Find the orthogonal trajectories of the family of curves given by $x^2 + y^2 + 2Cy =1$.












0












$begingroup$


Find the orthogonal trajectories of the family of curves given by $$x^2 + y^2 + 2Cy =1.$$



The ordinary differential equation for the family of curves is given by $y'=frac{2xy}{x^2-y^2-1}$.Therefore, the differential equation for the orthogonal curves is given by $y'=frac{1-x^2-y^2}{2xy}$.



This is an exact differential equation. So, solving by the standard method for exact differential equation gives $x-x^3/3+xy^2=C$. But this is not the correct answer according to the answers given at back of the book.



The answer given at the back of the book is $x^2 - y^2 - Cx +1 = 0$.



Can someone please find out at which step I am maing a mistake or provide a solution that leads to the correct answer?






calculus ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Wouldn't it be easier to complete the square first?
    $endgroup$
    – abiessu
    Jun 15 '14 at 14:26












  • $begingroup$
    Your equation for the family of curves is not correct. Can you show the steps of your solution?
    $endgroup$
    – Artem
    Jun 15 '14 at 17:23
















0












$begingroup$


Find the orthogonal trajectories of the family of curves given by $$x^2 + y^2 + 2Cy =1.$$



The ordinary differential equation for the family of curves is given by $y'=frac{2xy}{x^2-y^2-1}$.Therefore, the differential equation for the orthogonal curves is given by $y'=frac{1-x^2-y^2}{2xy}$.



This is an exact differential equation. So, solving by the standard method for exact differential equation gives $x-x^3/3+xy^2=C$. But this is not the correct answer according to the answers given at back of the book.



The answer given at the back of the book is $x^2 - y^2 - Cx +1 = 0$.



Can someone please find out at which step I am maing a mistake or provide a solution that leads to the correct answer?






calculus ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Wouldn't it be easier to complete the square first?
    $endgroup$
    – abiessu
    Jun 15 '14 at 14:26












  • $begingroup$
    Your equation for the family of curves is not correct. Can you show the steps of your solution?
    $endgroup$
    – Artem
    Jun 15 '14 at 17:23














0












0








0





$begingroup$


Find the orthogonal trajectories of the family of curves given by $$x^2 + y^2 + 2Cy =1.$$



The ordinary differential equation for the family of curves is given by $y'=frac{2xy}{x^2-y^2-1}$.Therefore, the differential equation for the orthogonal curves is given by $y'=frac{1-x^2-y^2}{2xy}$.



This is an exact differential equation. So, solving by the standard method for exact differential equation gives $x-x^3/3+xy^2=C$. But this is not the correct answer according to the answers given at back of the book.



The answer given at the back of the book is $x^2 - y^2 - Cx +1 = 0$.



Can someone please find out at which step I am maing a mistake or provide a solution that leads to the correct answer?






calculus ordinary-differential-equations







share|cite|improve this question









$endgroup$




Find the orthogonal trajectories of the family of curves given by $$x^2 + y^2 + 2Cy =1.$$



The ordinary differential equation for the family of curves is given by $y'=frac{2xy}{x^2-y^2-1}$.Therefore, the differential equation for the orthogonal curves is given by $y'=frac{1-x^2-y^2}{2xy}$.



This is an exact differential equation. So, solving by the standard method for exact differential equation gives $x-x^3/3+xy^2=C$. But this is not the correct answer according to the answers given at back of the book.



The answer given at the back of the book is $x^2 - y^2 - Cx +1 = 0$.



Can someone please find out at which step I am maing a mistake or provide a solution that leads to the correct answer?






calculus ordinary-differential-equations




calculus ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jun 15 '14 at 14:24









rockstar123rockstar123

641418




641418








  • 2




    $begingroup$
    Wouldn't it be easier to complete the square first?
    $endgroup$
    – abiessu
    Jun 15 '14 at 14:26












  • $begingroup$
    Your equation for the family of curves is not correct. Can you show the steps of your solution?
    $endgroup$
    – Artem
    Jun 15 '14 at 17:23














  • 2




    $begingroup$
    Wouldn't it be easier to complete the square first?
    $endgroup$
    – abiessu
    Jun 15 '14 at 14:26












  • $begingroup$
    Your equation for the family of curves is not correct. Can you show the steps of your solution?
    $endgroup$
    – Artem
    Jun 15 '14 at 17:23








2




2




$begingroup$
Wouldn't it be easier to complete the square first?
$endgroup$
– abiessu
Jun 15 '14 at 14:26






$begingroup$
Wouldn't it be easier to complete the square first?
$endgroup$
– abiessu
Jun 15 '14 at 14:26














$begingroup$
Your equation for the family of curves is not correct. Can you show the steps of your solution?
$endgroup$
– Artem
Jun 15 '14 at 17:23




$begingroup$
Your equation for the family of curves is not correct. Can you show the steps of your solution?
$endgroup$
– Artem
Jun 15 '14 at 17:23










1 Answer
1






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oldest

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0












$begingroup$

For the given family of curves
$$
x^2+y^2+2cy=1
$$
which can be written as
$$
frac{1-x^2-y^2}{2y} = c
$$
differentiating both sides
$$
frac{dy}{dx}=frac{2xy}{x^2-y^2-1}
$$
and the family of curves orthogonal to this will be
$$
-frac{dx}{dy}=frac{2xy}{x^2-y^2-1}
$$
or,
$$(x^2-y^2-1)dx+(2xy)dy=0
$$
this is not an exact DE, and hence will have to converted to one,
$$
frac{partial{M}}{partial{y}}=-2y; frac{partial{N}}{partial{x}}=2y
$$
$$
frac{1}{N}bigg(frac{partial{M}}{partial{y}}-frac{partial{N}}{partial{x}}bigg)=-frac{2}{x}
$$
$$
I.F=e^{int{-frac{2}{x}}dx}=frac{1}{x^2}
$$
after multiply with the Integrating Factor the DE becomes exact
$$
bigg(1-frac{y^2}{x^2}-frac{1}{x^2}bigg)dx+frac{2y}{x}dx=0
$$
solving for this,the orthogonal family of curves is
$$
x^2+y^2+1=cx
$$






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






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    active

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    active

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    0












    $begingroup$

    For the given family of curves
    $$
    x^2+y^2+2cy=1
    $$
    which can be written as
    $$
    frac{1-x^2-y^2}{2y} = c
    $$
    differentiating both sides
    $$
    frac{dy}{dx}=frac{2xy}{x^2-y^2-1}
    $$
    and the family of curves orthogonal to this will be
    $$
    -frac{dx}{dy}=frac{2xy}{x^2-y^2-1}
    $$
    or,
    $$(x^2-y^2-1)dx+(2xy)dy=0
    $$
    this is not an exact DE, and hence will have to converted to one,
    $$
    frac{partial{M}}{partial{y}}=-2y; frac{partial{N}}{partial{x}}=2y
    $$
    $$
    frac{1}{N}bigg(frac{partial{M}}{partial{y}}-frac{partial{N}}{partial{x}}bigg)=-frac{2}{x}
    $$
    $$
    I.F=e^{int{-frac{2}{x}}dx}=frac{1}{x^2}
    $$
    after multiply with the Integrating Factor the DE becomes exact
    $$
    bigg(1-frac{y^2}{x^2}-frac{1}{x^2}bigg)dx+frac{2y}{x}dx=0
    $$
    solving for this,the orthogonal family of curves is
    $$
    x^2+y^2+1=cx
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      For the given family of curves
      $$
      x^2+y^2+2cy=1
      $$
      which can be written as
      $$
      frac{1-x^2-y^2}{2y} = c
      $$
      differentiating both sides
      $$
      frac{dy}{dx}=frac{2xy}{x^2-y^2-1}
      $$
      and the family of curves orthogonal to this will be
      $$
      -frac{dx}{dy}=frac{2xy}{x^2-y^2-1}
      $$
      or,
      $$(x^2-y^2-1)dx+(2xy)dy=0
      $$
      this is not an exact DE, and hence will have to converted to one,
      $$
      frac{partial{M}}{partial{y}}=-2y; frac{partial{N}}{partial{x}}=2y
      $$
      $$
      frac{1}{N}bigg(frac{partial{M}}{partial{y}}-frac{partial{N}}{partial{x}}bigg)=-frac{2}{x}
      $$
      $$
      I.F=e^{int{-frac{2}{x}}dx}=frac{1}{x^2}
      $$
      after multiply with the Integrating Factor the DE becomes exact
      $$
      bigg(1-frac{y^2}{x^2}-frac{1}{x^2}bigg)dx+frac{2y}{x}dx=0
      $$
      solving for this,the orthogonal family of curves is
      $$
      x^2+y^2+1=cx
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        For the given family of curves
        $$
        x^2+y^2+2cy=1
        $$
        which can be written as
        $$
        frac{1-x^2-y^2}{2y} = c
        $$
        differentiating both sides
        $$
        frac{dy}{dx}=frac{2xy}{x^2-y^2-1}
        $$
        and the family of curves orthogonal to this will be
        $$
        -frac{dx}{dy}=frac{2xy}{x^2-y^2-1}
        $$
        or,
        $$(x^2-y^2-1)dx+(2xy)dy=0
        $$
        this is not an exact DE, and hence will have to converted to one,
        $$
        frac{partial{M}}{partial{y}}=-2y; frac{partial{N}}{partial{x}}=2y
        $$
        $$
        frac{1}{N}bigg(frac{partial{M}}{partial{y}}-frac{partial{N}}{partial{x}}bigg)=-frac{2}{x}
        $$
        $$
        I.F=e^{int{-frac{2}{x}}dx}=frac{1}{x^2}
        $$
        after multiply with the Integrating Factor the DE becomes exact
        $$
        bigg(1-frac{y^2}{x^2}-frac{1}{x^2}bigg)dx+frac{2y}{x}dx=0
        $$
        solving for this,the orthogonal family of curves is
        $$
        x^2+y^2+1=cx
        $$






        share|cite|improve this answer









        $endgroup$



        For the given family of curves
        $$
        x^2+y^2+2cy=1
        $$
        which can be written as
        $$
        frac{1-x^2-y^2}{2y} = c
        $$
        differentiating both sides
        $$
        frac{dy}{dx}=frac{2xy}{x^2-y^2-1}
        $$
        and the family of curves orthogonal to this will be
        $$
        -frac{dx}{dy}=frac{2xy}{x^2-y^2-1}
        $$
        or,
        $$(x^2-y^2-1)dx+(2xy)dy=0
        $$
        this is not an exact DE, and hence will have to converted to one,
        $$
        frac{partial{M}}{partial{y}}=-2y; frac{partial{N}}{partial{x}}=2y
        $$
        $$
        frac{1}{N}bigg(frac{partial{M}}{partial{y}}-frac{partial{N}}{partial{x}}bigg)=-frac{2}{x}
        $$
        $$
        I.F=e^{int{-frac{2}{x}}dx}=frac{1}{x^2}
        $$
        after multiply with the Integrating Factor the DE becomes exact
        $$
        bigg(1-frac{y^2}{x^2}-frac{1}{x^2}bigg)dx+frac{2y}{x}dx=0
        $$
        solving for this,the orthogonal family of curves is
        $$
        x^2+y^2+1=cx
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 25 '18 at 19:19









        Satyajit GhanaSatyajit Ghana

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