Solving $frac{mathrm dy}{mathrm dx}+frac y{x^2}=frac1{x^2}$












1












$begingroup$


Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$



My Attempt



Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.



Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
    $endgroup$
    – John Hughes
    Dec 26 '18 at 13:18






  • 5




    $begingroup$
    @blue_eyed_: This is a Separable Equation.
    $endgroup$
    – Moo
    Dec 26 '18 at 13:20
















1












$begingroup$


Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$



My Attempt



Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.



Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
    $endgroup$
    – John Hughes
    Dec 26 '18 at 13:18






  • 5




    $begingroup$
    @blue_eyed_: This is a Separable Equation.
    $endgroup$
    – Moo
    Dec 26 '18 at 13:20














1












1








1





$begingroup$


Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$



My Attempt



Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.



Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$






ordinary-differential-equations







share|cite|improve this question











$endgroup$




Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$



My Attempt



Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.



Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 15:16









Saad

19.7k92352




19.7k92352










asked Dec 26 '18 at 13:14









blue_eyed_...blue_eyed_...

3,25921649




3,25921649








  • 6




    $begingroup$
    OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
    $endgroup$
    – John Hughes
    Dec 26 '18 at 13:18






  • 5




    $begingroup$
    @blue_eyed_: This is a Separable Equation.
    $endgroup$
    – Moo
    Dec 26 '18 at 13:20














  • 6




    $begingroup$
    OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
    $endgroup$
    – John Hughes
    Dec 26 '18 at 13:18






  • 5




    $begingroup$
    @blue_eyed_: This is a Separable Equation.
    $endgroup$
    – Moo
    Dec 26 '18 at 13:20








6




6




$begingroup$
OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
$endgroup$
– John Hughes
Dec 26 '18 at 13:18




$begingroup$
OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
$endgroup$
– John Hughes
Dec 26 '18 at 13:18




5




5




$begingroup$
@blue_eyed_: This is a Separable Equation.
$endgroup$
– Moo
Dec 26 '18 at 13:20




$begingroup$
@blue_eyed_: This is a Separable Equation.
$endgroup$
– Moo
Dec 26 '18 at 13:20










4 Answers
4






active

oldest

votes


















4












$begingroup$

Hint: Write your equation in the form
$$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Great, now write it in the form
    $$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$



    We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.



    Hence $$frac{partial F}{partial y} = e^{-1/x}$$
    $$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$



    Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.



    Hence our equation is $$dF(x,y) = 0$$



    so $F(x,y) = C$ for some $C in mathbb{R}$. We get



    $$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        This is a linear DE so considering instead



        $$
        x^2y'+y=1
        $$



        $$
        y = y_h + y_p
        $$



        such that



        $$
        x^2y_h'+y_h = 0\
        x^2y_p' + y_p = 1
        $$



        for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives



        $$
        -C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
        $$



        and for the particular $y_p = 1$ then



        $$
        y = C_0 e^{frac 1x}+1
        $$






        share|cite|improve this answer











        $endgroup$













          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',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          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%2f3052931%2fsolving-frac-mathrm-dy-mathrm-dx-frac-yx2-frac1x2%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Hint: Write your equation in the form
          $$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$






          share|cite|improve this answer









          $endgroup$


















            4












            $begingroup$

            Hint: Write your equation in the form
            $$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$






            share|cite|improve this answer









            $endgroup$
















              4












              4








              4





              $begingroup$

              Hint: Write your equation in the form
              $$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$






              share|cite|improve this answer









              $endgroup$



              Hint: Write your equation in the form
              $$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 26 '18 at 13:29









              Dr. Sonnhard GraubnerDr. Sonnhard Graubner

              75.3k42865




              75.3k42865























                  2












                  $begingroup$

                  Great, now write it in the form
                  $$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$



                  We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.



                  Hence $$frac{partial F}{partial y} = e^{-1/x}$$
                  $$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$



                  Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.



                  Hence our equation is $$dF(x,y) = 0$$



                  so $F(x,y) = C$ for some $C in mathbb{R}$. We get



                  $$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    Great, now write it in the form
                    $$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$



                    We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.



                    Hence $$frac{partial F}{partial y} = e^{-1/x}$$
                    $$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$



                    Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.



                    Hence our equation is $$dF(x,y) = 0$$



                    so $F(x,y) = C$ for some $C in mathbb{R}$. We get



                    $$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Great, now write it in the form
                      $$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$



                      We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.



                      Hence $$frac{partial F}{partial y} = e^{-1/x}$$
                      $$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$



                      Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.



                      Hence our equation is $$dF(x,y) = 0$$



                      so $F(x,y) = C$ for some $C in mathbb{R}$. We get



                      $$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$






                      share|cite|improve this answer









                      $endgroup$



                      Great, now write it in the form
                      $$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$



                      We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.



                      Hence $$frac{partial F}{partial y} = e^{-1/x}$$
                      $$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$



                      Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.



                      Hence our equation is $$dF(x,y) = 0$$



                      so $F(x,y) = C$ for some $C in mathbb{R}$. We get



                      $$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 26 '18 at 13:33









                      mechanodroidmechanodroid

                      27.6k62447




                      27.6k62447























                          1












                          $begingroup$

                          I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.






                              share|cite|improve this answer









                              $endgroup$



                              I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 26 '18 at 13:44









                              J.G.J.G.

                              26.3k22541




                              26.3k22541























                                  1












                                  $begingroup$

                                  This is a linear DE so considering instead



                                  $$
                                  x^2y'+y=1
                                  $$



                                  $$
                                  y = y_h + y_p
                                  $$



                                  such that



                                  $$
                                  x^2y_h'+y_h = 0\
                                  x^2y_p' + y_p = 1
                                  $$



                                  for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives



                                  $$
                                  -C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
                                  $$



                                  and for the particular $y_p = 1$ then



                                  $$
                                  y = C_0 e^{frac 1x}+1
                                  $$






                                  share|cite|improve this answer











                                  $endgroup$


















                                    1












                                    $begingroup$

                                    This is a linear DE so considering instead



                                    $$
                                    x^2y'+y=1
                                    $$



                                    $$
                                    y = y_h + y_p
                                    $$



                                    such that



                                    $$
                                    x^2y_h'+y_h = 0\
                                    x^2y_p' + y_p = 1
                                    $$



                                    for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives



                                    $$
                                    -C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
                                    $$



                                    and for the particular $y_p = 1$ then



                                    $$
                                    y = C_0 e^{frac 1x}+1
                                    $$






                                    share|cite|improve this answer











                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      This is a linear DE so considering instead



                                      $$
                                      x^2y'+y=1
                                      $$



                                      $$
                                      y = y_h + y_p
                                      $$



                                      such that



                                      $$
                                      x^2y_h'+y_h = 0\
                                      x^2y_p' + y_p = 1
                                      $$



                                      for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives



                                      $$
                                      -C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
                                      $$



                                      and for the particular $y_p = 1$ then



                                      $$
                                      y = C_0 e^{frac 1x}+1
                                      $$






                                      share|cite|improve this answer











                                      $endgroup$



                                      This is a linear DE so considering instead



                                      $$
                                      x^2y'+y=1
                                      $$



                                      $$
                                      y = y_h + y_p
                                      $$



                                      such that



                                      $$
                                      x^2y_h'+y_h = 0\
                                      x^2y_p' + y_p = 1
                                      $$



                                      for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives



                                      $$
                                      -C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
                                      $$



                                      and for the particular $y_p = 1$ then



                                      $$
                                      y = C_0 e^{frac 1x}+1
                                      $$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 26 '18 at 14:53

























                                      answered Dec 26 '18 at 13:50









                                      CesareoCesareo

                                      8,8293516




                                      8,8293516






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052931%2fsolving-frac-mathrm-dy-mathrm-dx-frac-yx2-frac1x2%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          -

                                          Popular posts from this blog

                                          Bressuire

                                          Image
                                          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                                          Read more

                                          Cabo Verde

                                          Image
                                            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
                                          Read more

                                          Gyllenstierna

                                          Image
                                          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                                          Read more