$yz,dx-2xz,dy+(xy-y^3z),dz=0$
$begingroup$
My attempt at the question
What I don't know is do we integrate partially w.r.t. z to find the value of Φ (z) here?
$$
require{begingroup}
begingroup
newcommand{dd}{;mathrm{d}}$$
$$yzdd x-2xz dd y + (xy-y^3z) dd z=0.tag{1}$$
$P=yz$, $Q=-2xz$ and $R=xy-y^3z$
Now
begin{align*}
&Pleft(frac{partial Q}{partial z}-frac{partial R}{partial y}right)+
Qleft(frac{partial R}{partial x}-frac{partial P}{partial z}right)+
Rleft(frac{partial P}{partial y}-frac{partial Q}{partial z}right)\[.8em]
&=P(-2x-x+3y^2z)+Q(y-y)+R(z+2z)\
&=yz(-3x+3y^2z)+(-2xz)(0)+(xy-y^3z)(z+2z)\
&=-3xzy+3y^3z^2+0+3xyz-3y^3z^2\
&=0
end{align*}
Hence equation $(1)$ is integrable.
Let $z$ be a constant in $(1)$, then $dd z=0$.
Integrating both sides
begin{align*}
intfrac1{2xz}dd x-intfrac1{2yz}dd y&=0\
frac1{2z}ln x - frac1z ln y &= const\
frac{x^{1/2}}y &= Phi(z) tag{2}
end{align*}
Differentiating w.r.t $x$, $y$, $z$
$$frac{x^{-1/2}}{2y} dd x + left(-frac{x^{1/2}}{y^2}right)dd y+(-Phi'(z)) dd z = 0 tag{3}$$
From equations $(1)$ and $(3)$
$$frac{yz}{1/(2x^{1/2}y)}=frac{-2xz}{-x^{1/2}/y^2}=frac{xy-y^3z}{-Phi'(z)}.$$
$$endgroup$$
linear-algebra ordinary-differential-equations differential-topology
edited Dec 4 '18 at 9:55
LutzL
56.9k42054
asked Dec 1 '18 at 18:33
Muntaha MunawarMuntaha Munawar
131
$endgroup$
add a comment |
$begingroup$
My attempt at the question
What I don't know is do we integrate partially w.r.t. z to find the value of Φ (z) here?
$$
require{begingroup}
begingroup
newcommand{dd}{;mathrm{d}}$$
$$yzdd x-2xz dd y + (xy-y^3z) dd z=0.tag{1}$$
$P=yz$, $Q=-2xz$ and $R=xy-y^3z$
Now
begin{align*}
&Pleft(frac{partial Q}{partial z}-frac{partial R}{partial y}right)+
Qleft(frac{partial R}{partial x}-frac{partial P}{partial z}right)+
Rleft(frac{partial P}{partial y}-frac{partial Q}{partial z}right)\[.8em]
&=P(-2x-x+3y^2z)+Q(y-y)+R(z+2z)\
&=yz(-3x+3y^2z)+(-2xz)(0)+(xy-y^3z)(z+2z)\
&=-3xzy+3y^3z^2+0+3xyz-3y^3z^2\
&=0
end{align*}
Hence equation $(1)$ is integrable.
Let $z$ be a constant in $(1)$, then $dd z=0$.
Integrating both sides
begin{align*}
intfrac1{2xz}dd x-intfrac1{2yz}dd y&=0\
frac1{2z}ln x - frac1z ln y &= const\
frac{x^{1/2}}y &= Phi(z) tag{2}
end{align*}
Differentiating w.r.t $x$, $y$, $z$
$$frac{x^{-1/2}}{2y} dd x + left(-frac{x^{1/2}}{y^2}right)dd y+(-Phi'(z)) dd z = 0 tag{3}$$
From equations $(1)$ and $(3)$
$$frac{yz}{1/(2x^{1/2}y)}=frac{-2xz}{-x^{1/2}/y^2}=frac{xy-y^3z}{-Phi'(z)}.$$
$$endgroup$$
linear-algebra ordinary-differential-equations differential-topology
edited Dec 4 '18 at 9:55
LutzL
56.9k42054
asked Dec 1 '18 at 18:33
Muntaha MunawarMuntaha Munawar
131
$endgroup$
add a comment |
$begingroup$
My attempt at the question
What I don't know is do we integrate partially w.r.t. z to find the value of Φ (z) here?
$$
require{begingroup}
begingroup
newcommand{dd}{;mathrm{d}}$$
$$yzdd x-2xz dd y + (xy-y^3z) dd z=0.tag{1}$$
$P=yz$, $Q=-2xz$ and $R=xy-y^3z$
Now
begin{align*}
&Pleft(frac{partial Q}{partial z}-frac{partial R}{partial y}right)+
Qleft(frac{partial R}{partial x}-frac{partial P}{partial z}right)+
Rleft(frac{partial P}{partial y}-frac{partial Q}{partial z}right)\[.8em]
&=P(-2x-x+3y^2z)+Q(y-y)+R(z+2z)\
&=yz(-3x+3y^2z)+(-2xz)(0)+(xy-y^3z)(z+2z)\
&=-3xzy+3y^3z^2+0+3xyz-3y^3z^2\
&=0
end{align*}
Hence equation $(1)$ is integrable.
Let $z$ be a constant in $(1)$, then $dd z=0$.
Integrating both sides
begin{align*}
intfrac1{2xz}dd x-intfrac1{2yz}dd y&=0\
frac1{2z}ln x - frac1z ln y &= const\
frac{x^{1/2}}y &= Phi(z) tag{2}
end{align*}
Differentiating w.r.t $x$, $y$, $z$
$$frac{x^{-1/2}}{2y} dd x + left(-frac{x^{1/2}}{y^2}right)dd y+(-Phi'(z)) dd z = 0 tag{3}$$
From equations $(1)$ and $(3)$
$$frac{yz}{1/(2x^{1/2}y)}=frac{-2xz}{-x^{1/2}/y^2}=frac{xy-y^3z}{-Phi'(z)}.$$
$$endgroup$$
linear-algebra ordinary-differential-equations differential-topology
edited Dec 4 '18 at 9:55
LutzL
56.9k42054
asked Dec 1 '18 at 18:33
Muntaha MunawarMuntaha Munawar
131
$endgroup$
My attempt at the question
What I don't know is do we integrate partially w.r.t. z to find the value of Φ (z) here?
$$
require{begingroup}
begingroup
newcommand{dd}{;mathrm{d}}$$
$$yzdd x-2xz dd y + (xy-y^3z) dd z=0.tag{1}$$
$P=yz$, $Q=-2xz$ and $R=xy-y^3z$
Now
begin{align*}
&Pleft(frac{partial Q}{partial z}-frac{partial R}{partial y}right)+
Qleft(frac{partial R}{partial x}-frac{partial P}{partial z}right)+
Rleft(frac{partial P}{partial y}-frac{partial Q}{partial z}right)\[.8em]
&=P(-2x-x+3y^2z)+Q(y-y)+R(z+2z)\
&=yz(-3x+3y^2z)+(-2xz)(0)+(xy-y^3z)(z+2z)\
&=-3xzy+3y^3z^2+0+3xyz-3y^3z^2\
&=0
end{align*}
Hence equation $(1)$ is integrable.
Let $z$ be a constant in $(1)$, then $dd z=0$.
Integrating both sides
begin{align*}
intfrac1{2xz}dd x-intfrac1{2yz}dd y&=0\
frac1{2z}ln x - frac1z ln y &= const\
frac{x^{1/2}}y &= Phi(z) tag{2}
end{align*}
Differentiating w.r.t $x$, $y$, $z$
$$frac{x^{-1/2}}{2y} dd x + left(-frac{x^{1/2}}{y^2}right)dd y+(-Phi'(z)) dd z = 0 tag{3}$$
From equations $(1)$ and $(3)$
$$frac{yz}{1/(2x^{1/2}y)}=frac{-2xz}{-x^{1/2}/y^2}=frac{xy-y^3z}{-Phi'(z)}.$$
$$endgroup$$
linear-algebra ordinary-differential-equations differential-topology
linear-algebra ordinary-differential-equations differential-topology
edited Dec 4 '18 at 9:55
LutzL
56.9k42054
asked Dec 1 '18 at 18:33
Muntaha MunawarMuntaha Munawar
131
edited Dec 4 '18 at 9:55
LutzL
56.9k42054
asked Dec 1 '18 at 18:33
Muntaha MunawarMuntaha Munawar
131
edited Dec 4 '18 at 9:55
LutzL
56.9k42054
edited Dec 4 '18 at 9:55
LutzL
56.9k42054
edited Dec 4 '18 at 9:55
LutzL
56.9k42054
56.9k42054
asked Dec 1 '18 at 18:33
Muntaha MunawarMuntaha Munawar
131
asked Dec 1 '18 at 18:33
Muntaha MunawarMuntaha Munawar
131
asked Dec 1 '18 at 18:33
Muntaha MunawarMuntaha Munawar
131
131
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You need to find an integrating factor. If you treat this problem like a puzzle you can combine like terms to find
$$
y,d(xz)-2(xz),dy=y^3zdz
$$
and from that that you get an integrable expression by dividing by $y^3$, that is, the integrating factor should turn out to be $y^{-3}$.
In your solution, I'd have avoided the square root to get
$$
xy^{-2}=Phi(z)
$$
as the equation for the solution surface.
Then the derivative of that is
$$
y^{-2},dx-2xy^{-3},dy =Phi'(z),dz\
yz,dx - 2xz,dy = y^3zPhi'(z),dz\
implies
y^3zPhi'(z),dz=(y^3z-xy),dzimplies Phi'(z)=1-z^{-1}Phi(z)
$$
which can now be easily solved.
In your version with $Φ(z)=frac{x^{1/2}}y$, pick one of the relations and eliminate $y$ against $Φ$. The first two fractions are equal after simplification, the equality to the last term gives
$$
2x^{1/2}y^2z=y^3frac{Φ(z)^2-z}{-Φ'(z)}iff 2zΦ'(z)Φ(z)=z-Φ(z)^2implies zΦ(z)^2=frac12z^2+C
$$
answered Dec 1 '18 at 18:42
LutzLLutzL
56.9k42054
$endgroup$
$begingroup$
Thanks a lot ! :)
$endgroup$
– Muntaha Munawar
Dec 4 '18 at 9:26
add a comment |
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$begingroup$
You need to find an integrating factor. If you treat this problem like a puzzle you can combine like terms to find
$$
y,d(xz)-2(xz),dy=y^3zdz
$$
and from that that you get an integrable expression by dividing by $y^3$, that is, the integrating factor should turn out to be $y^{-3}$.
In your solution, I'd have avoided the square root to get
$$
xy^{-2}=Phi(z)
$$
as the equation for the solution surface.
Then the derivative of that is
$$
y^{-2},dx-2xy^{-3},dy =Phi'(z),dz\
yz,dx - 2xz,dy = y^3zPhi'(z),dz\
implies
y^3zPhi'(z),dz=(y^3z-xy),dzimplies Phi'(z)=1-z^{-1}Phi(z)
$$
which can now be easily solved.
In your version with $Φ(z)=frac{x^{1/2}}y$, pick one of the relations and eliminate $y$ against $Φ$. The first two fractions are equal after simplification, the equality to the last term gives
$$
2x^{1/2}y^2z=y^3frac{Φ(z)^2-z}{-Φ'(z)}iff 2zΦ'(z)Φ(z)=z-Φ(z)^2implies zΦ(z)^2=frac12z^2+C
$$
answered Dec 1 '18 at 18:42
LutzLLutzL
56.9k42054
$endgroup$
$begingroup$
Thanks a lot ! :)
$endgroup$
– Muntaha Munawar
Dec 4 '18 at 9:26
add a comment |
$begingroup$
You need to find an integrating factor. If you treat this problem like a puzzle you can combine like terms to find
$$
y,d(xz)-2(xz),dy=y^3zdz
$$
and from that that you get an integrable expression by dividing by $y^3$, that is, the integrating factor should turn out to be $y^{-3}$.
In your solution, I'd have avoided the square root to get
$$
xy^{-2}=Phi(z)
$$
as the equation for the solution surface.
Then the derivative of that is
$$
y^{-2},dx-2xy^{-3},dy =Phi'(z),dz\
yz,dx - 2xz,dy = y^3zPhi'(z),dz\
implies
y^3zPhi'(z),dz=(y^3z-xy),dzimplies Phi'(z)=1-z^{-1}Phi(z)
$$
which can now be easily solved.
In your version with $Φ(z)=frac{x^{1/2}}y$, pick one of the relations and eliminate $y$ against $Φ$. The first two fractions are equal after simplification, the equality to the last term gives
$$
2x^{1/2}y^2z=y^3frac{Φ(z)^2-z}{-Φ'(z)}iff 2zΦ'(z)Φ(z)=z-Φ(z)^2implies zΦ(z)^2=frac12z^2+C
$$
answered Dec 1 '18 at 18:42
LutzLLutzL
56.9k42054
$endgroup$
$begingroup$
Thanks a lot ! :)
$endgroup$
– Muntaha Munawar
Dec 4 '18 at 9:26
add a comment |
$begingroup$
You need to find an integrating factor. If you treat this problem like a puzzle you can combine like terms to find
$$
y,d(xz)-2(xz),dy=y^3zdz
$$
and from that that you get an integrable expression by dividing by $y^3$, that is, the integrating factor should turn out to be $y^{-3}$.
In your solution, I'd have avoided the square root to get
$$
xy^{-2}=Phi(z)
$$
as the equation for the solution surface.
Then the derivative of that is
$$
y^{-2},dx-2xy^{-3},dy =Phi'(z),dz\
yz,dx - 2xz,dy = y^3zPhi'(z),dz\
implies
y^3zPhi'(z),dz=(y^3z-xy),dzimplies Phi'(z)=1-z^{-1}Phi(z)
$$
which can now be easily solved.
In your version with $Φ(z)=frac{x^{1/2}}y$, pick one of the relations and eliminate $y$ against $Φ$. The first two fractions are equal after simplification, the equality to the last term gives
$$
2x^{1/2}y^2z=y^3frac{Φ(z)^2-z}{-Φ'(z)}iff 2zΦ'(z)Φ(z)=z-Φ(z)^2implies zΦ(z)^2=frac12z^2+C
$$
answered Dec 1 '18 at 18:42
LutzLLutzL
56.9k42054
$endgroup$
You need to find an integrating factor. If you treat this problem like a puzzle you can combine like terms to find
$$
y,d(xz)-2(xz),dy=y^3zdz
$$
and from that that you get an integrable expression by dividing by $y^3$, that is, the integrating factor should turn out to be $y^{-3}$.
In your solution, I'd have avoided the square root to get
$$
xy^{-2}=Phi(z)
$$
as the equation for the solution surface.
Then the derivative of that is
$$
y^{-2},dx-2xy^{-3},dy =Phi'(z),dz\
yz,dx - 2xz,dy = y^3zPhi'(z),dz\
implies
y^3zPhi'(z),dz=(y^3z-xy),dzimplies Phi'(z)=1-z^{-1}Phi(z)
$$
which can now be easily solved.
In your version with $Φ(z)=frac{x^{1/2}}y$, pick one of the relations and eliminate $y$ against $Φ$. The first two fractions are equal after simplification, the equality to the last term gives
$$
2x^{1/2}y^2z=y^3frac{Φ(z)^2-z}{-Φ'(z)}iff 2zΦ'(z)Φ(z)=z-Φ(z)^2implies zΦ(z)^2=frac12z^2+C
$$
answered Dec 1 '18 at 18:42
LutzLLutzL
56.9k42054
edited Dec 4 '18 at 10:05
answered Dec 1 '18 at 18:42
LutzLLutzL
56.9k42054
answered Dec 1 '18 at 18:42
LutzLLutzL
56.9k42054
answered Dec 1 '18 at 18:42
LutzLLutzL
56.9k42054
56.9k42054
$begingroup$
Thanks a lot ! :)
$endgroup$
– Muntaha Munawar
Dec 4 '18 at 9:26
add a comment |
$begingroup$
Thanks a lot ! :)
$endgroup$
– Muntaha Munawar
Dec 4 '18 at 9:26
$begingroup$
Thanks a lot ! :)
$endgroup$
– Muntaha Munawar
Dec 4 '18 at 9:26
$begingroup$
Thanks a lot ! :)
$endgroup$
– Muntaha Munawar
Dec 4 '18 at 9:26
add a comment |
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