Separation of variables for nonhomogeneous PDE
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I need to solve the equation below using separation of variables.
$$frac{partial f(x,y)}{partial x} - frac{partial f(x,y)}{partial y} = 2$$
The thing is, i've always done with $0$ after the equal sign. I'm really stuck with that $2$; when doing the separation I get $X'Y-XY'=2$ and can't separate X and Y after that.
differential-equations pde
asked Nov 15 at 21:01
Alexandre
183
add a comment |
up vote
3
down vote
favorite
I need to solve the equation below using separation of variables.
$$frac{partial f(x,y)}{partial x} - frac{partial f(x,y)}{partial y} = 2$$
The thing is, i've always done with $0$ after the equal sign. I'm really stuck with that $2$; when doing the separation I get $X'Y-XY'=2$ and can't separate X and Y after that.
differential-equations pde
asked Nov 15 at 21:01
Alexandre
183
Did try to substitute $f=h(x,y)-2y$ ?
– Isham
Nov 15 at 21:52
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I need to solve the equation below using separation of variables.
$$frac{partial f(x,y)}{partial x} - frac{partial f(x,y)}{partial y} = 2$$
The thing is, i've always done with $0$ after the equal sign. I'm really stuck with that $2$; when doing the separation I get $X'Y-XY'=2$ and can't separate X and Y after that.
differential-equations pde
asked Nov 15 at 21:01
Alexandre
183
I need to solve the equation below using separation of variables.
$$frac{partial f(x,y)}{partial x} - frac{partial f(x,y)}{partial y} = 2$$
The thing is, i've always done with $0$ after the equal sign. I'm really stuck with that $2$; when doing the separation I get $X'Y-XY'=2$ and can't separate X and Y after that.
differential-equations pde
differential-equations pde
asked Nov 15 at 21:01
Alexandre
183
asked Nov 15 at 21:01
Alexandre
183
asked Nov 15 at 21:01
Alexandre
183
asked Nov 15 at 21:01
Alexandre
183
asked Nov 15 at 21:01
Alexandre
183
183
Did try to substitute $f=h(x,y)-2y$ ?
– Isham
Nov 15 at 21:52
add a comment |
Did try to substitute $f=h(x,y)-2y$ ?
– Isham
Nov 15 at 21:52
Did try to substitute $f=h(x,y)-2y$ ?
– Isham
Nov 15 at 21:52
Did try to substitute $f=h(x,y)-2y$ ?
– Isham
Nov 15 at 21:52
add a comment |
2 Answers
2
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up vote
3
down vote
accepted
$$frac{partial f(x,y)}{partial x} - frac{partial f(x,y)}{partial y} = 2$$
Substitute $f(x,y)=h(x,y)+2x$
$$frac{partial h(x,y)}{partial x} - frac{partial h(x,y)}{partial y} =0 $$
answered Nov 15 at 22:00
Isham
12.7k3929
1
This. Now I'm able to solve the problem. Many thanks, my friend!
– Alexandre
Nov 16 at 12:39
@Alexandre yes you can solve it now you're welcomed
– Isham
Nov 16 at 12:42
add a comment |
up vote
1
down vote
Let $(a,b)$ be some vector such that $a-b=2$. Then,
$$f_x-f_y=2=(1,-1) cdot (a,b)$$
$$(1,-1) cdot (f_x,f_y)=(1,-1) cdot (a,b)$$
It follows that $(f_x-a,f_y-b)$ is orthogonal to $(1,-1)$. Let $g(x,y)=f(x,y)-ax-by$. Then $nabla g$ is orthogonal to $(1,-1)$. So $nabla g=lambda(1,1)$ and $g(x,y)=lambda (x+y)$. Hence $lambda(x+y)+ax+by=f(x,y)$. Or with $b=2-a$:
$$f(x,y)=lambda(x+y)+ax+(2-a)x$$
$$=lambda x+ lambda y+2 x$$
answered Nov 15 at 22:22
Ahmed S. Attaalla
14.7k12049
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$frac{partial f(x,y)}{partial x} - frac{partial f(x,y)}{partial y} = 2$$
Substitute $f(x,y)=h(x,y)+2x$
$$frac{partial h(x,y)}{partial x} - frac{partial h(x,y)}{partial y} =0 $$
answered Nov 15 at 22:00
Isham
12.7k3929
1
This. Now I'm able to solve the problem. Many thanks, my friend!
– Alexandre
Nov 16 at 12:39
@Alexandre yes you can solve it now you're welcomed
– Isham
Nov 16 at 12:42
add a comment |
up vote
3
down vote
accepted
$$frac{partial f(x,y)}{partial x} - frac{partial f(x,y)}{partial y} = 2$$
Substitute $f(x,y)=h(x,y)+2x$
$$frac{partial h(x,y)}{partial x} - frac{partial h(x,y)}{partial y} =0 $$
answered Nov 15 at 22:00
Isham
12.7k3929
1
This. Now I'm able to solve the problem. Many thanks, my friend!
– Alexandre
Nov 16 at 12:39
@Alexandre yes you can solve it now you're welcomed
– Isham
Nov 16 at 12:42
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$frac{partial f(x,y)}{partial x} - frac{partial f(x,y)}{partial y} = 2$$
Substitute $f(x,y)=h(x,y)+2x$
$$frac{partial h(x,y)}{partial x} - frac{partial h(x,y)}{partial y} =0 $$
answered Nov 15 at 22:00
Isham
12.7k3929
$$frac{partial f(x,y)}{partial x} - frac{partial f(x,y)}{partial y} = 2$$
Substitute $f(x,y)=h(x,y)+2x$
$$frac{partial h(x,y)}{partial x} - frac{partial h(x,y)}{partial y} =0 $$
answered Nov 15 at 22:00
Isham
12.7k3929
answered Nov 15 at 22:00
Isham
12.7k3929
answered Nov 15 at 22:00
Isham
12.7k3929
answered Nov 15 at 22:00
Isham
12.7k3929
12.7k3929
1
This. Now I'm able to solve the problem. Many thanks, my friend!
– Alexandre
Nov 16 at 12:39
@Alexandre yes you can solve it now you're welcomed
– Isham
Nov 16 at 12:42
add a comment |
1
This. Now I'm able to solve the problem. Many thanks, my friend!
– Alexandre
Nov 16 at 12:39
@Alexandre yes you can solve it now you're welcomed
– Isham
Nov 16 at 12:42
1
1
This. Now I'm able to solve the problem. Many thanks, my friend!
– Alexandre
Nov 16 at 12:39
This. Now I'm able to solve the problem. Many thanks, my friend!
– Alexandre
Nov 16 at 12:39
@Alexandre yes you can solve it now you're welcomed
– Isham
Nov 16 at 12:42
@Alexandre yes you can solve it now you're welcomed
– Isham
Nov 16 at 12:42
add a comment |
up vote
1
down vote
Let $(a,b)$ be some vector such that $a-b=2$. Then,
$$f_x-f_y=2=(1,-1) cdot (a,b)$$
$$(1,-1) cdot (f_x,f_y)=(1,-1) cdot (a,b)$$
It follows that $(f_x-a,f_y-b)$ is orthogonal to $(1,-1)$. Let $g(x,y)=f(x,y)-ax-by$. Then $nabla g$ is orthogonal to $(1,-1)$. So $nabla g=lambda(1,1)$ and $g(x,y)=lambda (x+y)$. Hence $lambda(x+y)+ax+by=f(x,y)$. Or with $b=2-a$:
$$f(x,y)=lambda(x+y)+ax+(2-a)x$$
$$=lambda x+ lambda y+2 x$$
answered Nov 15 at 22:22
Ahmed S. Attaalla
14.7k12049
add a comment |
up vote
1
down vote
Let $(a,b)$ be some vector such that $a-b=2$. Then,
$$f_x-f_y=2=(1,-1) cdot (a,b)$$
$$(1,-1) cdot (f_x,f_y)=(1,-1) cdot (a,b)$$
It follows that $(f_x-a,f_y-b)$ is orthogonal to $(1,-1)$. Let $g(x,y)=f(x,y)-ax-by$. Then $nabla g$ is orthogonal to $(1,-1)$. So $nabla g=lambda(1,1)$ and $g(x,y)=lambda (x+y)$. Hence $lambda(x+y)+ax+by=f(x,y)$. Or with $b=2-a$:
$$f(x,y)=lambda(x+y)+ax+(2-a)x$$
$$=lambda x+ lambda y+2 x$$
answered Nov 15 at 22:22
Ahmed S. Attaalla
14.7k12049
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $(a,b)$ be some vector such that $a-b=2$. Then,
$$f_x-f_y=2=(1,-1) cdot (a,b)$$
$$(1,-1) cdot (f_x,f_y)=(1,-1) cdot (a,b)$$
It follows that $(f_x-a,f_y-b)$ is orthogonal to $(1,-1)$. Let $g(x,y)=f(x,y)-ax-by$. Then $nabla g$ is orthogonal to $(1,-1)$. So $nabla g=lambda(1,1)$ and $g(x,y)=lambda (x+y)$. Hence $lambda(x+y)+ax+by=f(x,y)$. Or with $b=2-a$:
$$f(x,y)=lambda(x+y)+ax+(2-a)x$$
$$=lambda x+ lambda y+2 x$$
answered Nov 15 at 22:22
Ahmed S. Attaalla
14.7k12049
Let $(a,b)$ be some vector such that $a-b=2$. Then,
$$f_x-f_y=2=(1,-1) cdot (a,b)$$
$$(1,-1) cdot (f_x,f_y)=(1,-1) cdot (a,b)$$
It follows that $(f_x-a,f_y-b)$ is orthogonal to $(1,-1)$. Let $g(x,y)=f(x,y)-ax-by$. Then $nabla g$ is orthogonal to $(1,-1)$. So $nabla g=lambda(1,1)$ and $g(x,y)=lambda (x+y)$. Hence $lambda(x+y)+ax+by=f(x,y)$. Or with $b=2-a$:
$$f(x,y)=lambda(x+y)+ax+(2-a)x$$
$$=lambda x+ lambda y+2 x$$
answered Nov 15 at 22:22
Ahmed S. Attaalla
14.7k12049
answered Nov 15 at 22:22
Ahmed S. Attaalla
14.7k12049
answered Nov 15 at 22:22
Ahmed S. Attaalla
14.7k12049
answered Nov 15 at 22:22
Ahmed S. Attaalla
14.7k12049
14.7k12049
add a comment |
add a comment |
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Did try to substitute $f=h(x,y)-2y$ ?
– Isham
Nov 15 at 21:52