$yz,dx-2xz,dy+(xy-y^3z),dz=0$












0














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 differential-equations differential-topology







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    0














    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 differential-equations differential-topology







    share|cite|improve this question



























      0












      0








      0


      1





      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 differential-equations differential-topology







      share|cite|improve this question















      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 differential-equations differential-topology




      linear-algebra differential-equations differential-topology






      share|cite|improve this question















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      edited Dec 4 at 9:55









      LutzL

      55.8k42054




      55.8k42054










      asked Dec 1 at 18:33









      Muntaha Munawar

      131




      131






















          1 Answer
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          1














          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
          $$
          2x^{1/2}y^2z=y^3frac{Φ(z)^2-z}{-Φ'(z)}iff 2zΦ'(z)Φ(z)=z-Φ(z)^2implies zΦ(z)^2=frac12z^2+C
          $$






          share|cite|improve this answer























          • Thanks a lot ! :)
            – Muntaha Munawar
            Dec 4 at 9:26











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          active

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          1














          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
          $$
          2x^{1/2}y^2z=y^3frac{Φ(z)^2-z}{-Φ'(z)}iff 2zΦ'(z)Φ(z)=z-Φ(z)^2implies zΦ(z)^2=frac12z^2+C
          $$






          share|cite|improve this answer























          • Thanks a lot ! :)
            – Muntaha Munawar
            Dec 4 at 9:26
















          1














          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
          $$
          2x^{1/2}y^2z=y^3frac{Φ(z)^2-z}{-Φ'(z)}iff 2zΦ'(z)Φ(z)=z-Φ(z)^2implies zΦ(z)^2=frac12z^2+C
          $$






          share|cite|improve this answer























          • Thanks a lot ! :)
            – Muntaha Munawar
            Dec 4 at 9:26














          1












          1








          1






          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
          $$
          2x^{1/2}y^2z=y^3frac{Φ(z)^2-z}{-Φ'(z)}iff 2zΦ'(z)Φ(z)=z-Φ(z)^2implies zΦ(z)^2=frac12z^2+C
          $$






          share|cite|improve this answer














          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
          $$
          2x^{1/2}y^2z=y^3frac{Φ(z)^2-z}{-Φ'(z)}iff 2zΦ'(z)Φ(z)=z-Φ(z)^2implies zΦ(z)^2=frac12z^2+C
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 at 10:05

























          answered Dec 1 at 18:42









          LutzL

          55.8k42054




          55.8k42054












          • Thanks a lot ! :)
            – Muntaha Munawar
            Dec 4 at 9:26


















          • Thanks a lot ! :)
            – Muntaha Munawar
            Dec 4 at 9:26
















          Thanks a lot ! :)
          – Muntaha Munawar
          Dec 4 at 9:26




          Thanks a lot ! :)
          – Muntaha Munawar
          Dec 4 at 9:26


















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