Particular integral of Partial DIfferential Equation












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I've found this equation related Homogeneous PDE $$D^2-D'^2 = x-y$$
I tried to solve it using General Rule for PDE but i got different result $$P.I. = {1over (D+D')*(D-D')} cdot(x-y)$$ $$= {1over(D+D')}int_{}^{} (x-c+x)dx $$, where y = c - x $$= int -xy dx$$$$ =int -x(c+x)dx $$ where y = c + x



But in my textbook they solved it another way and got different result. Am i doing any wrong?






pde







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  • $begingroup$
    $D^2-D'^2=x-yquad$ is a non-sens since the operators $D$ and $D'$ are acting on no function.
    $endgroup$
    – JJacquelin
    Jan 10 '17 at 7:48


















0












$begingroup$


I've found this equation related Homogeneous PDE $$D^2-D'^2 = x-y$$
I tried to solve it using General Rule for PDE but i got different result $$P.I. = {1over (D+D')*(D-D')} cdot(x-y)$$ $$= {1over(D+D')}int_{}^{} (x-c+x)dx $$, where y = c - x $$= int -xy dx$$$$ =int -x(c+x)dx $$ where y = c + x



But in my textbook they solved it another way and got different result. Am i doing any wrong?






pde







share|cite|improve this question









$endgroup$












  • $begingroup$
    $D^2-D'^2=x-yquad$ is a non-sens since the operators $D$ and $D'$ are acting on no function.
    $endgroup$
    – JJacquelin
    Jan 10 '17 at 7:48
















0












0








0





$begingroup$


I've found this equation related Homogeneous PDE $$D^2-D'^2 = x-y$$
I tried to solve it using General Rule for PDE but i got different result $$P.I. = {1over (D+D')*(D-D')} cdot(x-y)$$ $$= {1over(D+D')}int_{}^{} (x-c+x)dx $$, where y = c - x $$= int -xy dx$$$$ =int -x(c+x)dx $$ where y = c + x



But in my textbook they solved it another way and got different result. Am i doing any wrong?






pde







share|cite|improve this question









$endgroup$




I've found this equation related Homogeneous PDE $$D^2-D'^2 = x-y$$
I tried to solve it using General Rule for PDE but i got different result $$P.I. = {1over (D+D')*(D-D')} cdot(x-y)$$ $$= {1over(D+D')}int_{}^{} (x-c+x)dx $$, where y = c - x $$= int -xy dx$$$$ =int -x(c+x)dx $$ where y = c + x



But in my textbook they solved it another way and got different result. Am i doing any wrong?






pde




pde






share|cite|improve this question













share|cite|improve this question











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asked Jan 9 '17 at 14:27









IAmBlakeIAmBlake

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70113












  • $begingroup$
    $D^2-D'^2=x-yquad$ is a non-sens since the operators $D$ and $D'$ are acting on no function.
    $endgroup$
    – JJacquelin
    Jan 10 '17 at 7:48




















  • $begingroup$
    $D^2-D'^2=x-yquad$ is a non-sens since the operators $D$ and $D'$ are acting on no function.
    $endgroup$
    – JJacquelin
    Jan 10 '17 at 7:48


















$begingroup$
$D^2-D'^2=x-yquad$ is a non-sens since the operators $D$ and $D'$ are acting on no function.
$endgroup$
– JJacquelin
Jan 10 '17 at 7:48






$begingroup$
$D^2-D'^2=x-yquad$ is a non-sens since the operators $D$ and $D'$ are acting on no function.
$endgroup$
– JJacquelin
Jan 10 '17 at 7:48












1 Answer
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Supposing that the PDE is :
$$frac{partial^2 z(x,y)}{partial x^2} -frac{partial^2 z(x,y)}{partial y^2}=x-y$$
HINT :



First solve the associated homogeneous PDE $quad frac{partial^2 Z}{partial x^2} -frac{partial^2 Z}{partial y^2}=0$
$$Z(x,y)=f(x-y)+g(x+y)$$
with any differentiable functions $f$ and $g$.



Second, find a particular solution of the inhomogeneous PDE. For example, thanks to the separation of variables method.



Finally : $quad z(x,y)=f(x-y)+g(x+y)+frac{1}{6}(x^3-y^3)$






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    1 Answer
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    $begingroup$

    Supposing that the PDE is :
    $$frac{partial^2 z(x,y)}{partial x^2} -frac{partial^2 z(x,y)}{partial y^2}=x-y$$
    HINT :



    First solve the associated homogeneous PDE $quad frac{partial^2 Z}{partial x^2} -frac{partial^2 Z}{partial y^2}=0$
    $$Z(x,y)=f(x-y)+g(x+y)$$
    with any differentiable functions $f$ and $g$.



    Second, find a particular solution of the inhomogeneous PDE. For example, thanks to the separation of variables method.



    Finally : $quad z(x,y)=f(x-y)+g(x+y)+frac{1}{6}(x^3-y^3)$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Supposing that the PDE is :
      $$frac{partial^2 z(x,y)}{partial x^2} -frac{partial^2 z(x,y)}{partial y^2}=x-y$$
      HINT :



      First solve the associated homogeneous PDE $quad frac{partial^2 Z}{partial x^2} -frac{partial^2 Z}{partial y^2}=0$
      $$Z(x,y)=f(x-y)+g(x+y)$$
      with any differentiable functions $f$ and $g$.



      Second, find a particular solution of the inhomogeneous PDE. For example, thanks to the separation of variables method.



      Finally : $quad z(x,y)=f(x-y)+g(x+y)+frac{1}{6}(x^3-y^3)$






      share|cite|improve this answer









      $endgroup$
















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        0





        $begingroup$

        Supposing that the PDE is :
        $$frac{partial^2 z(x,y)}{partial x^2} -frac{partial^2 z(x,y)}{partial y^2}=x-y$$
        HINT :



        First solve the associated homogeneous PDE $quad frac{partial^2 Z}{partial x^2} -frac{partial^2 Z}{partial y^2}=0$
        $$Z(x,y)=f(x-y)+g(x+y)$$
        with any differentiable functions $f$ and $g$.



        Second, find a particular solution of the inhomogeneous PDE. For example, thanks to the separation of variables method.



        Finally : $quad z(x,y)=f(x-y)+g(x+y)+frac{1}{6}(x^3-y^3)$






        share|cite|improve this answer









        $endgroup$



        Supposing that the PDE is :
        $$frac{partial^2 z(x,y)}{partial x^2} -frac{partial^2 z(x,y)}{partial y^2}=x-y$$
        HINT :



        First solve the associated homogeneous PDE $quad frac{partial^2 Z}{partial x^2} -frac{partial^2 Z}{partial y^2}=0$
        $$Z(x,y)=f(x-y)+g(x+y)$$
        with any differentiable functions $f$ and $g$.



        Second, find a particular solution of the inhomogeneous PDE. For example, thanks to the separation of variables method.



        Finally : $quad z(x,y)=f(x-y)+g(x+y)+frac{1}{6}(x^3-y^3)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 '17 at 7:49









        JJacquelinJJacquelin

        45.6k21857




        45.6k21857






























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