Find the general integral of the ODE $xy'^2+2xy'-y=0$












1












$begingroup$


I have to find the general integral of the following ODE
$$
xy'^2+2xy'-y=0
$$
The book says that, integrating w.r.t. $y'$, we get two homogeneous equations
$$
y'=-1+sqrt{1+frac{y}{x}},qquad y'=-1-sqrt{1+frac{y}{x}}
$$
defined for $x(x+y)>0$. Until this point all is clear. Now it says that the general integrals of such homogeneous equations are
$$
left(sqrt{1+frac{y}{x}}-1right)^2=frac{C}{x},qquad left(sqrt{1+frac{y}{x}}+1right)^2=frac{C}{x}.
$$
I do not understand how the book deduces such general integrals. Can someone help me?



Thank You






ordinary-differential-equations analysis







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  • $begingroup$
    What book are you reading?
    $endgroup$
    – user587192
    Dec 15 '18 at 14:44










  • $begingroup$
    @user587192 Cecconi, Stampacchia - Mathematical Analysis 2 (it is an italian book)
    $endgroup$
    – Jeji
    Dec 15 '18 at 16:30


















1












$begingroup$


I have to find the general integral of the following ODE
$$
xy'^2+2xy'-y=0
$$
The book says that, integrating w.r.t. $y'$, we get two homogeneous equations
$$
y'=-1+sqrt{1+frac{y}{x}},qquad y'=-1-sqrt{1+frac{y}{x}}
$$
defined for $x(x+y)>0$. Until this point all is clear. Now it says that the general integrals of such homogeneous equations are
$$
left(sqrt{1+frac{y}{x}}-1right)^2=frac{C}{x},qquad left(sqrt{1+frac{y}{x}}+1right)^2=frac{C}{x}.
$$
I do not understand how the book deduces such general integrals. Can someone help me?



Thank You






ordinary-differential-equations analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    What book are you reading?
    $endgroup$
    – user587192
    Dec 15 '18 at 14:44










  • $begingroup$
    @user587192 Cecconi, Stampacchia - Mathematical Analysis 2 (it is an italian book)
    $endgroup$
    – Jeji
    Dec 15 '18 at 16:30
















1












1








1





$begingroup$


I have to find the general integral of the following ODE
$$
xy'^2+2xy'-y=0
$$
The book says that, integrating w.r.t. $y'$, we get two homogeneous equations
$$
y'=-1+sqrt{1+frac{y}{x}},qquad y'=-1-sqrt{1+frac{y}{x}}
$$
defined for $x(x+y)>0$. Until this point all is clear. Now it says that the general integrals of such homogeneous equations are
$$
left(sqrt{1+frac{y}{x}}-1right)^2=frac{C}{x},qquad left(sqrt{1+frac{y}{x}}+1right)^2=frac{C}{x}.
$$
I do not understand how the book deduces such general integrals. Can someone help me?



Thank You






ordinary-differential-equations analysis







share|cite|improve this question









$endgroup$




I have to find the general integral of the following ODE
$$
xy'^2+2xy'-y=0
$$
The book says that, integrating w.r.t. $y'$, we get two homogeneous equations
$$
y'=-1+sqrt{1+frac{y}{x}},qquad y'=-1-sqrt{1+frac{y}{x}}
$$
defined for $x(x+y)>0$. Until this point all is clear. Now it says that the general integrals of such homogeneous equations are
$$
left(sqrt{1+frac{y}{x}}-1right)^2=frac{C}{x},qquad left(sqrt{1+frac{y}{x}}+1right)^2=frac{C}{x}.
$$
I do not understand how the book deduces such general integrals. Can someone help me?



Thank You






ordinary-differential-equations analysis




ordinary-differential-equations analysis






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share|cite|improve this question











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asked Dec 15 '18 at 14:27









JejiJeji

32918




32918












  • $begingroup$
    What book are you reading?
    $endgroup$
    – user587192
    Dec 15 '18 at 14:44










  • $begingroup$
    @user587192 Cecconi, Stampacchia - Mathematical Analysis 2 (it is an italian book)
    $endgroup$
    – Jeji
    Dec 15 '18 at 16:30




















  • $begingroup$
    What book are you reading?
    $endgroup$
    – user587192
    Dec 15 '18 at 14:44










  • $begingroup$
    @user587192 Cecconi, Stampacchia - Mathematical Analysis 2 (it is an italian book)
    $endgroup$
    – Jeji
    Dec 15 '18 at 16:30


















$begingroup$
What book are you reading?
$endgroup$
– user587192
Dec 15 '18 at 14:44




$begingroup$
What book are you reading?
$endgroup$
– user587192
Dec 15 '18 at 14:44












$begingroup$
@user587192 Cecconi, Stampacchia - Mathematical Analysis 2 (it is an italian book)
$endgroup$
– Jeji
Dec 15 '18 at 16:30






$begingroup$
@user587192 Cecconi, Stampacchia - Mathematical Analysis 2 (it is an italian book)
$endgroup$
– Jeji
Dec 15 '18 at 16:30












3 Answers
3






active

oldest

votes


















1












$begingroup$

Hint. Let $u(x):=sqrt{1+frac{y(x)}{x}}>0$ then $y(x)=xu^2(x)-x$ and $$y'(x)=u^2(x)+2xu'(x)u(x)-1.$$
Hence
$$y'(x)=-1pmsqrt{1+frac{y(x)}{x}}$$
becomes
$$u^2(x)+2xu'(x)u(x)-1=-1pm u(x)$$
that is
$$u^2(x)+2xu'(x)u(x)=pm u(x)$$
or
$$u(x)+2xu'(x)=pm 1$$
which is a linear ODE of first order. Can you take it from here?






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Substitute $$y(x)=xv(x)$$ then we get $$xleft(xfrac{dv(x)}{dx}+v(x)right)^2+2xleft(xfrac{dv(x)}{dx}+v(x)right)-xv(x)=0$$ and we get
    $$frac{dv(x)}{dx}=frac{-x-sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
    $$frac{dv(x)}{dx}=frac{-x+sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
    Can you finish?






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Set $y=u-x$, then $y'=u'-1$ and
      $$
      u=y+x=x(y'^2+2y'+1)=xu'^2tag1
      $$
      which has a derivative
      $$
      u'=u'^2+2xu'u''implies u'(2xu''+u'-1)=0
      $$
      Thus you get linear pieces $u=C$ and solutions of the linear ODE $2xu''+u'=1$. They can switch from one to the other where both factors are zero at the same time.



      For $x>0$ the second equation solves as
      $$
      (sqrt xu')'=frac{2xu''+u'}{2sqrt x}=frac1{2sqrt x}implies sqrt xu'=sqrt{x}+C
      $$
      and the original equation in form (1) gives
      $$
      u(x)=xu'^2=x+2Csqrt x+C^2
      $$






      share|cite|improve this answer











      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Hint. Let $u(x):=sqrt{1+frac{y(x)}{x}}>0$ then $y(x)=xu^2(x)-x$ and $$y'(x)=u^2(x)+2xu'(x)u(x)-1.$$
        Hence
        $$y'(x)=-1pmsqrt{1+frac{y(x)}{x}}$$
        becomes
        $$u^2(x)+2xu'(x)u(x)-1=-1pm u(x)$$
        that is
        $$u^2(x)+2xu'(x)u(x)=pm u(x)$$
        or
        $$u(x)+2xu'(x)=pm 1$$
        which is a linear ODE of first order. Can you take it from here?






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          Hint. Let $u(x):=sqrt{1+frac{y(x)}{x}}>0$ then $y(x)=xu^2(x)-x$ and $$y'(x)=u^2(x)+2xu'(x)u(x)-1.$$
          Hence
          $$y'(x)=-1pmsqrt{1+frac{y(x)}{x}}$$
          becomes
          $$u^2(x)+2xu'(x)u(x)-1=-1pm u(x)$$
          that is
          $$u^2(x)+2xu'(x)u(x)=pm u(x)$$
          or
          $$u(x)+2xu'(x)=pm 1$$
          which is a linear ODE of first order. Can you take it from here?






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            Hint. Let $u(x):=sqrt{1+frac{y(x)}{x}}>0$ then $y(x)=xu^2(x)-x$ and $$y'(x)=u^2(x)+2xu'(x)u(x)-1.$$
            Hence
            $$y'(x)=-1pmsqrt{1+frac{y(x)}{x}}$$
            becomes
            $$u^2(x)+2xu'(x)u(x)-1=-1pm u(x)$$
            that is
            $$u^2(x)+2xu'(x)u(x)=pm u(x)$$
            or
            $$u(x)+2xu'(x)=pm 1$$
            which is a linear ODE of first order. Can you take it from here?






            share|cite|improve this answer











            $endgroup$



            Hint. Let $u(x):=sqrt{1+frac{y(x)}{x}}>0$ then $y(x)=xu^2(x)-x$ and $$y'(x)=u^2(x)+2xu'(x)u(x)-1.$$
            Hence
            $$y'(x)=-1pmsqrt{1+frac{y(x)}{x}}$$
            becomes
            $$u^2(x)+2xu'(x)u(x)-1=-1pm u(x)$$
            that is
            $$u^2(x)+2xu'(x)u(x)=pm u(x)$$
            or
            $$u(x)+2xu'(x)=pm 1$$
            which is a linear ODE of first order. Can you take it from here?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 15 '18 at 16:16

























            answered Dec 15 '18 at 14:40









            Robert ZRobert Z

            94.4k1063134




            94.4k1063134























                1












                $begingroup$

                Substitute $$y(x)=xv(x)$$ then we get $$xleft(xfrac{dv(x)}{dx}+v(x)right)^2+2xleft(xfrac{dv(x)}{dx}+v(x)right)-xv(x)=0$$ and we get
                $$frac{dv(x)}{dx}=frac{-x-sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
                $$frac{dv(x)}{dx}=frac{-x+sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
                Can you finish?






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Substitute $$y(x)=xv(x)$$ then we get $$xleft(xfrac{dv(x)}{dx}+v(x)right)^2+2xleft(xfrac{dv(x)}{dx}+v(x)right)-xv(x)=0$$ and we get
                  $$frac{dv(x)}{dx}=frac{-x-sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
                  $$frac{dv(x)}{dx}=frac{-x+sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
                  Can you finish?






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Substitute $$y(x)=xv(x)$$ then we get $$xleft(xfrac{dv(x)}{dx}+v(x)right)^2+2xleft(xfrac{dv(x)}{dx}+v(x)right)-xv(x)=0$$ and we get
                    $$frac{dv(x)}{dx}=frac{-x-sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
                    $$frac{dv(x)}{dx}=frac{-x+sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
                    Can you finish?






                    share|cite|improve this answer









                    $endgroup$



                    Substitute $$y(x)=xv(x)$$ then we get $$xleft(xfrac{dv(x)}{dx}+v(x)right)^2+2xleft(xfrac{dv(x)}{dx}+v(x)right)-xv(x)=0$$ and we get
                    $$frac{dv(x)}{dx}=frac{-x-sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
                    $$frac{dv(x)}{dx}=frac{-x+sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
                    Can you finish?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 15 '18 at 14:42









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    73.7k42864




                    73.7k42864























                        1












                        $begingroup$

                        Set $y=u-x$, then $y'=u'-1$ and
                        $$
                        u=y+x=x(y'^2+2y'+1)=xu'^2tag1
                        $$
                        which has a derivative
                        $$
                        u'=u'^2+2xu'u''implies u'(2xu''+u'-1)=0
                        $$
                        Thus you get linear pieces $u=C$ and solutions of the linear ODE $2xu''+u'=1$. They can switch from one to the other where both factors are zero at the same time.



                        For $x>0$ the second equation solves as
                        $$
                        (sqrt xu')'=frac{2xu''+u'}{2sqrt x}=frac1{2sqrt x}implies sqrt xu'=sqrt{x}+C
                        $$
                        and the original equation in form (1) gives
                        $$
                        u(x)=xu'^2=x+2Csqrt x+C^2
                        $$






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Set $y=u-x$, then $y'=u'-1$ and
                          $$
                          u=y+x=x(y'^2+2y'+1)=xu'^2tag1
                          $$
                          which has a derivative
                          $$
                          u'=u'^2+2xu'u''implies u'(2xu''+u'-1)=0
                          $$
                          Thus you get linear pieces $u=C$ and solutions of the linear ODE $2xu''+u'=1$. They can switch from one to the other where both factors are zero at the same time.



                          For $x>0$ the second equation solves as
                          $$
                          (sqrt xu')'=frac{2xu''+u'}{2sqrt x}=frac1{2sqrt x}implies sqrt xu'=sqrt{x}+C
                          $$
                          and the original equation in form (1) gives
                          $$
                          u(x)=xu'^2=x+2Csqrt x+C^2
                          $$






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Set $y=u-x$, then $y'=u'-1$ and
                            $$
                            u=y+x=x(y'^2+2y'+1)=xu'^2tag1
                            $$
                            which has a derivative
                            $$
                            u'=u'^2+2xu'u''implies u'(2xu''+u'-1)=0
                            $$
                            Thus you get linear pieces $u=C$ and solutions of the linear ODE $2xu''+u'=1$. They can switch from one to the other where both factors are zero at the same time.



                            For $x>0$ the second equation solves as
                            $$
                            (sqrt xu')'=frac{2xu''+u'}{2sqrt x}=frac1{2sqrt x}implies sqrt xu'=sqrt{x}+C
                            $$
                            and the original equation in form (1) gives
                            $$
                            u(x)=xu'^2=x+2Csqrt x+C^2
                            $$






                            share|cite|improve this answer











                            $endgroup$



                            Set $y=u-x$, then $y'=u'-1$ and
                            $$
                            u=y+x=x(y'^2+2y'+1)=xu'^2tag1
                            $$
                            which has a derivative
                            $$
                            u'=u'^2+2xu'u''implies u'(2xu''+u'-1)=0
                            $$
                            Thus you get linear pieces $u=C$ and solutions of the linear ODE $2xu''+u'=1$. They can switch from one to the other where both factors are zero at the same time.



                            For $x>0$ the second equation solves as
                            $$
                            (sqrt xu')'=frac{2xu''+u'}{2sqrt x}=frac1{2sqrt x}implies sqrt xu'=sqrt{x}+C
                            $$
                            and the original equation in form (1) gives
                            $$
                            u(x)=xu'^2=x+2Csqrt x+C^2
                            $$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 15 '18 at 14:55

























                            answered Dec 15 '18 at 14:49









                            LutzLLutzL

                            57.1k42054




                            57.1k42054






























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