Solve Differential Equation 2: $fracd^2ydx^2-2xfracdydx+2y=0$ [closed]

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solve $fracd^2ydx^2-2xfracdydx+2y=0$
I tried to solve through linear differential equation but it has variable with x in first derivative .




differential-equations




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closed as off-topic by Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus Jul 21 at 0:03


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
















    up vote
    -2
    down vote

    favorite
    1












    solve $fracd^2ydx^2-2xfracdydx+2y=0$
    I tried to solve through linear differential equation but it has variable with x in first derivative .




    differential-equations




    share|cite|improve this question













    closed as off-topic by Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus Jul 21 at 0:03


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus
    If this question can be reworded to fit the rules in the help center, please edit the question.














      up vote
      -2
      down vote

      favorite
      1









      up vote
      -2
      down vote

      favorite
      1






      1





      solve $fracd^2ydx^2-2xfracdydx+2y=0$
      I tried to solve through linear differential equation but it has variable with x in first derivative .




      differential-equations




      share|cite|improve this question













      solve $fracd^2ydx^2-2xfracdydx+2y=0$
      I tried to solve through linear differential equation but it has variable with x in first derivative .





      differential-equations





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 20 at 18:57









      Nosrati

      19.5k41544




      19.5k41544









      asked Jul 20 at 18:55









      ravi yadav

      295




      295




      closed as off-topic by Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus Jul 21 at 0:03


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus
      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus Jul 21 at 0:03


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          2 Answers
          2






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          oldest

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          up vote
          2
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          accepted










          Substituting $$y=xv$$ then we get $$y'=xv'+v$$ and $$y''=xv''+2v'$$ then our equation will be



          $$xv''-2v'(x^2-1)=0$$ now we Substitute



          $$v'=u$$ then we get $$fracu'u=-frac2x+2x
          $$



          and we get



          $$intfracu'udx=intleft(-frac2x+2xright)dx$$



          this gives



          $$log(u(x))=x^2-log(x)+C$$
          Can you finish?






          share|cite|improve this answer





















          • yes , thank you very much
            – ravi yadav
            Jul 20 at 19:12






          • 1




            Then good luck Mr. Ravi (like Ravi Substitution?)
            – Dr. Sonnhard Graubner
            Jul 20 at 19:13

















          up vote
          2
          down vote













          $y=c_1x$ is a solution.



          You can try to integrate the differential by reduction of order



          Try



          $$y_1=v(x)x$$
          The original equation
          $$y''-2xy'+2y=0$$
          becomes
          $$v''x+2v'(1-x^2)=0$$
          Substitute $w=v'$
          $$w'x+2w(1-x^2)=0$$



          It's a first order,differential equation
          $$(ln w)'=2x-frac 2x$$
          $$ln w=x^2-2ln x + K$$
          $$w=C_1frac e^x^2x^2$$
          $$v'=C_1frac e^x^2x^2$$
          $$v(x)=C_1int frac e^x^2x^2dx+C_2$$
          $$y(x)=C_1xint frac e^x^2x^2dx+C_2x$$
          $$boxedy(x)=C_1(xsqrt pitext erfi(x)-e^x^2)+C_2x$$






          share|cite|improve this answer






























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            Substituting $$y=xv$$ then we get $$y'=xv'+v$$ and $$y''=xv''+2v'$$ then our equation will be



            $$xv''-2v'(x^2-1)=0$$ now we Substitute



            $$v'=u$$ then we get $$fracu'u=-frac2x+2x
            $$



            and we get



            $$intfracu'udx=intleft(-frac2x+2xright)dx$$



            this gives



            $$log(u(x))=x^2-log(x)+C$$
            Can you finish?






            share|cite|improve this answer





















            • yes , thank you very much
              – ravi yadav
              Jul 20 at 19:12






            • 1




              Then good luck Mr. Ravi (like Ravi Substitution?)
              – Dr. Sonnhard Graubner
              Jul 20 at 19:13














            up vote
            2
            down vote



            accepted










            Substituting $$y=xv$$ then we get $$y'=xv'+v$$ and $$y''=xv''+2v'$$ then our equation will be



            $$xv''-2v'(x^2-1)=0$$ now we Substitute



            $$v'=u$$ then we get $$fracu'u=-frac2x+2x
            $$



            and we get



            $$intfracu'udx=intleft(-frac2x+2xright)dx$$



            this gives



            $$log(u(x))=x^2-log(x)+C$$
            Can you finish?






            share|cite|improve this answer





















            • yes , thank you very much
              – ravi yadav
              Jul 20 at 19:12






            • 1




              Then good luck Mr. Ravi (like Ravi Substitution?)
              – Dr. Sonnhard Graubner
              Jul 20 at 19:13












            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Substituting $$y=xv$$ then we get $$y'=xv'+v$$ and $$y''=xv''+2v'$$ then our equation will be



            $$xv''-2v'(x^2-1)=0$$ now we Substitute



            $$v'=u$$ then we get $$fracu'u=-frac2x+2x
            $$



            and we get



            $$intfracu'udx=intleft(-frac2x+2xright)dx$$



            this gives



            $$log(u(x))=x^2-log(x)+C$$
            Can you finish?






            share|cite|improve this answer













            Substituting $$y=xv$$ then we get $$y'=xv'+v$$ and $$y''=xv''+2v'$$ then our equation will be



            $$xv''-2v'(x^2-1)=0$$ now we Substitute



            $$v'=u$$ then we get $$fracu'u=-frac2x+2x
            $$



            and we get



            $$intfracu'udx=intleft(-frac2x+2xright)dx$$



            this gives



            $$log(u(x))=x^2-log(x)+C$$
            Can you finish?







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Jul 20 at 19:10









            Dr. Sonnhard Graubner

            66.8k32659




            66.8k32659











            • yes , thank you very much
              – ravi yadav
              Jul 20 at 19:12






            • 1




              Then good luck Mr. Ravi (like Ravi Substitution?)
              – Dr. Sonnhard Graubner
              Jul 20 at 19:13
















            • yes , thank you very much
              – ravi yadav
              Jul 20 at 19:12






            • 1




              Then good luck Mr. Ravi (like Ravi Substitution?)
              – Dr. Sonnhard Graubner
              Jul 20 at 19:13















            yes , thank you very much
            – ravi yadav
            Jul 20 at 19:12




            yes , thank you very much
            – ravi yadav
            Jul 20 at 19:12




            1




            1




            Then good luck Mr. Ravi (like Ravi Substitution?)
            – Dr. Sonnhard Graubner
            Jul 20 at 19:13




            Then good luck Mr. Ravi (like Ravi Substitution?)
            – Dr. Sonnhard Graubner
            Jul 20 at 19:13










            up vote
            2
            down vote













            $y=c_1x$ is a solution.



            You can try to integrate the differential by reduction of order



            Try



            $$y_1=v(x)x$$
            The original equation
            $$y''-2xy'+2y=0$$
            becomes
            $$v''x+2v'(1-x^2)=0$$
            Substitute $w=v'$
            $$w'x+2w(1-x^2)=0$$



            It's a first order,differential equation
            $$(ln w)'=2x-frac 2x$$
            $$ln w=x^2-2ln x + K$$
            $$w=C_1frac e^x^2x^2$$
            $$v'=C_1frac e^x^2x^2$$
            $$v(x)=C_1int frac e^x^2x^2dx+C_2$$
            $$y(x)=C_1xint frac e^x^2x^2dx+C_2x$$
            $$boxedy(x)=C_1(xsqrt pitext erfi(x)-e^x^2)+C_2x$$






            share|cite|improve this answer



























              up vote
              2
              down vote













              $y=c_1x$ is a solution.



              You can try to integrate the differential by reduction of order



              Try



              $$y_1=v(x)x$$
              The original equation
              $$y''-2xy'+2y=0$$
              becomes
              $$v''x+2v'(1-x^2)=0$$
              Substitute $w=v'$
              $$w'x+2w(1-x^2)=0$$



              It's a first order,differential equation
              $$(ln w)'=2x-frac 2x$$
              $$ln w=x^2-2ln x + K$$
              $$w=C_1frac e^x^2x^2$$
              $$v'=C_1frac e^x^2x^2$$
              $$v(x)=C_1int frac e^x^2x^2dx+C_2$$
              $$y(x)=C_1xint frac e^x^2x^2dx+C_2x$$
              $$boxedy(x)=C_1(xsqrt pitext erfi(x)-e^x^2)+C_2x$$






              share|cite|improve this answer

























                up vote
                2
                down vote










                up vote
                2
                down vote









                $y=c_1x$ is a solution.



                You can try to integrate the differential by reduction of order



                Try



                $$y_1=v(x)x$$
                The original equation
                $$y''-2xy'+2y=0$$
                becomes
                $$v''x+2v'(1-x^2)=0$$
                Substitute $w=v'$
                $$w'x+2w(1-x^2)=0$$



                It's a first order,differential equation
                $$(ln w)'=2x-frac 2x$$
                $$ln w=x^2-2ln x + K$$
                $$w=C_1frac e^x^2x^2$$
                $$v'=C_1frac e^x^2x^2$$
                $$v(x)=C_1int frac e^x^2x^2dx+C_2$$
                $$y(x)=C_1xint frac e^x^2x^2dx+C_2x$$
                $$boxedy(x)=C_1(xsqrt pitext erfi(x)-e^x^2)+C_2x$$






                share|cite|improve this answer















                $y=c_1x$ is a solution.



                You can try to integrate the differential by reduction of order



                Try



                $$y_1=v(x)x$$
                The original equation
                $$y''-2xy'+2y=0$$
                becomes
                $$v''x+2v'(1-x^2)=0$$
                Substitute $w=v'$
                $$w'x+2w(1-x^2)=0$$



                It's a first order,differential equation
                $$(ln w)'=2x-frac 2x$$
                $$ln w=x^2-2ln x + K$$
                $$w=C_1frac e^x^2x^2$$
                $$v'=C_1frac e^x^2x^2$$
                $$v(x)=C_1int frac e^x^2x^2dx+C_2$$
                $$y(x)=C_1xint frac e^x^2x^2dx+C_2x$$
                $$boxedy(x)=C_1(xsqrt pitext erfi(x)-e^x^2)+C_2x$$







                share|cite|improve this answer















                share|cite|improve this answer



                share|cite|improve this answer








                edited Jul 20 at 19:52


























                answered Jul 20 at 19:07









                Isham

                10.6k3829




                10.6k3829












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