader back to the main body of text and these links for fine.   Edited MWE  After a lot of deletion, I have managed to reduce the document to the following lines. One might say that I have found the problem, since the removal of \usepackage[hang, flushmargin]{footmisc} stops the problem, but I would still like to have non-indented footnotes.  \documentclass{article}  \usepackage[usenames,dvipsnames]{color} \usepackage[colorlinks=true]{hyperref}  \usepackage[hang, flushmargin]{footmisc}         %Problem line. \usepackage{footnotebackref}  \begin{document} \tableofcontents             \newpage                 \section{Section} This is some text\footnote{This is a footnote.}. \end{document} Why do not we ask negative questions without a contraction on the not after the verb? I have found multiple questions touching on this but not a single one that has a comprehensive answer. The information is all there but in little bits.     etc.   Exactly why is "Do not you sleep?" ungrammatical (except in poetry/with artistic license)? I understand that not can't contract after you: "Do you not sleep?" because not can only contract after a verb, and that it always does after do: "Don't you sleep?" but why does it (why must it) contract if it's placed after the verb in a question?   Don't you sleep? (common form) Do not you sleep? (ungrammatical) Do you not sleep? (less common form) \@seccntformat only for section How do I get label of the selected item from a lightning:select drop down list? I would like to get the value of the label of the selected drop down item.  In this case, the label is the Name of the Contact.  &lt;aura:component&gt;  &lt;aura:attribute name="contacts" type="Contact[]"/&gt;  &lt;lightning:select label="Contact" aura:id="contactList" name="contactList"&gt;     &lt;aura:iteration items="{!v.contacts}" var="c" indexVar="index"&gt;        &lt;option value="{!c.Id}"&gt;{!c.Name}&lt;/option&gt;    &lt;/aura:iteration&gt;  &lt;/lightning:select&gt;  &lt;/aura:component&gt;   I know how to get the value:  component.find("contactList").get("v.value");   But how do I get the label? How to add video to iphone simulator I've seen many variants of this problem online, but not quite the same as this, so I don't believe this is a duplicate.  The famous differential equation problem models a skydiver jumping out of a plane as  $$m \frac{dv}{dt} = -mg-cv$$ where m is the mass of the skydiver, g is acceleration due to gravity (9.81 m/s) and c is the coefficient of air resistance.  Here is my question:     Suppose the skydiver exits the stationary helicopter at an altitude of 2000m and opens the parachute at 500m. For this exercise, the parachute opens instantaneously. Suppose m= 70kgs, the coefficient of air resistance is c = 14 for a free falling body and c = 105 after the parachute opens. When does the skydiver open the parachute? When does the Skydiver hit the ground?   So my attempt at a solution kinda starts like this, but this is where I get stuck:  Let's assume that s(t) is the distance the body falls in time t from its initial point of release, then $v = \frac{ds}{dt}$ and $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$. Thus we have a second order differential equation: $m \frac{d^2s}{dt^2} = -mg-c\frac{ds}{dt}$. By simply manipulation, we get $m \frac{d^2s}{dt^2} + c\frac{ds}{dt} = mg$. I originally started thinking of finding the integrating factor, however, this seems more complicated and won't help produce the equation that I need.  Would anyone be able to help? Much thanks in advanced! Is there a way to kill a process tree using the terminal?  By killing a process tree, I mean killing all descendants of a process along with the process it self.  I tagged this question htop because I am using that utility a lot these days and it would be cool if there was an easy way to kill a process tree inside htop (something better than just selecting all the processes in the subtree). Feynman diagram for attractive forces