Solving a differential equation with natural log The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Help solving a differential equationproof of a solution to a differential equationWhat method I should use to solve this differantial equation?Solving an Exact Differential EquationHelp solving third-order differential equationSolving this kind of differential equationHelp with solving linear differential equations using an integrating factorSolving differential equation with the Dirac Delta FunctionHow to solve this Linear Differential Equation?

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Solving a differential equation with natural log



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Help solving a differential equationproof of a solution to a differential equationWhat method I should use to solve this differantial equation?Solving an Exact Differential EquationHelp solving third-order differential equationSolving this kind of differential equationHelp with solving linear differential equations using an integrating factorSolving differential equation with the Dirac Delta FunctionHow to solve this Linear Differential Equation?










4












$begingroup$


I am given:



$xdfracdydx=dfrac1y^3$



After separating and integrating, I have:



$y^4/4=ln x+C$



I am supposed to solve this equation, but I'm stuck here. Should I solve explicitly so I can keep $C$?



EDIT:



A solution I came up with last night was:



$y=(4ln x+C)^1/4$










share|cite|improve this question











$endgroup$











  • $begingroup$
    don't forget the negative fourth root.
    $endgroup$
    – Stefan Smith
    Jan 16 '14 at 2:46















4












$begingroup$


I am given:



$xdfracdydx=dfrac1y^3$



After separating and integrating, I have:



$y^4/4=ln x+C$



I am supposed to solve this equation, but I'm stuck here. Should I solve explicitly so I can keep $C$?



EDIT:



A solution I came up with last night was:



$y=(4ln x+C)^1/4$










share|cite|improve this question











$endgroup$











  • $begingroup$
    don't forget the negative fourth root.
    $endgroup$
    – Stefan Smith
    Jan 16 '14 at 2:46













4












4








4





$begingroup$


I am given:



$xdfracdydx=dfrac1y^3$



After separating and integrating, I have:



$y^4/4=ln x+C$



I am supposed to solve this equation, but I'm stuck here. Should I solve explicitly so I can keep $C$?



EDIT:



A solution I came up with last night was:



$y=(4ln x+C)^1/4$










share|cite|improve this question











$endgroup$




I am given:



$xdfracdydx=dfrac1y^3$



After separating and integrating, I have:



$y^4/4=ln x+C$



I am supposed to solve this equation, but I'm stuck here. Should I solve explicitly so I can keep $C$?



EDIT:



A solution I came up with last night was:



$y=(4ln x+C)^1/4$







ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 '14 at 20:45









BaronVT

11.6k11337




11.6k11337










asked Jan 15 '14 at 20:36









westhe32ndwesthe32nd

198213




198213











  • $begingroup$
    don't forget the negative fourth root.
    $endgroup$
    – Stefan Smith
    Jan 16 '14 at 2:46
















  • $begingroup$
    don't forget the negative fourth root.
    $endgroup$
    – Stefan Smith
    Jan 16 '14 at 2:46















$begingroup$
don't forget the negative fourth root.
$endgroup$
– Stefan Smith
Jan 16 '14 at 2:46




$begingroup$
don't forget the negative fourth root.
$endgroup$
– Stefan Smith
Jan 16 '14 at 2:46










1 Answer
1






active

oldest

votes


















1












$begingroup$

Try differentiating to see if you got the correct solution!



You can compute
$$
fracdydx = frac14left(4 ln x + Cright)^-3/4left(frac4xright) = frac1xleft(4 ln x + Cright)^-3/4
$$



so
$$
xfracdydx = left(4 ln x + Cright)^-3/4.
$$



Is this equal to $dfrac1y^3$?






share|cite|improve this answer









$endgroup$













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    1 Answer
    1






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    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Try differentiating to see if you got the correct solution!



    You can compute
    $$
    fracdydx = frac14left(4 ln x + Cright)^-3/4left(frac4xright) = frac1xleft(4 ln x + Cright)^-3/4
    $$



    so
    $$
    xfracdydx = left(4 ln x + Cright)^-3/4.
    $$



    Is this equal to $dfrac1y^3$?






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      Try differentiating to see if you got the correct solution!



      You can compute
      $$
      fracdydx = frac14left(4 ln x + Cright)^-3/4left(frac4xright) = frac1xleft(4 ln x + Cright)^-3/4
      $$



      so
      $$
      xfracdydx = left(4 ln x + Cright)^-3/4.
      $$



      Is this equal to $dfrac1y^3$?






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        Try differentiating to see if you got the correct solution!



        You can compute
        $$
        fracdydx = frac14left(4 ln x + Cright)^-3/4left(frac4xright) = frac1xleft(4 ln x + Cright)^-3/4
        $$



        so
        $$
        xfracdydx = left(4 ln x + Cright)^-3/4.
        $$



        Is this equal to $dfrac1y^3$?






        share|cite|improve this answer









        $endgroup$



        Try differentiating to see if you got the correct solution!



        You can compute
        $$
        fracdydx = frac14left(4 ln x + Cright)^-3/4left(frac4xright) = frac1xleft(4 ln x + Cright)^-3/4
        $$



        so
        $$
        xfracdydx = left(4 ln x + Cright)^-3/4.
        $$



        Is this equal to $dfrac1y^3$?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 '14 at 20:49









        BaronVTBaronVT

        11.6k11337




        11.6k11337



























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