How to determine an integrating factor which is both function of x and y but not an exponential function e.g.($x^my^n$)
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I was watching https://m.youtube.com/watch?v=efBdnK1q504&list=PLj7p5OoL6vGykv4JM5MpY3I5j6mkLyKQU&index=8&t=0s video.
Now I would wonder what if special integrating factor had been a function of $x$ and $y$ other than some exponential function of $x$ and $y$. Then how should we get that?
Suppose we have this differential $$(2y^3sec y + 2tan y)dx + left(3xy^2sec y + dfrac3y^2 sec yx+ xright)dy= 0$$ What is the perfect procedure to find its integrating factor.
differential-equations differential-forms line-integrals
edited Aug 15 at 11:53
Nash J.
1,092315
asked Aug 15 at 10:53
user187604
2286
add a comment |Â
up vote
0
down vote
favorite
I was watching https://m.youtube.com/watch?v=efBdnK1q504&list=PLj7p5OoL6vGykv4JM5MpY3I5j6mkLyKQU&index=8&t=0s video.
Now I would wonder what if special integrating factor had been a function of $x$ and $y$ other than some exponential function of $x$ and $y$. Then how should we get that?
Suppose we have this differential $$(2y^3sec y + 2tan y)dx + left(3xy^2sec y + dfrac3y^2 sec yx+ xright)dy= 0$$ What is the perfect procedure to find its integrating factor.
differential-equations differential-forms line-integrals
edited Aug 15 at 11:53
Nash J.
1,092315
asked Aug 15 at 10:53
user187604
2286
1
If I had to face this equation, the only thing I could do is to cry !
– Claude Leibovici
Aug 15 at 12:00
1
You asked this exact question before and accepted an answer. Why ask it again?
– Julián Aguirre
Aug 15 at 15:38
@JuliánAguirre note the differential equation is different.
– user187604
Aug 15 at 15:38
@user187604 : The method is exactly the same.
– JJacquelin
Aug 15 at 17:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was watching https://m.youtube.com/watch?v=efBdnK1q504&list=PLj7p5OoL6vGykv4JM5MpY3I5j6mkLyKQU&index=8&t=0s video.
Now I would wonder what if special integrating factor had been a function of $x$ and $y$ other than some exponential function of $x$ and $y$. Then how should we get that?
Suppose we have this differential $$(2y^3sec y + 2tan y)dx + left(3xy^2sec y + dfrac3y^2 sec yx+ xright)dy= 0$$ What is the perfect procedure to find its integrating factor.
differential-equations differential-forms line-integrals
edited Aug 15 at 11:53
Nash J.
1,092315
asked Aug 15 at 10:53
user187604
2286
I was watching https://m.youtube.com/watch?v=efBdnK1q504&list=PLj7p5OoL6vGykv4JM5MpY3I5j6mkLyKQU&index=8&t=0s video.
Now I would wonder what if special integrating factor had been a function of $x$ and $y$ other than some exponential function of $x$ and $y$. Then how should we get that?
Suppose we have this differential $$(2y^3sec y + 2tan y)dx + left(3xy^2sec y + dfrac3y^2 sec yx+ xright)dy= 0$$ What is the perfect procedure to find its integrating factor.
differential-equations differential-forms line-integrals
edited Aug 15 at 11:53
Nash J.
1,092315
asked Aug 15 at 10:53
user187604
2286
edited Aug 15 at 11:53
Nash J.
1,092315
edited Aug 15 at 11:53
Nash J.
1,092315
edited Aug 15 at 11:53
Nash J.
1,092315
1,092315
asked Aug 15 at 10:53
user187604
2286
asked Aug 15 at 10:53
user187604
2286
asked Aug 15 at 10:53
user187604
2286
2286
1
If I had to face this equation, the only thing I could do is to cry !
– Claude Leibovici
Aug 15 at 12:00
1
You asked this exact question before and accepted an answer. Why ask it again?
– Julián Aguirre
Aug 15 at 15:38
@JuliánAguirre note the differential equation is different.
– user187604
Aug 15 at 15:38
@user187604 : The method is exactly the same.
– JJacquelin
Aug 15 at 17:28
add a comment |Â
1
If I had to face this equation, the only thing I could do is to cry !
– Claude Leibovici
Aug 15 at 12:00
1
You asked this exact question before and accepted an answer. Why ask it again?
– Julián Aguirre
Aug 15 at 15:38
@JuliánAguirre note the differential equation is different.
– user187604
Aug 15 at 15:38
@user187604 : The method is exactly the same.
– JJacquelin
Aug 15 at 17:28
1
1
If I had to face this equation, the only thing I could do is to cry !
– Claude Leibovici
Aug 15 at 12:00
If I had to face this equation, the only thing I could do is to cry !
– Claude Leibovici
Aug 15 at 12:00
1
1
You asked this exact question before and accepted an answer. Why ask it again?
– Julián Aguirre
Aug 15 at 15:38
You asked this exact question before and accepted an answer. Why ask it again?
– Julián Aguirre
Aug 15 at 15:38
@JuliánAguirre note the differential equation is different.
– user187604
Aug 15 at 15:38
@JuliánAguirre note the differential equation is different.
– user187604
Aug 15 at 15:38
@user187604 : The method is exactly the same.
– JJacquelin
Aug 15 at 17:28
@user187604 : The method is exactly the same.
– JJacquelin
Aug 15 at 17:28
add a comment |Â
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
$$(2y^3sec y + 2tan y)dx + left(3xy^2sec y + dfrac3y^2 sec yx+ xright)dy= 0$$
$$left(frac2y^3cos(y) + frac2sin(y)cos(y)right)dx + left(frac3xy^2cos(y) + dfrac3y^2xcos(y)+ xright)dy= 0$$
Reducing to the common denominator $xcos(y)$ leads to :
$$left(2xy^3 + 2xsin(y)right)dx + left(3x^2y^2 + 3y^2+ x^2cos(y)right)dy= 0$$
Which is obviously exact:
$$dleft(x^2y^3+y^3 + x^2sin(y)right)=0$$
So, the integrating factor for the ODE in it's initial form was $xcos(y)$ .
And the final result is :
$$x^2y^3+y^3 + x^2sin(y)=c$$
A much more tedious method was to look for an integrating factor on the form $f(x)g(y)$ .
NOTE AFTER THE OP's COMMENT :
For the general ODE : $M(x,y)dx+N(x,y)dy=0$ if the integrating factor if function of $y$ only, then the I.F. is $f(y)=expleft( int fracN_x-M_yMdyright)$. If the I.F. is function of $x$ only then it is $g(x)=expleft( int fracN_x-M_yNdyright)$. The formula is different for each form of I.F.
Using such ready-made formula is a short-cut possible only if the formula exists, that is if someone already solved the related kind of ODEs. As far as I know there is no ready-made formula for the general case. So, if no formula is available in the case we are considering, we have to solve it by ourselves without the help of a ready-made formula.
In case of academic exercises, generally the I.F. can be found by simple inspection, without arduous calculus. The ODE considered in your question is such an example of simple academic case.
answered Aug 15 at 15:39
JJacquelin
40.6k21650
My question is that do i I only have one way to estimate the function or I can have some specific formula to get the integrating factor like if it had been a function of y then it had been $$intN_x-M_y/Mdx$$
– user187604
Aug 15 at 22:46
@user187604 : See the note added to my answer.
– JJacquelin
Aug 16 at 8:08
thanks for the help.
– user187604
Aug 16 at 8:11
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$(2y^3sec y + 2tan y)dx + left(3xy^2sec y + dfrac3y^2 sec yx+ xright)dy= 0$$
$$left(frac2y^3cos(y) + frac2sin(y)cos(y)right)dx + left(frac3xy^2cos(y) + dfrac3y^2xcos(y)+ xright)dy= 0$$
Reducing to the common denominator $xcos(y)$ leads to :
$$left(2xy^3 + 2xsin(y)right)dx + left(3x^2y^2 + 3y^2+ x^2cos(y)right)dy= 0$$
Which is obviously exact:
$$dleft(x^2y^3+y^3 + x^2sin(y)right)=0$$
So, the integrating factor for the ODE in it's initial form was $xcos(y)$ .
And the final result is :
$$x^2y^3+y^3 + x^2sin(y)=c$$
A much more tedious method was to look for an integrating factor on the form $f(x)g(y)$ .
NOTE AFTER THE OP's COMMENT :
For the general ODE : $M(x,y)dx+N(x,y)dy=0$ if the integrating factor if function of $y$ only, then the I.F. is $f(y)=expleft( int fracN_x-M_yMdyright)$. If the I.F. is function of $x$ only then it is $g(x)=expleft( int fracN_x-M_yNdyright)$. The formula is different for each form of I.F.
Using such ready-made formula is a short-cut possible only if the formula exists, that is if someone already solved the related kind of ODEs. As far as I know there is no ready-made formula for the general case. So, if no formula is available in the case we are considering, we have to solve it by ourselves without the help of a ready-made formula.
In case of academic exercises, generally the I.F. can be found by simple inspection, without arduous calculus. The ODE considered in your question is such an example of simple academic case.
answered Aug 15 at 15:39
JJacquelin
40.6k21650
My question is that do i I only have one way to estimate the function or I can have some specific formula to get the integrating factor like if it had been a function of y then it had been $$intN_x-M_y/Mdx$$
– user187604
Aug 15 at 22:46
@user187604 : See the note added to my answer.
– JJacquelin
Aug 16 at 8:08
thanks for the help.
– user187604
Aug 16 at 8:11
add a comment |Â
up vote
2
down vote
accepted
$$(2y^3sec y + 2tan y)dx + left(3xy^2sec y + dfrac3y^2 sec yx+ xright)dy= 0$$
$$left(frac2y^3cos(y) + frac2sin(y)cos(y)right)dx + left(frac3xy^2cos(y) + dfrac3y^2xcos(y)+ xright)dy= 0$$
Reducing to the common denominator $xcos(y)$ leads to :
$$left(2xy^3 + 2xsin(y)right)dx + left(3x^2y^2 + 3y^2+ x^2cos(y)right)dy= 0$$
Which is obviously exact:
$$dleft(x^2y^3+y^3 + x^2sin(y)right)=0$$
So, the integrating factor for the ODE in it's initial form was $xcos(y)$ .
And the final result is :
$$x^2y^3+y^3 + x^2sin(y)=c$$
A much more tedious method was to look for an integrating factor on the form $f(x)g(y)$ .
NOTE AFTER THE OP's COMMENT :
For the general ODE : $M(x,y)dx+N(x,y)dy=0$ if the integrating factor if function of $y$ only, then the I.F. is $f(y)=expleft( int fracN_x-M_yMdyright)$. If the I.F. is function of $x$ only then it is $g(x)=expleft( int fracN_x-M_yNdyright)$. The formula is different for each form of I.F.
Using such ready-made formula is a short-cut possible only if the formula exists, that is if someone already solved the related kind of ODEs. As far as I know there is no ready-made formula for the general case. So, if no formula is available in the case we are considering, we have to solve it by ourselves without the help of a ready-made formula.
In case of academic exercises, generally the I.F. can be found by simple inspection, without arduous calculus. The ODE considered in your question is such an example of simple academic case.
answered Aug 15 at 15:39
JJacquelin
40.6k21650
My question is that do i I only have one way to estimate the function or I can have some specific formula to get the integrating factor like if it had been a function of y then it had been $$intN_x-M_y/Mdx$$
– user187604
Aug 15 at 22:46
@user187604 : See the note added to my answer.
– JJacquelin
Aug 16 at 8:08
thanks for the help.
– user187604
Aug 16 at 8:11
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$(2y^3sec y + 2tan y)dx + left(3xy^2sec y + dfrac3y^2 sec yx+ xright)dy= 0$$
$$left(frac2y^3cos(y) + frac2sin(y)cos(y)right)dx + left(frac3xy^2cos(y) + dfrac3y^2xcos(y)+ xright)dy= 0$$
Reducing to the common denominator $xcos(y)$ leads to :
$$left(2xy^3 + 2xsin(y)right)dx + left(3x^2y^2 + 3y^2+ x^2cos(y)right)dy= 0$$
Which is obviously exact:
$$dleft(x^2y^3+y^3 + x^2sin(y)right)=0$$
So, the integrating factor for the ODE in it's initial form was $xcos(y)$ .
And the final result is :
$$x^2y^3+y^3 + x^2sin(y)=c$$
A much more tedious method was to look for an integrating factor on the form $f(x)g(y)$ .
NOTE AFTER THE OP's COMMENT :
For the general ODE : $M(x,y)dx+N(x,y)dy=0$ if the integrating factor if function of $y$ only, then the I.F. is $f(y)=expleft( int fracN_x-M_yMdyright)$. If the I.F. is function of $x$ only then it is $g(x)=expleft( int fracN_x-M_yNdyright)$. The formula is different for each form of I.F.
Using such ready-made formula is a short-cut possible only if the formula exists, that is if someone already solved the related kind of ODEs. As far as I know there is no ready-made formula for the general case. So, if no formula is available in the case we are considering, we have to solve it by ourselves without the help of a ready-made formula.
In case of academic exercises, generally the I.F. can be found by simple inspection, without arduous calculus. The ODE considered in your question is such an example of simple academic case.
answered Aug 15 at 15:39
JJacquelin
40.6k21650
$$(2y^3sec y + 2tan y)dx + left(3xy^2sec y + dfrac3y^2 sec yx+ xright)dy= 0$$
$$left(frac2y^3cos(y) + frac2sin(y)cos(y)right)dx + left(frac3xy^2cos(y) + dfrac3y^2xcos(y)+ xright)dy= 0$$
Reducing to the common denominator $xcos(y)$ leads to :
$$left(2xy^3 + 2xsin(y)right)dx + left(3x^2y^2 + 3y^2+ x^2cos(y)right)dy= 0$$
Which is obviously exact:
$$dleft(x^2y^3+y^3 + x^2sin(y)right)=0$$
So, the integrating factor for the ODE in it's initial form was $xcos(y)$ .
And the final result is :
$$x^2y^3+y^3 + x^2sin(y)=c$$
A much more tedious method was to look for an integrating factor on the form $f(x)g(y)$ .
NOTE AFTER THE OP's COMMENT :
For the general ODE : $M(x,y)dx+N(x,y)dy=0$ if the integrating factor if function of $y$ only, then the I.F. is $f(y)=expleft( int fracN_x-M_yMdyright)$. If the I.F. is function of $x$ only then it is $g(x)=expleft( int fracN_x-M_yNdyright)$. The formula is different for each form of I.F.
Using such ready-made formula is a short-cut possible only if the formula exists, that is if someone already solved the related kind of ODEs. As far as I know there is no ready-made formula for the general case. So, if no formula is available in the case we are considering, we have to solve it by ourselves without the help of a ready-made formula.
In case of academic exercises, generally the I.F. can be found by simple inspection, without arduous calculus. The ODE considered in your question is such an example of simple academic case.
answered Aug 15 at 15:39
JJacquelin
40.6k21650
edited Aug 16 at 8:10
answered Aug 15 at 15:39
JJacquelin
40.6k21650
answered Aug 15 at 15:39
JJacquelin
40.6k21650
answered Aug 15 at 15:39
JJacquelin
40.6k21650
40.6k21650
My question is that do i I only have one way to estimate the function or I can have some specific formula to get the integrating factor like if it had been a function of y then it had been $$intN_x-M_y/Mdx$$
– user187604
Aug 15 at 22:46
@user187604 : See the note added to my answer.
– JJacquelin
Aug 16 at 8:08
thanks for the help.
– user187604
Aug 16 at 8:11
add a comment |Â
My question is that do i I only have one way to estimate the function or I can have some specific formula to get the integrating factor like if it had been a function of y then it had been $$intN_x-M_y/Mdx$$
– user187604
Aug 15 at 22:46
@user187604 : See the note added to my answer.
– JJacquelin
Aug 16 at 8:08
thanks for the help.
– user187604
Aug 16 at 8:11
My question is that do i I only have one way to estimate the function or I can have some specific formula to get the integrating factor like if it had been a function of y then it had been $$intN_x-M_y/Mdx$$
– user187604
Aug 15 at 22:46
My question is that do i I only have one way to estimate the function or I can have some specific formula to get the integrating factor like if it had been a function of y then it had been $$intN_x-M_y/Mdx$$
– user187604
Aug 15 at 22:46
@user187604 : See the note added to my answer.
– JJacquelin
Aug 16 at 8:08
@user187604 : See the note added to my answer.
– JJacquelin
Aug 16 at 8:08
thanks for the help.
– user187604
Aug 16 at 8:11
thanks for the help.
– user187604
Aug 16 at 8:11
add a comment |Â
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1
If I had to face this equation, the only thing I could do is to cry !
– Claude Leibovici
Aug 15 at 12:00
1
You asked this exact question before and accepted an answer. Why ask it again?
– Julián Aguirre
Aug 15 at 15:38
@JuliánAguirre note the differential equation is different.
– user187604
Aug 15 at 15:38
@user187604 : The method is exactly the same.
– JJacquelin
Aug 15 at 17:28