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
asked Jan 9 '17 at 14:27
IAmBlakeIAmBlake
70113
$endgroup$
add a comment |
$begingroup$
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
asked Jan 9 '17 at 14:27
IAmBlakeIAmBlake
70113
$endgroup$
$begingroup$
$D^2-D'^2=x-yquad$ is a non-sens since the operators $D$ and $D'$ are acting on no function.
$endgroup$
– JJacquelin
Jan 10 '17 at 7:48
add a comment |
$begingroup$
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
asked Jan 9 '17 at 14:27
IAmBlakeIAmBlake
70113
$endgroup$
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
asked Jan 9 '17 at 14:27
IAmBlakeIAmBlake
70113
asked Jan 9 '17 at 14:27
IAmBlakeIAmBlake
70113
asked Jan 9 '17 at 14:27
IAmBlakeIAmBlake
70113
asked Jan 9 '17 at 14:27
IAmBlakeIAmBlake
70113
asked Jan 9 '17 at 14:27
IAmBlakeIAmBlake
70113
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
add a comment |
$begingroup$
$D^2-D'^2=x-yquad$ 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
add a comment |
1 Answer
1
active
oldest
votes
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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 : $quad z(x,y)=f(x-y)+g(x+y)+frac{1}{6}(x^3-y^3)$
answered Jan 10 '17 at 7:49
JJacquelinJJacquelin
45.6k21857
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add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
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 : $quad z(x,y)=f(x-y)+g(x+y)+frac{1}{6}(x^3-y^3)$
answered Jan 10 '17 at 7:49
JJacquelinJJacquelin
45.6k21857
$endgroup$
add a comment |
$begingroup$
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 : $quad z(x,y)=f(x-y)+g(x+y)+frac{1}{6}(x^3-y^3)$
answered Jan 10 '17 at 7:49
JJacquelinJJacquelin
45.6k21857
$endgroup$
add a comment |
$begingroup$
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 : $quad z(x,y)=f(x-y)+g(x+y)+frac{1}{6}(x^3-y^3)$
answered Jan 10 '17 at 7:49
JJacquelinJJacquelin
45.6k21857
$endgroup$
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 : $quad z(x,y)=f(x-y)+g(x+y)+frac{1}{6}(x^3-y^3)$
answered Jan 10 '17 at 7:49
JJacquelinJJacquelin
45.6k21857
answered Jan 10 '17 at 7:49
JJacquelinJJacquelin
45.6k21857
answered Jan 10 '17 at 7:49
JJacquelinJJacquelin
45.6k21857
answered Jan 10 '17 at 7:49
JJacquelinJJacquelin
45.6k21857
45.6k21857
add a comment |
add a comment |
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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