Differential equation with square root, how to use $\sgn()$?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












I know I should do some substitution and then somehow use $textsign(cdot)$ function to solve this, but I have no idea. My problem is the $textsign(cdot)$ function. I found this answer Differential equation, a square root and substitution, but I don't know how he extracted $textsign(cdot)$ under the square root.



$$x y' - y = sqrtx^2+y^2$$




differential-equations




share|cite|improve this question


























    up vote
    0
    down vote

    favorite












    I know I should do some substitution and then somehow use $textsign(cdot)$ function to solve this, but I have no idea. My problem is the $textsign(cdot)$ function. I found this answer Differential equation, a square root and substitution, but I don't know how he extracted $textsign(cdot)$ under the square root.



    $$x y' - y = sqrtx^2+y^2$$




    differential-equations




    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I know I should do some substitution and then somehow use $textsign(cdot)$ function to solve this, but I have no idea. My problem is the $textsign(cdot)$ function. I found this answer Differential equation, a square root and substitution, but I don't know how he extracted $textsign(cdot)$ under the square root.



      $$x y' - y = sqrtx^2+y^2$$




      differential-equations




      share|cite|improve this question














      I know I should do some substitution and then somehow use $textsign(cdot)$ function to solve this, but I have no idea. My problem is the $textsign(cdot)$ function. I found this answer Differential equation, a square root and substitution, but I don't know how he extracted $textsign(cdot)$ under the square root.



      $$x y' - y = sqrtx^2+y^2$$





      differential-equations





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 29 at 9:15









      Mattos

      2,66721121




      2,66721121










      asked Aug 29 at 9:08









      Igor

      1




      1




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          0
          down vote













          Following the similar path the link takes, let $w=fracyx$. Then
          beginalign
          wx &= y \
          implies xw'+w &= y'
          endalign
          hence your ODE becomes
          beginalign
          x(w+xw') - wx &= sqrtx^2left(1+w^2right) \
          implies x^2 w' &= operatornamesgn(x)sqrt1+w^2
          endalign
          Where the Signum funciton is defined as
          $$operatornamesgn(x) := begincases
          -1 & textif x < 0, \
          0 & textif x = 0, \
          1 & textif x > 0. endcases$$






          share|cite|improve this answer




















          • Thanxs, but It looked so unnatural for me that only extracting $\x^2$ would be enough. Can you tell me, what rule allows me to subtract it out as $\sign(x)$.
            – Igor
            Aug 29 at 9:33











          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2898133%2fdifferential-equation-with-square-root-how-to-use-sgn%23new-answer', 'question_page');

          );

          Post as a guest

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          0
          down vote













          Following the similar path the link takes, let $w=fracyx$. Then
          beginalign
          wx &= y \
          implies xw'+w &= y'
          endalign
          hence your ODE becomes
          beginalign
          x(w+xw') - wx &= sqrtx^2left(1+w^2right) \
          implies x^2 w' &= operatornamesgn(x)sqrt1+w^2
          endalign
          Where the Signum funciton is defined as
          $$operatornamesgn(x) := begincases
          -1 & textif x < 0, \
          0 & textif x = 0, \
          1 & textif x > 0. endcases$$






          share|cite|improve this answer




















          • Thanxs, but It looked so unnatural for me that only extracting $\x^2$ would be enough. Can you tell me, what rule allows me to subtract it out as $\sign(x)$.
            – Igor
            Aug 29 at 9:33















          up vote
          0
          down vote













          Following the similar path the link takes, let $w=fracyx$. Then
          beginalign
          wx &= y \
          implies xw'+w &= y'
          endalign
          hence your ODE becomes
          beginalign
          x(w+xw') - wx &= sqrtx^2left(1+w^2right) \
          implies x^2 w' &= operatornamesgn(x)sqrt1+w^2
          endalign
          Where the Signum funciton is defined as
          $$operatornamesgn(x) := begincases
          -1 & textif x < 0, \
          0 & textif x = 0, \
          1 & textif x > 0. endcases$$






          share|cite|improve this answer




















          • Thanxs, but It looked so unnatural for me that only extracting $\x^2$ would be enough. Can you tell me, what rule allows me to subtract it out as $\sign(x)$.
            – Igor
            Aug 29 at 9:33













          up vote
          0
          down vote










          up vote
          0
          down vote









          Following the similar path the link takes, let $w=fracyx$. Then
          beginalign
          wx &= y \
          implies xw'+w &= y'
          endalign
          hence your ODE becomes
          beginalign
          x(w+xw') - wx &= sqrtx^2left(1+w^2right) \
          implies x^2 w' &= operatornamesgn(x)sqrt1+w^2
          endalign
          Where the Signum funciton is defined as
          $$operatornamesgn(x) := begincases
          -1 & textif x < 0, \
          0 & textif x = 0, \
          1 & textif x > 0. endcases$$






          share|cite|improve this answer












          Following the similar path the link takes, let $w=fracyx$. Then
          beginalign
          wx &= y \
          implies xw'+w &= y'
          endalign
          hence your ODE becomes
          beginalign
          x(w+xw') - wx &= sqrtx^2left(1+w^2right) \
          implies x^2 w' &= operatornamesgn(x)sqrt1+w^2
          endalign
          Where the Signum funciton is defined as
          $$operatornamesgn(x) := begincases
          -1 & textif x < 0, \
          0 & textif x = 0, \
          1 & textif x > 0. endcases$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 29 at 9:24









          Kevin

          5,138722




          5,138722











          • Thanxs, but It looked so unnatural for me that only extracting $\x^2$ would be enough. Can you tell me, what rule allows me to subtract it out as $\sign(x)$.
            – Igor
            Aug 29 at 9:33

















          • Thanxs, but It looked so unnatural for me that only extracting $\x^2$ would be enough. Can you tell me, what rule allows me to subtract it out as $\sign(x)$.
            – Igor
            Aug 29 at 9:33
















          Thanxs, but It looked so unnatural for me that only extracting $\x^2$ would be enough. Can you tell me, what rule allows me to subtract it out as $\sign(x)$.
          – Igor
          Aug 29 at 9:33





          Thanxs, but It looked so unnatural for me that only extracting $\x^2$ would be enough. Can you tell me, what rule allows me to subtract it out as $\sign(x)$.
          – Igor
          Aug 29 at 9:33


















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2898133%2fdifferential-equation-with-square-root-how-to-use-sgn%23new-answer', 'question_page');

          );

          Post as a guest
































































          -

          這個網誌中的熱門文章

          How to combine Bézier curves to a surface?

          圖片
          Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
          閱讀完整內容

          Mutual Information Always Non-negative

          圖片
          Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
          閱讀完整內容

          Why am i infinitely getting the same tweet with the Twitter Search API?

          圖片
          up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
          閱讀完整內容