First Order Differential equation - separable form

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I am trying to evaluate the equation:



$$y'=yleft(y^2-frac12right)$$



I multiplied the y over and tried to solve it in seperable form (M and N). The partial deritives did not work out to be equal to eachother so I am now stuck finding an integrating factor. Is this the right approach?






differential-equations







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    up vote
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    favorite
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    I am trying to evaluate the equation:



    $$y'=yleft(y^2-frac12right)$$



    I multiplied the y over and tried to solve it in seperable form (M and N). The partial deritives did not work out to be equal to eachother so I am now stuck finding an integrating factor. Is this the right approach?






    differential-equations







    share|cite|improve this question

























      up vote
      4
      down vote

      favorite
      1









      up vote
      4
      down vote

      favorite
      1






      1





      I am trying to evaluate the equation:



      $$y'=yleft(y^2-frac12right)$$



      I multiplied the y over and tried to solve it in seperable form (M and N). The partial deritives did not work out to be equal to eachother so I am now stuck finding an integrating factor. Is this the right approach?






      differential-equations







      share|cite|improve this question















      I am trying to evaluate the equation:



      $$y'=yleft(y^2-frac12right)$$



      I multiplied the y over and tried to solve it in seperable form (M and N). The partial deritives did not work out to be equal to eachother so I am now stuck finding an integrating factor. Is this the right approach?






      differential-equations




      differential-equations






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      edited 3 hours ago









      Parcly Taxel

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      Jytrex

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          Separation will work, just that partial fractions have to be dealt with:
          $$intfrac1y(y^2-1/2),dy=int1,dx$$
          $$intleft(frac4y2y^2-1-frac2yright),dy=int1,dx$$
          $$ln(2y^2-1)-2ln y=lnfrac2y^2-1y^2=x+K$$
          $$frac2y^2-1y^2=2-frac1y^2=Ae^x$$
          $$y=pmfrac1sqrt2-Ae^x$$






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            Separation will work, just that partial fractions have to be dealt with:
            $$intfrac1y(y^2-1/2),dy=int1,dx$$
            $$intleft(frac4y2y^2-1-frac2yright),dy=int1,dx$$
            $$ln(2y^2-1)-2ln y=lnfrac2y^2-1y^2=x+K$$
            $$frac2y^2-1y^2=2-frac1y^2=Ae^x$$
            $$y=pmfrac1sqrt2-Ae^x$$






            share|cite|improve this answer
























              up vote
              4
              down vote













              Separation will work, just that partial fractions have to be dealt with:
              $$intfrac1y(y^2-1/2),dy=int1,dx$$
              $$intleft(frac4y2y^2-1-frac2yright),dy=int1,dx$$
              $$ln(2y^2-1)-2ln y=lnfrac2y^2-1y^2=x+K$$
              $$frac2y^2-1y^2=2-frac1y^2=Ae^x$$
              $$y=pmfrac1sqrt2-Ae^x$$






              share|cite|improve this answer






















                up vote
                4
                down vote










                up vote
                4
                down vote









                Separation will work, just that partial fractions have to be dealt with:
                $$intfrac1y(y^2-1/2),dy=int1,dx$$
                $$intleft(frac4y2y^2-1-frac2yright),dy=int1,dx$$
                $$ln(2y^2-1)-2ln y=lnfrac2y^2-1y^2=x+K$$
                $$frac2y^2-1y^2=2-frac1y^2=Ae^x$$
                $$y=pmfrac1sqrt2-Ae^x$$






                share|cite|improve this answer












                Separation will work, just that partial fractions have to be dealt with:
                $$intfrac1y(y^2-1/2),dy=int1,dx$$
                $$intleft(frac4y2y^2-1-frac2yright),dy=int1,dx$$
                $$ln(2y^2-1)-2ln y=lnfrac2y^2-1y^2=x+K$$
                $$frac2y^2-1y^2=2-frac1y^2=Ae^x$$
                $$y=pmfrac1sqrt2-Ae^x$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Parcly Taxel

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