Solving 2nd order differential

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3
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I'm trying to solve:




$ fracd^2dx^2 - left(1/xfracdydxright) = 0$




I've been following a tutorial and they say the answer is:




$y = frac12Cx^2 + D$




They've used:




$fracdydx = Cx$




How did they get that result??






calculus integration differential-equations







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  • Could you please type the math into LaTex for easier reading. It improves the chances of people contributing to your question.
    – Gaby Boy Analysis
    58 mins ago














up vote
3
down vote

favorite












I'm trying to solve:




$ fracd^2dx^2 - left(1/xfracdydxright) = 0$




I've been following a tutorial and they say the answer is:




$y = frac12Cx^2 + D$




They've used:




$fracdydx = Cx$




How did they get that result??






calculus integration differential-equations







share|cite|improve this question























  • Could you please type the math into LaTex for easier reading. It improves the chances of people contributing to your question.
    – Gaby Boy Analysis
    58 mins ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











I'm trying to solve:




$ fracd^2dx^2 - left(1/xfracdydxright) = 0$




I've been following a tutorial and they say the answer is:




$y = frac12Cx^2 + D$




They've used:




$fracdydx = Cx$




How did they get that result??






calculus integration differential-equations







share|cite|improve this question















I'm trying to solve:




$ fracd^2dx^2 - left(1/xfracdydxright) = 0$




I've been following a tutorial and they say the answer is:




$y = frac12Cx^2 + D$




They've used:




$fracdydx = Cx$




How did they get that result??






calculus integration differential-equations




calculus integration differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 56 mins ago









José Carlos Santos

126k17102189




126k17102189










asked 1 hour ago









Hews

685




685











  • Could you please type the math into LaTex for easier reading. It improves the chances of people contributing to your question.
    – Gaby Boy Analysis
    58 mins ago
















  • Could you please type the math into LaTex for easier reading. It improves the chances of people contributing to your question.
    – Gaby Boy Analysis
    58 mins ago















Could you please type the math into LaTex for easier reading. It improves the chances of people contributing to your question.
– Gaby Boy Analysis
58 mins ago




Could you please type the math into LaTex for easier reading. It improves the chances of people contributing to your question.
– Gaby Boy Analysis
58 mins ago










2 Answers
2






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up vote
5
down vote



accepted










I think that the equation read as follows:



$fracd^2 ydx^2 - left(1/xfracdydxright) = 0$. Put $z: =y'$ then we have



$z'=frac1zz$. The last equation has the general solution $z(x)=Cx$. Hence $y'(x)=Cx$, thus $y(x) = frac12Cx^2 + D$.






share|cite|improve this answer





























    up vote
    1
    down vote













    Let us consider your differential equation:
    $$y''(x)-frac1xcdot y'(x)=0.$$
    Add $frac1xcdot y'(x)$ on both sides:
    $$y''(x)=frac1xcdot y'(x).$$
    Divide by $y'(x)$ on both sides:
    $$fracy''(x)y'(x)=frac1x.$$
    Integrate with respect to $x$ on both sides:
    $$int fracy''(x) dxy'(x)=int fracdxx.$$
    It follows that
    $$lnleft(y'(x)right)=C+ln(x).$$
    An expression for $y'(x)$ is given by
    $$y'(x)=e^C+ln(x)=e^Ccdot e^ln(x)=e^Ccdot x.$$
    Redefine $e^C$ as $C$, since it is an arbitrary constant:
    $$y'(x)=Ccdot x.$$
    Integrate with respect to $x$ on both sides:
    $$int y'(x) dx=int Ccdot x dx.$$
    An expression for $y(x)$ is given by
    $$y(x)=frac12cdot Ccdot x^2 + D.$$






    share|cite|improve this answer




















      Your Answer




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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

      votes









      active

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      active

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      up vote
      5
      down vote



      accepted










      I think that the equation read as follows:



      $fracd^2 ydx^2 - left(1/xfracdydxright) = 0$. Put $z: =y'$ then we have



      $z'=frac1zz$. The last equation has the general solution $z(x)=Cx$. Hence $y'(x)=Cx$, thus $y(x) = frac12Cx^2 + D$.






      share|cite|improve this answer


























        up vote
        5
        down vote



        accepted










        I think that the equation read as follows:



        $fracd^2 ydx^2 - left(1/xfracdydxright) = 0$. Put $z: =y'$ then we have



        $z'=frac1zz$. The last equation has the general solution $z(x)=Cx$. Hence $y'(x)=Cx$, thus $y(x) = frac12Cx^2 + D$.






        share|cite|improve this answer
























          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          I think that the equation read as follows:



          $fracd^2 ydx^2 - left(1/xfracdydxright) = 0$. Put $z: =y'$ then we have



          $z'=frac1zz$. The last equation has the general solution $z(x)=Cx$. Hence $y'(x)=Cx$, thus $y(x) = frac12Cx^2 + D$.






          share|cite|improve this answer














          I think that the equation read as follows:



          $fracd^2 ydx^2 - left(1/xfracdydxright) = 0$. Put $z: =y'$ then we have



          $z'=frac1zz$. The last equation has the general solution $z(x)=Cx$. Hence $y'(x)=Cx$, thus $y(x) = frac12Cx^2 + D$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 40 mins ago

























          answered 50 mins ago









          Fred

          39.4k1238




          39.4k1238




















              up vote
              1
              down vote













              Let us consider your differential equation:
              $$y''(x)-frac1xcdot y'(x)=0.$$
              Add $frac1xcdot y'(x)$ on both sides:
              $$y''(x)=frac1xcdot y'(x).$$
              Divide by $y'(x)$ on both sides:
              $$fracy''(x)y'(x)=frac1x.$$
              Integrate with respect to $x$ on both sides:
              $$int fracy''(x) dxy'(x)=int fracdxx.$$
              It follows that
              $$lnleft(y'(x)right)=C+ln(x).$$
              An expression for $y'(x)$ is given by
              $$y'(x)=e^C+ln(x)=e^Ccdot e^ln(x)=e^Ccdot x.$$
              Redefine $e^C$ as $C$, since it is an arbitrary constant:
              $$y'(x)=Ccdot x.$$
              Integrate with respect to $x$ on both sides:
              $$int y'(x) dx=int Ccdot x dx.$$
              An expression for $y(x)$ is given by
              $$y(x)=frac12cdot Ccdot x^2 + D.$$






              share|cite|improve this answer
























                up vote
                1
                down vote













                Let us consider your differential equation:
                $$y''(x)-frac1xcdot y'(x)=0.$$
                Add $frac1xcdot y'(x)$ on both sides:
                $$y''(x)=frac1xcdot y'(x).$$
                Divide by $y'(x)$ on both sides:
                $$fracy''(x)y'(x)=frac1x.$$
                Integrate with respect to $x$ on both sides:
                $$int fracy''(x) dxy'(x)=int fracdxx.$$
                It follows that
                $$lnleft(y'(x)right)=C+ln(x).$$
                An expression for $y'(x)$ is given by
                $$y'(x)=e^C+ln(x)=e^Ccdot e^ln(x)=e^Ccdot x.$$
                Redefine $e^C$ as $C$, since it is an arbitrary constant:
                $$y'(x)=Ccdot x.$$
                Integrate with respect to $x$ on both sides:
                $$int y'(x) dx=int Ccdot x dx.$$
                An expression for $y(x)$ is given by
                $$y(x)=frac12cdot Ccdot x^2 + D.$$






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Let us consider your differential equation:
                  $$y''(x)-frac1xcdot y'(x)=0.$$
                  Add $frac1xcdot y'(x)$ on both sides:
                  $$y''(x)=frac1xcdot y'(x).$$
                  Divide by $y'(x)$ on both sides:
                  $$fracy''(x)y'(x)=frac1x.$$
                  Integrate with respect to $x$ on both sides:
                  $$int fracy''(x) dxy'(x)=int fracdxx.$$
                  It follows that
                  $$lnleft(y'(x)right)=C+ln(x).$$
                  An expression for $y'(x)$ is given by
                  $$y'(x)=e^C+ln(x)=e^Ccdot e^ln(x)=e^Ccdot x.$$
                  Redefine $e^C$ as $C$, since it is an arbitrary constant:
                  $$y'(x)=Ccdot x.$$
                  Integrate with respect to $x$ on both sides:
                  $$int y'(x) dx=int Ccdot x dx.$$
                  An expression for $y(x)$ is given by
                  $$y(x)=frac12cdot Ccdot x^2 + D.$$






                  share|cite|improve this answer












                  Let us consider your differential equation:
                  $$y''(x)-frac1xcdot y'(x)=0.$$
                  Add $frac1xcdot y'(x)$ on both sides:
                  $$y''(x)=frac1xcdot y'(x).$$
                  Divide by $y'(x)$ on both sides:
                  $$fracy''(x)y'(x)=frac1x.$$
                  Integrate with respect to $x$ on both sides:
                  $$int fracy''(x) dxy'(x)=int fracdxx.$$
                  It follows that
                  $$lnleft(y'(x)right)=C+ln(x).$$
                  An expression for $y'(x)$ is given by
                  $$y'(x)=e^C+ln(x)=e^Ccdot e^ln(x)=e^Ccdot x.$$
                  Redefine $e^C$ as $C$, since it is an arbitrary constant:
                  $$y'(x)=Ccdot x.$$
                  Integrate with respect to $x$ on both sides:
                  $$int y'(x) dx=int Ccdot x dx.$$
                  An expression for $y(x)$ is given by
                  $$y(x)=frac12cdot Ccdot x^2 + D.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 17 mins ago









                  Stijn Dietz

                  1369




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