Solve partial differential equation by Charpit's equations

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Solve partial differential equation, $$F=xp^2-ypq+y^2q-y^2z=0,;;;; p = fracpartial zpartial x,;;;;q = fracpartial zpartial y$$




My attempt:



Charpit's Equations:



$$fracdxyq-2px = fracdyy(p-y) = fracdpp(p-y^2) = fracdqq(2y-p)-yq(p-y)-2yz $$



Taking $dp$, $dy$ terms: $$(p^2-py^2)dy-(py-y^2)dp=0 \ p(-pdy+ydp)+y^2(pdy-dp) =0\d(p/y)+dy-fracdpp=0 \ e^fracpye^y=p c_1 \$$



where $c_1$ is some arbitrary constant



Now how to proceed further?




pde characteristics




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    Solve partial differential equation, $$F=xp^2-ypq+y^2q-y^2z=0,;;;; p = fracpartial zpartial x,;;;;q = fracpartial zpartial y$$




    My attempt:



    Charpit's Equations:



    $$fracdxyq-2px = fracdyy(p-y) = fracdpp(p-y^2) = fracdqq(2y-p)-yq(p-y)-2yz $$



    Taking $dp$, $dy$ terms: $$(p^2-py^2)dy-(py-y^2)dp=0 \ p(-pdy+ydp)+y^2(pdy-dp) =0\d(p/y)+dy-fracdpp=0 \ e^fracpye^y=p c_1 \$$



    where $c_1$ is some arbitrary constant



    Now how to proceed further?




    pde characteristics




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      Solve partial differential equation, $$F=xp^2-ypq+y^2q-y^2z=0,;;;; p = fracpartial zpartial x,;;;;q = fracpartial zpartial y$$




      My attempt:



      Charpit's Equations:



      $$fracdxyq-2px = fracdyy(p-y) = fracdpp(p-y^2) = fracdqq(2y-p)-yq(p-y)-2yz $$



      Taking $dp$, $dy$ terms: $$(p^2-py^2)dy-(py-y^2)dp=0 \ p(-pdy+ydp)+y^2(pdy-dp) =0\d(p/y)+dy-fracdpp=0 \ e^fracpye^y=p c_1 \$$



      where $c_1$ is some arbitrary constant



      Now how to proceed further?




      pde characteristics




      share|cite|improve this question














      Solve partial differential equation, $$F=xp^2-ypq+y^2q-y^2z=0,;;;; p = fracpartial zpartial x,;;;;q = fracpartial zpartial y$$




      My attempt:



      Charpit's Equations:



      $$fracdxyq-2px = fracdyy(p-y) = fracdpp(p-y^2) = fracdqq(2y-p)-yq(p-y)-2yz $$



      Taking $dp$, $dy$ terms: $$(p^2-py^2)dy-(py-y^2)dp=0 \ p(-pdy+ydp)+y^2(pdy-dp) =0\d(p/y)+dy-fracdpp=0 \ e^fracpye^y=p c_1 \$$



      where $c_1$ is some arbitrary constant



      Now how to proceed further?





      pde characteristics





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited 6 hours ago









      Davide Morgante

      1,566118




      1,566118









      asked 7 hours ago









      Magneto

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