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







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  • $begingroup$
    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






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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






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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









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    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









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      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






























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