Find the orthogonal trajectories of the family of curves given by $x^2 + y^2 + 2Cy =1$.
$begingroup$
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
asked Jun 15 '14 at 14:24
rockstar123rockstar123
641418
$endgroup$
add a comment |
$begingroup$
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
asked Jun 15 '14 at 14:24
rockstar123rockstar123
641418
$endgroup$
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
add a comment |
$begingroup$
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
asked Jun 15 '14 at 14:24
rockstar123rockstar123
641418
$endgroup$
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
asked Jun 15 '14 at 14:24
rockstar123rockstar123
641418
asked Jun 15 '14 at 14:24
rockstar123rockstar123
641418
asked Jun 15 '14 at 14:24
rockstar123rockstar123
641418
asked Jun 15 '14 at 14:24
rockstar123rockstar123
641418
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$
x^2+y^2+1=cx
$$
answered Feb 25 '18 at 19:19
Satyajit GhanaSatyajit Ghana
1
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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
$$
x^2+y^2+1=cx
$$
answered Feb 25 '18 at 19:19
Satyajit GhanaSatyajit Ghana
1
$endgroup$
add a comment |
$begingroup$
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
$$
x^2+y^2+1=cx
$$
answered Feb 25 '18 at 19:19
Satyajit GhanaSatyajit Ghana
1
$endgroup$
add a comment |
$begingroup$
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
$$
x^2+y^2+1=cx
$$
answered Feb 25 '18 at 19:19
Satyajit GhanaSatyajit Ghana
1
$endgroup$
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
$$
x^2+y^2+1=cx
$$
answered Feb 25 '18 at 19:19
Satyajit GhanaSatyajit Ghana
1
answered Feb 25 '18 at 19:19
Satyajit GhanaSatyajit Ghana
1
answered Feb 25 '18 at 19:19
Satyajit GhanaSatyajit Ghana
1
answered Feb 25 '18 at 19:19
Satyajit GhanaSatyajit Ghana
1
1
add a comment |
add a comment |
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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