Non-linear partial differential equation of order 1












2












$begingroup$


Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$



where $displaystyle p = frac{partial z}{partial x}$, $,displaystyle q = frac{partial z}{partial y}$



$f_p=2q+2x\ f_q=2p+2x\ f_z=0\ f_x=2q+2x\ f_y=2p+2y$



Using Charpit's method



$$frac{dp}{2q+2x}=frac{dq}{2p+2y}=frac{dz}{-4pq-2py-2qx}=frac{dx}{-2q-2y}=frac{dy}{-2p-2x}$$



I deduce $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$



I am not sure how to proceed from here. Kinda stuck.






pde







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  • 2




    $begingroup$
    Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
    $endgroup$
    – Dylan
    Jan 15 at 15:54






  • 1




    $begingroup$
    I will keep that in mind in future.
    $endgroup$
    – Piyush Divyanakar
    Jan 15 at 16:11
















2












$begingroup$


Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$



where $displaystyle p = frac{partial z}{partial x}$, $,displaystyle q = frac{partial z}{partial y}$



$f_p=2q+2x\ f_q=2p+2x\ f_z=0\ f_x=2q+2x\ f_y=2p+2y$



Using Charpit's method



$$frac{dp}{2q+2x}=frac{dq}{2p+2y}=frac{dz}{-4pq-2py-2qx}=frac{dx}{-2q-2y}=frac{dy}{-2p-2x}$$



I deduce $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$



I am not sure how to proceed from here. Kinda stuck.






pde







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
    $endgroup$
    – Dylan
    Jan 15 at 15:54






  • 1




    $begingroup$
    I will keep that in mind in future.
    $endgroup$
    – Piyush Divyanakar
    Jan 15 at 16:11














2












2








2


2



$begingroup$


Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$



where $displaystyle p = frac{partial z}{partial x}$, $,displaystyle q = frac{partial z}{partial y}$



$f_p=2q+2x\ f_q=2p+2x\ f_z=0\ f_x=2q+2x\ f_y=2p+2y$



Using Charpit's method



$$frac{dp}{2q+2x}=frac{dq}{2p+2y}=frac{dz}{-4pq-2py-2qx}=frac{dx}{-2q-2y}=frac{dy}{-2p-2x}$$



I deduce $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$



I am not sure how to proceed from here. Kinda stuck.






pde







share|cite|improve this question











$endgroup$




Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$



where $displaystyle p = frac{partial z}{partial x}$, $,displaystyle q = frac{partial z}{partial y}$



$f_p=2q+2x\ f_q=2p+2x\ f_z=0\ f_x=2q+2x\ f_y=2p+2y$



Using Charpit's method



$$frac{dp}{2q+2x}=frac{dq}{2p+2y}=frac{dz}{-4pq-2py-2qx}=frac{dx}{-2q-2y}=frac{dy}{-2p-2x}$$



I deduce $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$



I am not sure how to proceed from here. Kinda stuck.






pde




pde






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 17:25









El borito

714216




714216










asked Jan 15 at 7:17









Piyush DivyanakarPiyush Divyanakar

3,338325




3,338325








  • 2




    $begingroup$
    Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
    $endgroup$
    – Dylan
    Jan 15 at 15:54






  • 1




    $begingroup$
    I will keep that in mind in future.
    $endgroup$
    – Piyush Divyanakar
    Jan 15 at 16:11














  • 2




    $begingroup$
    Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
    $endgroup$
    – Dylan
    Jan 15 at 15:54






  • 1




    $begingroup$
    I will keep that in mind in future.
    $endgroup$
    – Piyush Divyanakar
    Jan 15 at 16:11








2




2




$begingroup$
Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
$endgroup$
– Dylan
Jan 15 at 15:54




$begingroup$
Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
$endgroup$
– Dylan
Jan 15 at 15:54




1




1




$begingroup$
I will keep that in mind in future.
$endgroup$
– Piyush Divyanakar
Jan 15 at 16:11




$begingroup$
I will keep that in mind in future.
$endgroup$
– Piyush Divyanakar
Jan 15 at 16:11










1 Answer
1






active

oldest

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1












$begingroup$

$$2(pq+py+qx)+x^2+y^2=0$$
You got to :
$$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I checked and agree.



HINT :



Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$.



For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$.



Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.



$z=int p(x,y)dx+F(y)$ and $z=int q(x,y)dy+G(x)$



Determine $F(y)$ and $G(x)$ to make them consistent.



The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :



$$z(x,y)=u(x,y)-frac12 x^2-frac12 y^2$$



$p=z_x=u_x-x=P-x$



$q=z_y=u_y-y=Q-y$



$$2left((P-x)(Q-y)+(P-x)y+(Q-y)xright)+x^2+y^2=0$$
After simplification :
$$PQ+frac12(x-y)^2=0$$



$$u_xu_y+frac12(x-y)^2=0$$
$$frac{dx}{Q}=frac{dy}{P}=frac{du}{2PQ}=frac{dP}{y-x}=frac{dQ}{x-y}=ds$$
$$begin{cases}
P+Q=c_1 \
PQ=-frac12(x-y)^2
end{cases}$$
etc.



Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .



Note: It is a pity that no boundary condition be specified.






share|cite|improve this answer











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    1












    $begingroup$

    $$2(pq+py+qx)+x^2+y^2=0$$
    You got to :
    $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
    I checked and agree.



    HINT :



    Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$.



    For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$.



    Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.



    $z=int p(x,y)dx+F(y)$ and $z=int q(x,y)dy+G(x)$



    Determine $F(y)$ and $G(x)$ to make them consistent.



    The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :



    $$z(x,y)=u(x,y)-frac12 x^2-frac12 y^2$$



    $p=z_x=u_x-x=P-x$



    $q=z_y=u_y-y=Q-y$



    $$2left((P-x)(Q-y)+(P-x)y+(Q-y)xright)+x^2+y^2=0$$
    After simplification :
    $$PQ+frac12(x-y)^2=0$$



    $$u_xu_y+frac12(x-y)^2=0$$
    $$frac{dx}{Q}=frac{dy}{P}=frac{du}{2PQ}=frac{dP}{y-x}=frac{dQ}{x-y}=ds$$
    $$begin{cases}
    P+Q=c_1 \
    PQ=-frac12(x-y)^2
    end{cases}$$
    etc.



    Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .



    Note: It is a pity that no boundary condition be specified.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      $$2(pq+py+qx)+x^2+y^2=0$$
      You got to :
      $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
      I checked and agree.



      HINT :



      Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$.



      For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$.



      Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.



      $z=int p(x,y)dx+F(y)$ and $z=int q(x,y)dy+G(x)$



      Determine $F(y)$ and $G(x)$ to make them consistent.



      The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :



      $$z(x,y)=u(x,y)-frac12 x^2-frac12 y^2$$



      $p=z_x=u_x-x=P-x$



      $q=z_y=u_y-y=Q-y$



      $$2left((P-x)(Q-y)+(P-x)y+(Q-y)xright)+x^2+y^2=0$$
      After simplification :
      $$PQ+frac12(x-y)^2=0$$



      $$u_xu_y+frac12(x-y)^2=0$$
      $$frac{dx}{Q}=frac{dy}{P}=frac{du}{2PQ}=frac{dP}{y-x}=frac{dQ}{x-y}=ds$$
      $$begin{cases}
      P+Q=c_1 \
      PQ=-frac12(x-y)^2
      end{cases}$$
      etc.



      Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .



      Note: It is a pity that no boundary condition be specified.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        $$2(pq+py+qx)+x^2+y^2=0$$
        You got to :
        $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
        I checked and agree.



        HINT :



        Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$.



        For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$.



        Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.



        $z=int p(x,y)dx+F(y)$ and $z=int q(x,y)dy+G(x)$



        Determine $F(y)$ and $G(x)$ to make them consistent.



        The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :



        $$z(x,y)=u(x,y)-frac12 x^2-frac12 y^2$$



        $p=z_x=u_x-x=P-x$



        $q=z_y=u_y-y=Q-y$



        $$2left((P-x)(Q-y)+(P-x)y+(Q-y)xright)+x^2+y^2=0$$
        After simplification :
        $$PQ+frac12(x-y)^2=0$$



        $$u_xu_y+frac12(x-y)^2=0$$
        $$frac{dx}{Q}=frac{dy}{P}=frac{du}{2PQ}=frac{dP}{y-x}=frac{dQ}{x-y}=ds$$
        $$begin{cases}
        P+Q=c_1 \
        PQ=-frac12(x-y)^2
        end{cases}$$
        etc.



        Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .



        Note: It is a pity that no boundary condition be specified.






        share|cite|improve this answer











        $endgroup$



        $$2(pq+py+qx)+x^2+y^2=0$$
        You got to :
        $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
        I checked and agree.



        HINT :



        Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$.



        For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$.



        Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.



        $z=int p(x,y)dx+F(y)$ and $z=int q(x,y)dy+G(x)$



        Determine $F(y)$ and $G(x)$ to make them consistent.



        The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :



        $$z(x,y)=u(x,y)-frac12 x^2-frac12 y^2$$



        $p=z_x=u_x-x=P-x$



        $q=z_y=u_y-y=Q-y$



        $$2left((P-x)(Q-y)+(P-x)y+(Q-y)xright)+x^2+y^2=0$$
        After simplification :
        $$PQ+frac12(x-y)^2=0$$



        $$u_xu_y+frac12(x-y)^2=0$$
        $$frac{dx}{Q}=frac{dy}{P}=frac{du}{2PQ}=frac{dP}{y-x}=frac{dQ}{x-y}=ds$$
        $$begin{cases}
        P+Q=c_1 \
        PQ=-frac12(x-y)^2
        end{cases}$$
        etc.



        Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .



        Note: It is a pity that no boundary condition be specified.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 20 at 11:18

























        answered Jan 20 at 11:07









        JJacquelinJJacquelin

        43.5k21853




        43.5k21853






























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