$$x = y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^2$$ I think it can be converted into a Clairaut. form but not able to do it
I've recently been learning about differential equations, and my teacher has been giving some particularly difficult examples out for those of us who finish early. He gave us the following differential equation: $$x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$$ With the hint "Differentiate with respect to $y$, then let $\frac{dy}{dx}=p$". I've solved it after the lesson and would like someone to check whether or not I've done so correctly.    Following the hint, $$\frac{dx}{dy}=\frac{1}{p}=p+y\frac{dp}{dy}-2p\frac{dp}{dy}$$ Which rearranges to $$p-\frac{1}{p}+y\frac{dp}{dy}=2p\frac{dp}{dy}$$ Multiplying through by $\frac{dy}{dp}$, $$\left(p-\frac{1}{p}\right)\frac{dy}{dp}+y=2p$$ Rearranging into the form $\frac{dy}{dp}+f(p)y=g(p)$, and writing $p-\frac{1}{p}=\frac{p^{2}-1}{p}$ $$\frac{dy}{dp}+\frac{p}{p^{2}-1}y=\frac{2p^{2}}{p^{2}-1}$$ Our integrating factor is $$e^{\int\frac{p}{p^{2}-1}dp}=e^{\frac{1}{2}\ln(p^{2}-1)}=\sqrt{p^{2}-1}$$ So we have $$\sqrt{p^{2}-1}\frac{dy}{dp}+\frac{p}{\sqrt{p^{2}-1}}y=\frac{2p^{2}}{\sqrt{p^{2}-1}}$$ The left hand side is, by design, $\frac{d}{dp}(y\sqrt{p^{2}-1})$, and the right hand side can be integrated as follows: $$\int \frac{2p^{2}}{\sqrt{p^{2}-1}}dp=\int\frac{2(p^{2}-1)+2}{\sqrt{p^{2}-1}}dp=\int 2\sqrt{p^{2}-1}dp+\int \frac{2}{\sqrt{p^{2}-1}} dp$$ These can both be solved using the substitution $p=\cosh(u)$, and gives $$y\sqrt{p^{2}-1}=p\sqrt{p^{2}+1}-\cosh^{-1}(p)+\alpha$$ Where $\alpha$ is an arbitrary constant. Dividing, we finally have $$y=p+\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}$$  Here I was stuck for a long time, until I realised that we also have the equation we began with: $x=py-p^{2}$, which we can solve as a quadratic in $p$ to also get $$p=\frac{y\pm \sqrt{y^{2}-4x}}{2} \implies p^{2}=\frac{y^{2}+y^{2}-4x \pm 2y\sqrt{y^{2}-4x}}{4}$$  Thus, $$y=\frac{y\pm \sqrt{y^{2}-4x}}{2}+\frac{\cosh^{-1}\left(\frac{y\pm \sqrt{y^{2}-4x}}{2}\right)+\alpha}{\sqrt{\frac{y^{2}-2x \pm y\sqrt{y^{2}-4x}}{2}-1}}$$  This is as good as I can do - we have a relation between $y$ and $x$ - am I done? Is this the correct answer? Is there an easier way?     Thank you for your time.    Following the suggestion by @Valentin $$\frac{dy}{dp}=1+\frac{\frac{1}{\sqrt{p^{2}-1}}\sqrt{p^{2}-1}-p\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}=1+p\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}$$ Hence, $$dx=\frac{dy}{p}=dp\left(\frac{1}{p}+\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}\right)$$
Using Apple Script to Manage Sound Output Selection I am brand new here and I have an additional clarification or need additional help for an issue from this thread -   In my preference pane, in addition to Headphones and Digital out, I also have 2 apple monitors and would like the option of selecting one of them as out put (namely the first one in the list).   Basically I need help with adjust his solution to be able to select the third row or first monitor as sound output.  Or a way to accomplish this without knowing how to edit AppleScript for a different setup. Irreducible polynomial in field extension I have this problem      Let $L/K$ be a finite field extension and $|L:K|=m$. Let $f(x)$ be an irreducible polynomial of degree $n$ in $K[x]$ and $\gcd(m,n)=1$. Prove that $f$ is irreducible over $L$.   I'm just a beginner so I'm really confused with the $\gcd(m,n)=1$, I don't know how to use it. So help me with this problem. Thank you Testing equality of coefficients from two different regressions This seems to be a basic issue, but I just realized that I actually don't know how to test equality of coefficients from two different regressions. Can anyone shed some light on this?   More formally, suppose I ran the following two regressions:  $$ y_1 = X_1\beta_1 + \epsilon_1 $$ and  $$ y_2 = X_2\beta_2 + \epsilon_2 $$ where $X_i$ refers to the design matrix of regression $i$, and $\beta_i$ to the vector of coefficients in regression $i$. Note that $X_1$ and $X_2$ are potentially very different, with different dimensions etc. I am interested in for instance whether or not $\hat\beta_{11} \neq \hat\beta_{21}$.   If these came from the same regression, this would be trivial. But since they come from different ones, I am not quite sure how to do it. Does anyone have an idea or can give me some pointers?   My problem in detail: My first intuition was to look at the confidence intervals, and if they overlap, then I would say they are essentially the same. This procedure does not come with the correct size of the test, though (i.e. each individual confidence interval has $\alpha=0.05$, say, but looking at them jointly will not have the same probability). My "second" intuition was to conduct a normal t-test. That is, take   $$ \frac{\beta_{11}-\beta_{21}}{sd(\beta_{11})} $$  where $\beta_{21}$ is taken as the value of my null hypothesis. This does not take into account the estimation uncertainty of $\beta_{21}$, though, and the answer may depend on the order of the regressions (which one I call 1 and 2).   My third idea was to do it as in a standard test for equality of two coefficients from the same regression, that is take  $$ \frac{\beta_{11}-\beta_{21}}{sd(\beta_{11}-\beta_{21})} $$   The complication arises due to the fact that both come from different regressions. Note that   $$ Var(\beta_{11}-\beta_{21}) = Var(\beta_{11}) + Var(\beta_{21}) -2 Cov(\beta_{11},\beta_{21}) $$ but since they are from different regressions, how would I get $Cov(\beta_{11},\beta_{21})$?  This led me to ask this question here. This must be a standard procedure / standard test, but I cound not find anything that was sufficiently similar to this problem. So, if anyone can point me to the correct procedure, I would be very grateful! Question about arithmetic–geometric mean We have two sequences: $$a_{n+1}=\sqrt{a_nb_n}$$ $$b_{n+1}=\frac{a_n+b_n}{2}$$ I need to prove that those are making Cantor's Lemma.(At the end I shold get that: $\lim_{n\to \infty}a_n=\lim_{n\to \infty}b_n$ by Cantor's Lemma)  Any ideas how?  Thank you. How do you determine using stat() whether a file is a symbolic link? I'm looking for the noun form of "person with intermediate skill". For example, in the context of a particular activity, "person with no skill" might be designated a novice, and "person with much skill" might be designated an expert. However I know of no such word in between these two extremes.   My only thought is amateur, which has a distinct meaning from describing level of skill, although skill level can often be inferred. I'm looking for a better solution.  EDIT: due to the attention this question has received, I'll try to offer some clarification in order to reduce repeat questions.   "Journeyman" would certainly not work. I am a software developer by profession. However the word I'm looking for should hopefully be profession-agnostic. I am trying to designate the skill level of the person that some learning material might be appropriate for. The word that I'm looking for follows the preposition "for".   For Beginners  For [word for intermediate-skilled person] For Experts  Sorry, but even after all the great answers, I'm still learning towards "amateur" even though it is technically incorrect. Also, the word should not be so esoteric that only people with English degrees know what it means. Keep the ideas coming! What should I do if I want to be able to run a given program regardless of my current directory? Should I create a symbolic link to the program in the /bin folder? Why don’t photons interact with the Higgs field and hence remain massless? I am looking for tutorials, articles or books containing theory of  in context of mathematical physics. Please suggest. Probability that median falls within a range? If I draw 5 samples from a distribution that is cont., what is the probability that the median of the population is between the xmin and xmax of the draw samples?  What is the formula to apply here? Could someone please check if I'm saying something that make sense? I am trying to check the following:     If $f$ is a measurable function, then $\text{supp}(f)$ is measurable set?   Is it simply because, if we consider $E_1 = (0,\infty)$ and $E_2 = (-\infty,0)$, since they are open, they are measurable. So their union $\mathbb{R}-\{0\}$ is measurable as well. Since $f$ is measurable, then the preimage $f^{-1}(\mathbb{R}-\{0\})$ is measurable. But the preimage is just the support? checking if a number is a prime I was reading Wikipedia, and it was given that "all primes are of the form 6k ± 1"  (other than 2 and 3), where k = 1,2,3,4,...  Is this statement correct? If yes, can we use this to test if a given number is a prime number? For instance, we can say that 41 is a prime number, since there exists an integer K (K >= 1), such that 6k - 1 = 41 ==> k = 7.  I am confused why this test cannot be used to test if a number is prime?  Thanks, Sekhar An infinite subset of a countable set is countable In my book, it proves that an infinite subset of a coutnable set is countable. But not all the details are filled in, and I've tried to fill in all the details below. Could someone tell me if what I wrote below is valid?  Let $S$ be an infinite subset of a countable set $T$. Since $T$ is countable, there exists a bijection $f: \mathbb{N} \rightarrow T$. And $S \subseteq T = \{ f(n) \ | \ n \in \mathbb{N} \}$. Let $n_{1}$ be the smallest positive integer such that $f(n_{1}) \in S$. And continue where $n_{k}$ is the smallest positive integer greater than $n_{k-1}$ such that $f(n_{k}) \in S$. And because $S$ is infinite, we continue forever. Now consider the function $\beta : \mathbb{N} \rightarrow S$ which sends $k \rightarrow f(n_{k})$.   So in order for $S$ to be countable, $\beta$ would have to be a bijection. $\beta$ is injective since if $f(n_{k}) = f(n_{j})$, then $n_{k} = n_{j}$ because $f$ is injective. We also can conclude $k = j$ because the various $n_{i}$ chosen were strictly increasing.   Edit: $\beta$ also needs to be surjective. So for every $r \in S$ there exists a $q \in \mathbb{N}$ such that $\beta(q) = f(n_{q}) = r$. We know that $f^{-1}(r) \in \{n_{1}, n_{2}, ..., n_{k} ... \}$, so we can let $f^{-1}(r) = n_{d}$. Thus $\beta(d) = f(n_{d}) = r$. Easy way to use variables of enum types as string in C? Here's what I am trying to do:    typedef enum { ONE, TWO, THREE } Numbers;   I am trying to write a function that would do a switch case similar to the following:   char num_str[10]; int process_numbers_str(Numbers num) {   switch(num) {     case ONE:     case TWO:     case THREE:     {       strcpy(num_str, num); //some way to get the symbolic constant name in here?     } break;     default:       return 0; //no match   return 1; }   Instead of defining at every case, is there a way to set it using the enum variable like I am trying to do above? Facebook changed how their API serves profile pictures that are requested with a non-application scope UID. Now because before Graph API 2 all applications shared the original UID, if users like me has joined Stack Exchange websites before  2, rolling out their profile pictures are not shown properly. Facebook suggests to use an access token to query those profile pictures, either an application access token or a user access token. A user access token is not a reliable option at all, because it would only work for users who joined website with Facebook and are currently logged in. Application access tokens, by the way, can be used by creating sort of a proxy that would retrieve a profile picture from Facebook and display it. Yet still it's not perfect. I think the best solution is to communicate with Facebook and ask them to offer an opportunity to convert old Graph API 1 UID's to application scope user IDs. Is there a convergent, alternating series that fails the AST? The alternating series test (AST) says, briefly, that if   $a_k&gt;0$ $a_k \geq a_{k+1}$, and $a_k \to 0$ as $k \to \infty$   then $\sum_k (-1)^k a_k$ converges.  This seems to be a one-way test (that is, if an alternating series fails the test, we don't know that it diverges).   document says so explicitly. It then gives an example of an alternating series which fails the AST. That doesn't prove the series is divergent (but it turns out to be divergent, anyway, by $n$th term test).  Is there a convergent, alternating series that fails the AST?  Of course, if the series fails condition 3, then it also fails the $n$th term test, and must diverge. And if it fails the first condition, then it's not strictly alternating, anyway.  So it must be a series that is not getting smaller (condition 2), but still converges. How to force Wine into acting like 32-bit Windows on 64-bit Ubuntu? How to Find Nearby iOS Devices (GPS locations) using same app Enter from password field doesn't submit SE login form On  page for a site where I don't yet have an account, if I select the "Log in using Stack Exchange" box, when I hit enter from the password form, nothing happens.  Minor, but... Why was C-3PO's memory wiped, but not R2-D2's? In Revenge of the Sith after Luke and Leia are born, they are sent to separate parts of the galaxy. The droids are given to Captain Antilles, with the instructions to clean them up, and have C-3PO's memory wiped. Why just C-3PO's and not also R2-D2's?   Initially I thought it was due to R2-D2's inability to communicate verbally, but he not only can communicate via text message (as seen in The Empire Strikes Back I believe), but we see other characters understanding his blips and bleeps as well.