General solution of $x^2 frac{d^2y}{dx^2}+xfrac{dy}{dx}-y=xln x$












0












$begingroup$


Find the general solution of $x^2 frac{d^2y}{dx^2}+xfrac{dy}{dx}-y=xln x$ .



Attempt :



With $x=e^z$, then the equation becomes :
$$[D(D-1)+D-1]y=ze^z$$



This complementary function comes from
$$m(m-1)+m-1=m^2-1=0$$



then $m=1$ and $m=-1$
which gives
$$y_c=c_1e^z+c_2e^{-z}$$
For $g(D)ze^z=0, g(D)$ must be ${(D-1)^2}$. The characteristic is the then
$$(m^prime-1)=0$$
then get $m^prime=1,1$
. Therefore the particular solution is of the form :
$$y_p=c_3ze^z+c_4z^2e^z$$



Substituting it back in the differential equation



$$[D(D-1)+D-1](c_3ze^z+c_4z^2e^z)=ze^z$$



I am a bit confused how to find the value of $c_3$ and $c_4$. Any help ?






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Simplify that last expression, and see what $c_3$ and $c_4$ have to be to make the left side equal to the right side.
    $endgroup$
    – Robert Israel
    Dec 17 '18 at 1:59










  • $begingroup$
    Have you tried taking the derivatives of $y_p$ and plugging the into the ODE?
    $endgroup$
    – Dylan
    Dec 17 '18 at 3:04










  • $begingroup$
    yes, but still not know clearly.
    $endgroup$
    – Verse
    Dec 17 '18 at 3:23
















0












$begingroup$


Find the general solution of $x^2 frac{d^2y}{dx^2}+xfrac{dy}{dx}-y=xln x$ .



Attempt :



With $x=e^z$, then the equation becomes :
$$[D(D-1)+D-1]y=ze^z$$



This complementary function comes from
$$m(m-1)+m-1=m^2-1=0$$



then $m=1$ and $m=-1$
which gives
$$y_c=c_1e^z+c_2e^{-z}$$
For $g(D)ze^z=0, g(D)$ must be ${(D-1)^2}$. The characteristic is the then
$$(m^prime-1)=0$$
then get $m^prime=1,1$
. Therefore the particular solution is of the form :
$$y_p=c_3ze^z+c_4z^2e^z$$



Substituting it back in the differential equation



$$[D(D-1)+D-1](c_3ze^z+c_4z^2e^z)=ze^z$$



I am a bit confused how to find the value of $c_3$ and $c_4$. Any help ?






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Simplify that last expression, and see what $c_3$ and $c_4$ have to be to make the left side equal to the right side.
    $endgroup$
    – Robert Israel
    Dec 17 '18 at 1:59










  • $begingroup$
    Have you tried taking the derivatives of $y_p$ and plugging the into the ODE?
    $endgroup$
    – Dylan
    Dec 17 '18 at 3:04










  • $begingroup$
    yes, but still not know clearly.
    $endgroup$
    – Verse
    Dec 17 '18 at 3:23














0












0








0





$begingroup$


Find the general solution of $x^2 frac{d^2y}{dx^2}+xfrac{dy}{dx}-y=xln x$ .



Attempt :



With $x=e^z$, then the equation becomes :
$$[D(D-1)+D-1]y=ze^z$$



This complementary function comes from
$$m(m-1)+m-1=m^2-1=0$$



then $m=1$ and $m=-1$
which gives
$$y_c=c_1e^z+c_2e^{-z}$$
For $g(D)ze^z=0, g(D)$ must be ${(D-1)^2}$. The characteristic is the then
$$(m^prime-1)=0$$
then get $m^prime=1,1$
. Therefore the particular solution is of the form :
$$y_p=c_3ze^z+c_4z^2e^z$$



Substituting it back in the differential equation



$$[D(D-1)+D-1](c_3ze^z+c_4z^2e^z)=ze^z$$



I am a bit confused how to find the value of $c_3$ and $c_4$. Any help ?






ordinary-differential-equations







share|cite|improve this question











$endgroup$




Find the general solution of $x^2 frac{d^2y}{dx^2}+xfrac{dy}{dx}-y=xln x$ .



Attempt :



With $x=e^z$, then the equation becomes :
$$[D(D-1)+D-1]y=ze^z$$



This complementary function comes from
$$m(m-1)+m-1=m^2-1=0$$



then $m=1$ and $m=-1$
which gives
$$y_c=c_1e^z+c_2e^{-z}$$
For $g(D)ze^z=0, g(D)$ must be ${(D-1)^2}$. The characteristic is the then
$$(m^prime-1)=0$$
then get $m^prime=1,1$
. Therefore the particular solution is of the form :
$$y_p=c_3ze^z+c_4z^2e^z$$



Substituting it back in the differential equation



$$[D(D-1)+D-1](c_3ze^z+c_4z^2e^z)=ze^z$$



I am a bit confused how to find the value of $c_3$ and $c_4$. Any help ?






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 17 '18 at 3:53









Tianlalu

3,09421138




3,09421138










asked Dec 17 '18 at 1:44









VerseVerse

605




605












  • $begingroup$
    Simplify that last expression, and see what $c_3$ and $c_4$ have to be to make the left side equal to the right side.
    $endgroup$
    – Robert Israel
    Dec 17 '18 at 1:59










  • $begingroup$
    Have you tried taking the derivatives of $y_p$ and plugging the into the ODE?
    $endgroup$
    – Dylan
    Dec 17 '18 at 3:04










  • $begingroup$
    yes, but still not know clearly.
    $endgroup$
    – Verse
    Dec 17 '18 at 3:23


















  • $begingroup$
    Simplify that last expression, and see what $c_3$ and $c_4$ have to be to make the left side equal to the right side.
    $endgroup$
    – Robert Israel
    Dec 17 '18 at 1:59










  • $begingroup$
    Have you tried taking the derivatives of $y_p$ and plugging the into the ODE?
    $endgroup$
    – Dylan
    Dec 17 '18 at 3:04










  • $begingroup$
    yes, but still not know clearly.
    $endgroup$
    – Verse
    Dec 17 '18 at 3:23
















$begingroup$
Simplify that last expression, and see what $c_3$ and $c_4$ have to be to make the left side equal to the right side.
$endgroup$
– Robert Israel
Dec 17 '18 at 1:59




$begingroup$
Simplify that last expression, and see what $c_3$ and $c_4$ have to be to make the left side equal to the right side.
$endgroup$
– Robert Israel
Dec 17 '18 at 1:59












$begingroup$
Have you tried taking the derivatives of $y_p$ and plugging the into the ODE?
$endgroup$
– Dylan
Dec 17 '18 at 3:04




$begingroup$
Have you tried taking the derivatives of $y_p$ and plugging the into the ODE?
$endgroup$
– Dylan
Dec 17 '18 at 3:04












$begingroup$
yes, but still not know clearly.
$endgroup$
– Verse
Dec 17 '18 at 3:23




$begingroup$
yes, but still not know clearly.
$endgroup$
– Verse
Dec 17 '18 at 3:23










2 Answers
2






active

oldest

votes


















1












$begingroup$

Reading the post, I had the feeling that making $x=e^z$ was complicating something.



I think that I would have started using $y=frac z x$ which makes the equation to be
$$x z''-z'=xlog(x)$$ Now, reduction of order $p=z'$ gives
$$x p' -p = x log(x)$$ The homogeneous equation $x p' -p=0$ gives $p=C x$ and the variation of parameters leads to
$$x^2 C'=x log(x)implies C'=frac {log(x)}ximplies C=frac{1}{2}log ^2(x)+c_1$$ So, we end with
$$z'=p=frac{1}{2}xlog ^2(x)+c_1 x$$ which does not make too much problems using integration by parts
$$z=left(frac{c_1}{2}+frac 18right)x^2+frac{1}{4} x^2 log ^2(x)-frac{1}{4} x^2 log (x)+c_2$$ In other words
$$y=frac z x=c_1 x+frac{1}{4} x log ^2(x)-frac{1}{4} xlog (x)+frac{c_2} x$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Let us make the transformation $x=e^t$. We get
    $$y'-y=te^t.$$
    Particular solution we find in form
    $$y_p=t(At+B)e^t$$
    We find
    $$A=frac14,quad B=-frac14.$$
    Then
    $$y=C_1e^t+C_2e^{-t}+frac{(t^2-t)e^t}{4}.$$
    After substitution $;t=ln x;$ final solution is
    $$y=C_1x+frac{C_2}{x}+frac{x(ln^2x-ln x)}{4}.$$






    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%2f3043450%2fgeneral-solution-of-x2-fracd2ydx2x-fracdydx-y-x-ln-x%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Reading the post, I had the feeling that making $x=e^z$ was complicating something.



      I think that I would have started using $y=frac z x$ which makes the equation to be
      $$x z''-z'=xlog(x)$$ Now, reduction of order $p=z'$ gives
      $$x p' -p = x log(x)$$ The homogeneous equation $x p' -p=0$ gives $p=C x$ and the variation of parameters leads to
      $$x^2 C'=x log(x)implies C'=frac {log(x)}ximplies C=frac{1}{2}log ^2(x)+c_1$$ So, we end with
      $$z'=p=frac{1}{2}xlog ^2(x)+c_1 x$$ which does not make too much problems using integration by parts
      $$z=left(frac{c_1}{2}+frac 18right)x^2+frac{1}{4} x^2 log ^2(x)-frac{1}{4} x^2 log (x)+c_2$$ In other words
      $$y=frac z x=c_1 x+frac{1}{4} x log ^2(x)-frac{1}{4} xlog (x)+frac{c_2} x$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Reading the post, I had the feeling that making $x=e^z$ was complicating something.



        I think that I would have started using $y=frac z x$ which makes the equation to be
        $$x z''-z'=xlog(x)$$ Now, reduction of order $p=z'$ gives
        $$x p' -p = x log(x)$$ The homogeneous equation $x p' -p=0$ gives $p=C x$ and the variation of parameters leads to
        $$x^2 C'=x log(x)implies C'=frac {log(x)}ximplies C=frac{1}{2}log ^2(x)+c_1$$ So, we end with
        $$z'=p=frac{1}{2}xlog ^2(x)+c_1 x$$ which does not make too much problems using integration by parts
        $$z=left(frac{c_1}{2}+frac 18right)x^2+frac{1}{4} x^2 log ^2(x)-frac{1}{4} x^2 log (x)+c_2$$ In other words
        $$y=frac z x=c_1 x+frac{1}{4} x log ^2(x)-frac{1}{4} xlog (x)+frac{c_2} x$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Reading the post, I had the feeling that making $x=e^z$ was complicating something.



          I think that I would have started using $y=frac z x$ which makes the equation to be
          $$x z''-z'=xlog(x)$$ Now, reduction of order $p=z'$ gives
          $$x p' -p = x log(x)$$ The homogeneous equation $x p' -p=0$ gives $p=C x$ and the variation of parameters leads to
          $$x^2 C'=x log(x)implies C'=frac {log(x)}ximplies C=frac{1}{2}log ^2(x)+c_1$$ So, we end with
          $$z'=p=frac{1}{2}xlog ^2(x)+c_1 x$$ which does not make too much problems using integration by parts
          $$z=left(frac{c_1}{2}+frac 18right)x^2+frac{1}{4} x^2 log ^2(x)-frac{1}{4} x^2 log (x)+c_2$$ In other words
          $$y=frac z x=c_1 x+frac{1}{4} x log ^2(x)-frac{1}{4} xlog (x)+frac{c_2} x$$






          share|cite|improve this answer









          $endgroup$



          Reading the post, I had the feeling that making $x=e^z$ was complicating something.



          I think that I would have started using $y=frac z x$ which makes the equation to be
          $$x z''-z'=xlog(x)$$ Now, reduction of order $p=z'$ gives
          $$x p' -p = x log(x)$$ The homogeneous equation $x p' -p=0$ gives $p=C x$ and the variation of parameters leads to
          $$x^2 C'=x log(x)implies C'=frac {log(x)}ximplies C=frac{1}{2}log ^2(x)+c_1$$ So, we end with
          $$z'=p=frac{1}{2}xlog ^2(x)+c_1 x$$ which does not make too much problems using integration by parts
          $$z=left(frac{c_1}{2}+frac 18right)x^2+frac{1}{4} x^2 log ^2(x)-frac{1}{4} x^2 log (x)+c_2$$ In other words
          $$y=frac z x=c_1 x+frac{1}{4} x log ^2(x)-frac{1}{4} xlog (x)+frac{c_2} x$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 5:12









          Claude LeiboviciClaude Leibovici

          124k1157135




          124k1157135























              1












              $begingroup$

              Let us make the transformation $x=e^t$. We get
              $$y'-y=te^t.$$
              Particular solution we find in form
              $$y_p=t(At+B)e^t$$
              We find
              $$A=frac14,quad B=-frac14.$$
              Then
              $$y=C_1e^t+C_2e^{-t}+frac{(t^2-t)e^t}{4}.$$
              After substitution $;t=ln x;$ final solution is
              $$y=C_1x+frac{C_2}{x}+frac{x(ln^2x-ln x)}{4}.$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let us make the transformation $x=e^t$. We get
                $$y'-y=te^t.$$
                Particular solution we find in form
                $$y_p=t(At+B)e^t$$
                We find
                $$A=frac14,quad B=-frac14.$$
                Then
                $$y=C_1e^t+C_2e^{-t}+frac{(t^2-t)e^t}{4}.$$
                After substitution $;t=ln x;$ final solution is
                $$y=C_1x+frac{C_2}{x}+frac{x(ln^2x-ln x)}{4}.$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let us make the transformation $x=e^t$. We get
                  $$y'-y=te^t.$$
                  Particular solution we find in form
                  $$y_p=t(At+B)e^t$$
                  We find
                  $$A=frac14,quad B=-frac14.$$
                  Then
                  $$y=C_1e^t+C_2e^{-t}+frac{(t^2-t)e^t}{4}.$$
                  After substitution $;t=ln x;$ final solution is
                  $$y=C_1x+frac{C_2}{x}+frac{x(ln^2x-ln x)}{4}.$$






                  share|cite|improve this answer









                  $endgroup$



                  Let us make the transformation $x=e^t$. We get
                  $$y'-y=te^t.$$
                  Particular solution we find in form
                  $$y_p=t(At+B)e^t$$
                  We find
                  $$A=frac14,quad B=-frac14.$$
                  Then
                  $$y=C_1e^t+C_2e^{-t}+frac{(t^2-t)e^t}{4}.$$
                  After substitution $;t=ln x;$ final solution is
                  $$y=C_1x+frac{C_2}{x}+frac{x(ln^2x-ln x)}{4}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 7:53









                  Aleksas DomarkasAleksas Domarkas

                  1,52816




                  1,52816






























                      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%2f3043450%2fgeneral-solution-of-x2-fracd2ydx2x-fracdydx-y-x-ln-x%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

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                      Read more

                      Puebla de Zaragoza

                      Image
                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                      Read more

                      Musa

                      Image
                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                      Read more