Simultaneous P.D.E. problem Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)P.D.E question. Constant Dirichlet'e EnergyHelp needed in solving a differential equationUsing a multivariate chain rule to solve a partial differential equationA system of simultaneous equationseasy simultaneous equation, please solveFinding X or Y first in simultaneous equationSimultaneous equation involving reciprocalsHow do I proceed from here solving partial differential equation with boundary conditionsSolving quasilinear p.d.e. with method of charactersticsP.D.E. problem Jacobi's Method
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Simultaneous P.D.E. problem
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)P.D.E question. Constant Dirichlet'e EnergyHelp needed in solving a differential equationUsing a multivariate chain rule to solve a partial differential equationA system of simultaneous equationseasy simultaneous equation, please solveFinding X or Y first in simultaneous equationSimultaneous equation involving reciprocalsHow do I proceed from here solving partial differential equation with boundary conditionsSolving quasilinear p.d.e. with method of charactersticsP.D.E. problem Jacobi's Method
$begingroup$
Solve the following differential equation:
$$ fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2) $$
I have so far got
$x^2dx = y^2dy $ which implies that $(x-y)(x^2 +xy + y^2) = a$ ...(1)
Then $$fracdx+ dy(x-y) = fracdzz$$
Using the equation (1), I get
$frac1a (x^2+xy+y^2)(dx+dy) = fracdzz$
How do I proceed from this point? I could have solved this question easily if the third fraction in the given P.D.E. were
$fracdzz(x^2 - y^2)$ but I need help for this problem. Please help me out.
pde systems-of-equations
edited Apr 3 at 17:28
Harry49
8,90331346
asked Apr 2 at 9:44
Vaibhav RamaniVaibhav Ramani
164
$endgroup$
add a comment |
$begingroup$
Solve the following differential equation:
$$ fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2) $$
I have so far got
$x^2dx = y^2dy $ which implies that $(x-y)(x^2 +xy + y^2) = a$ ...(1)
Then $$fracdx+ dy(x-y) = fracdzz$$
Using the equation (1), I get
$frac1a (x^2+xy+y^2)(dx+dy) = fracdzz$
How do I proceed from this point? I could have solved this question easily if the third fraction in the given P.D.E. were
$fracdzz(x^2 - y^2)$ but I need help for this problem. Please help me out.
pde systems-of-equations
edited Apr 3 at 17:28
Harry49
8,90331346
asked Apr 2 at 9:44
Vaibhav RamaniVaibhav Ramani
164
$endgroup$
$begingroup$
This appeared in my assignment in the subject of differential equations. The question as given in it is: Solve the following differential equation: $fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2)$
$endgroup$
– Vaibhav Ramani
Apr 3 at 15:06
add a comment |
$begingroup$
Solve the following differential equation:
$$ fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2) $$
I have so far got
$x^2dx = y^2dy $ which implies that $(x-y)(x^2 +xy + y^2) = a$ ...(1)
Then $$fracdx+ dy(x-y) = fracdzz$$
Using the equation (1), I get
$frac1a (x^2+xy+y^2)(dx+dy) = fracdzz$
How do I proceed from this point? I could have solved this question easily if the third fraction in the given P.D.E. were
$fracdzz(x^2 - y^2)$ but I need help for this problem. Please help me out.
pde systems-of-equations
edited Apr 3 at 17:28
Harry49
8,90331346
asked Apr 2 at 9:44
Vaibhav RamaniVaibhav Ramani
164
$endgroup$
Solve the following differential equation:
$$ fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2) $$
I have so far got
$x^2dx = y^2dy $ which implies that $(x-y)(x^2 +xy + y^2) = a$ ...(1)
Then $$fracdx+ dy(x-y) = fracdzz$$
Using the equation (1), I get
$frac1a (x^2+xy+y^2)(dx+dy) = fracdzz$
How do I proceed from this point? I could have solved this question easily if the third fraction in the given P.D.E. were
$fracdzz(x^2 - y^2)$ but I need help for this problem. Please help me out.
pde systems-of-equations
pde systems-of-equations
edited Apr 3 at 17:28
Harry49
8,90331346
asked Apr 2 at 9:44
Vaibhav RamaniVaibhav Ramani
164
edited Apr 3 at 17:28
Harry49
8,90331346
asked Apr 2 at 9:44
Vaibhav RamaniVaibhav Ramani
164
edited Apr 3 at 17:28
Harry49
8,90331346
edited Apr 3 at 17:28
Harry49
8,90331346
edited Apr 3 at 17:28
Harry49
8,90331346
8,90331346
asked Apr 2 at 9:44
Vaibhav RamaniVaibhav Ramani
164
asked Apr 2 at 9:44
Vaibhav RamaniVaibhav Ramani
164
asked Apr 2 at 9:44
Vaibhav RamaniVaibhav Ramani
164
164
$begingroup$
This appeared in my assignment in the subject of differential equations. The question as given in it is: Solve the following differential equation: $fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2)$
$endgroup$
– Vaibhav Ramani
Apr 3 at 15:06
add a comment |
$begingroup$
This appeared in my assignment in the subject of differential equations. The question as given in it is: Solve the following differential equation: $fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2)$
$endgroup$
– Vaibhav Ramani
Apr 3 at 15:06
$begingroup$
This appeared in my assignment in the subject of differential equations. The question as given in it is: Solve the following differential equation: $fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2)$
$endgroup$
– Vaibhav Ramani
Apr 3 at 15:06
$begingroup$
This appeared in my assignment in the subject of differential equations. The question as given in it is: Solve the following differential equation: $fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2)$
$endgroup$
– Vaibhav Ramani
Apr 3 at 15:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$ fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2) $$
This system of ODEs might come from solving the PDE :
$$y^2(x-y)fracpartial zpartial x+x^2(x-y)fracpartial zpartial y=(x^2+y^2)z$$
A first characteristic equation comes from $fracdxy^2(x-y) = fracdyx^2(x-y) $ which solution is :
$$y^3-x^3=c_1$$
This is what you rightly found : your equation (1).
A second characteristic equation comes from $fracdxy^2(x-y) = fracdzz(x^2 + y^2)$
$$fracdzz=fracx^2+y^2y^2(x-y)dx$$
with $y=(c_1+x^3)^1/3$
$$fracdzz=fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1-x^3)^1/3)dx$$
$$z=c_2expleft(int fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1+x^3)^1/3)dx right)$$
The difficulty is to integrate for a closed form. If we could do it we would have a function $I(c_1,x)$ such as :
$$I(c_1,x)=int fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1+x^3)^1/3)dx $$
and a second characteristic equation would be :
$$z,exp(-Ileft(c_1,x right) )=c_2$$
$$z,exp(-I(y^3-x^3,x) )=c_2$$
General solution of the PDE expressed on implicit form $c_2=F(c_1)$ :
$$z=exp(I(y^3-x^3,x) ) Fleft(y^3-x^3 right)$$
where $F$ is an arbitrary function to be determined according to some conditions (not specified in the wording of the question).
Nevertheless the solution is not obtained on a closed form, due to the weird integral.
answered Apr 4 at 9:59
JJacquelinJJacquelin
45.8k21858
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Vaibhav Ramani
Apr 5 at 2:13
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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oldest
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active
oldest
votes
$begingroup$
$$ fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2) $$
This system of ODEs might come from solving the PDE :
$$y^2(x-y)fracpartial zpartial x+x^2(x-y)fracpartial zpartial y=(x^2+y^2)z$$
A first characteristic equation comes from $fracdxy^2(x-y) = fracdyx^2(x-y) $ which solution is :
$$y^3-x^3=c_1$$
This is what you rightly found : your equation (1).
A second characteristic equation comes from $fracdxy^2(x-y) = fracdzz(x^2 + y^2)$
$$fracdzz=fracx^2+y^2y^2(x-y)dx$$
with $y=(c_1+x^3)^1/3$
$$fracdzz=fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1-x^3)^1/3)dx$$
$$z=c_2expleft(int fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1+x^3)^1/3)dx right)$$
The difficulty is to integrate for a closed form. If we could do it we would have a function $I(c_1,x)$ such as :
$$I(c_1,x)=int fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1+x^3)^1/3)dx $$
and a second characteristic equation would be :
$$z,exp(-Ileft(c_1,x right) )=c_2$$
$$z,exp(-I(y^3-x^3,x) )=c_2$$
General solution of the PDE expressed on implicit form $c_2=F(c_1)$ :
$$z=exp(I(y^3-x^3,x) ) Fleft(y^3-x^3 right)$$
where $F$ is an arbitrary function to be determined according to some conditions (not specified in the wording of the question).
Nevertheless the solution is not obtained on a closed form, due to the weird integral.
answered Apr 4 at 9:59
JJacquelinJJacquelin
45.8k21858
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Vaibhav Ramani
Apr 5 at 2:13
add a comment |
$begingroup$
$$ fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2) $$
This system of ODEs might come from solving the PDE :
$$y^2(x-y)fracpartial zpartial x+x^2(x-y)fracpartial zpartial y=(x^2+y^2)z$$
A first characteristic equation comes from $fracdxy^2(x-y) = fracdyx^2(x-y) $ which solution is :
$$y^3-x^3=c_1$$
This is what you rightly found : your equation (1).
A second characteristic equation comes from $fracdxy^2(x-y) = fracdzz(x^2 + y^2)$
$$fracdzz=fracx^2+y^2y^2(x-y)dx$$
with $y=(c_1+x^3)^1/3$
$$fracdzz=fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1-x^3)^1/3)dx$$
$$z=c_2expleft(int fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1+x^3)^1/3)dx right)$$
The difficulty is to integrate for a closed form. If we could do it we would have a function $I(c_1,x)$ such as :
$$I(c_1,x)=int fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1+x^3)^1/3)dx $$
and a second characteristic equation would be :
$$z,exp(-Ileft(c_1,x right) )=c_2$$
$$z,exp(-I(y^3-x^3,x) )=c_2$$
General solution of the PDE expressed on implicit form $c_2=F(c_1)$ :
$$z=exp(I(y^3-x^3,x) ) Fleft(y^3-x^3 right)$$
where $F$ is an arbitrary function to be determined according to some conditions (not specified in the wording of the question).
Nevertheless the solution is not obtained on a closed form, due to the weird integral.
answered Apr 4 at 9:59
JJacquelinJJacquelin
45.8k21858
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Vaibhav Ramani
Apr 5 at 2:13
add a comment |
$begingroup$
$$ fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2) $$
This system of ODEs might come from solving the PDE :
$$y^2(x-y)fracpartial zpartial x+x^2(x-y)fracpartial zpartial y=(x^2+y^2)z$$
A first characteristic equation comes from $fracdxy^2(x-y) = fracdyx^2(x-y) $ which solution is :
$$y^3-x^3=c_1$$
This is what you rightly found : your equation (1).
A second characteristic equation comes from $fracdxy^2(x-y) = fracdzz(x^2 + y^2)$
$$fracdzz=fracx^2+y^2y^2(x-y)dx$$
with $y=(c_1+x^3)^1/3$
$$fracdzz=fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1-x^3)^1/3)dx$$
$$z=c_2expleft(int fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1+x^3)^1/3)dx right)$$
The difficulty is to integrate for a closed form. If we could do it we would have a function $I(c_1,x)$ such as :
$$I(c_1,x)=int fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1+x^3)^1/3)dx $$
and a second characteristic equation would be :
$$z,exp(-Ileft(c_1,x right) )=c_2$$
$$z,exp(-I(y^3-x^3,x) )=c_2$$
General solution of the PDE expressed on implicit form $c_2=F(c_1)$ :
$$z=exp(I(y^3-x^3,x) ) Fleft(y^3-x^3 right)$$
where $F$ is an arbitrary function to be determined according to some conditions (not specified in the wording of the question).
Nevertheless the solution is not obtained on a closed form, due to the weird integral.
answered Apr 4 at 9:59
JJacquelinJJacquelin
45.8k21858
$endgroup$
$$ fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2) $$
This system of ODEs might come from solving the PDE :
$$y^2(x-y)fracpartial zpartial x+x^2(x-y)fracpartial zpartial y=(x^2+y^2)z$$
A first characteristic equation comes from $fracdxy^2(x-y) = fracdyx^2(x-y) $ which solution is :
$$y^3-x^3=c_1$$
This is what you rightly found : your equation (1).
A second characteristic equation comes from $fracdxy^2(x-y) = fracdzz(x^2 + y^2)$
$$fracdzz=fracx^2+y^2y^2(x-y)dx$$
with $y=(c_1+x^3)^1/3$
$$fracdzz=fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1-x^3)^1/3)dx$$
$$z=c_2expleft(int fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1+x^3)^1/3)dx right)$$
The difficulty is to integrate for a closed form. If we could do it we would have a function $I(c_1,x)$ such as :
$$I(c_1,x)=int fracx^2+ (c_1+x^3)^2/3 (c_1+x^3)^2/3(x-(c_1+x^3)^1/3)dx $$
and a second characteristic equation would be :
$$z,exp(-Ileft(c_1,x right) )=c_2$$
$$z,exp(-I(y^3-x^3,x) )=c_2$$
General solution of the PDE expressed on implicit form $c_2=F(c_1)$ :
$$z=exp(I(y^3-x^3,x) ) Fleft(y^3-x^3 right)$$
where $F$ is an arbitrary function to be determined according to some conditions (not specified in the wording of the question).
Nevertheless the solution is not obtained on a closed form, due to the weird integral.
answered Apr 4 at 9:59
JJacquelinJJacquelin
45.8k21858
edited Apr 5 at 5:52
answered Apr 4 at 9:59
JJacquelinJJacquelin
45.8k21858
answered Apr 4 at 9:59
JJacquelinJJacquelin
45.8k21858
answered Apr 4 at 9:59
JJacquelinJJacquelin
45.8k21858
45.8k21858
$begingroup$
Thank you very much!
$endgroup$
– Vaibhav Ramani
Apr 5 at 2:13
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– Vaibhav Ramani
Apr 5 at 2:13
$begingroup$
Thank you very much!
$endgroup$
– Vaibhav Ramani
Apr 5 at 2:13
$begingroup$
Thank you very much!
$endgroup$
– Vaibhav Ramani
Apr 5 at 2:13
add a comment |
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$begingroup$
This appeared in my assignment in the subject of differential equations. The question as given in it is: Solve the following differential equation: $fracdxy^2(x-y) = fracdyx^2(x-y) = fracdzz(x^2 + y^2)$
$endgroup$
– Vaibhav Ramani
Apr 3 at 15:06