Solve: $(3xy-2ay^2)dx+(x^2-2axy)dy=0$
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Solve: $(3xy-2ay^2)dx+(x^2-2axy)dy=0$
The given differential equation is not exact.
The given differential equation is homogeneous.
Therefore,
$I.F=frac1Mx+Ny$
$=frac1(3xy-2ay^2)x+(x^2-2axy)y$
$=frac1xy(4x-4ay)$
With Integrating Factor of this form (ugly value), it is difficult to compute further integration, that is involved in solving the equation.
So, can we not solve the equation using the rule:
If in the differential eqn $Mdx+Ndy=0$, provided $ Mx+Ny ne 0$, andM
and N ar eboth homogeneous then
$I.F=frac1Mx+Ny$
differential-equations
add a comment |Â
up vote
0
down vote
favorite
Solve: $(3xy-2ay^2)dx+(x^2-2axy)dy=0$
The given differential equation is not exact.
The given differential equation is homogeneous.
Therefore,
$I.F=frac1Mx+Ny$
$=frac1(3xy-2ay^2)x+(x^2-2axy)y$
$=frac1xy(4x-4ay)$
With Integrating Factor of this form (ugly value), it is difficult to compute further integration, that is involved in solving the equation.
So, can we not solve the equation using the rule:
If in the differential eqn $Mdx+Ndy=0$, provided $ Mx+Ny ne 0$, andM
and N ar eboth homogeneous then
$I.F=frac1Mx+Ny$
differential-equations
put $y=vx,$ .............
â Magneto
Aug 8 at 9:44
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Solve: $(3xy-2ay^2)dx+(x^2-2axy)dy=0$
The given differential equation is not exact.
The given differential equation is homogeneous.
Therefore,
$I.F=frac1Mx+Ny$
$=frac1(3xy-2ay^2)x+(x^2-2axy)y$
$=frac1xy(4x-4ay)$
With Integrating Factor of this form (ugly value), it is difficult to compute further integration, that is involved in solving the equation.
So, can we not solve the equation using the rule:
If in the differential eqn $Mdx+Ndy=0$, provided $ Mx+Ny ne 0$, andM
and N ar eboth homogeneous then
$I.F=frac1Mx+Ny$
differential-equations
Solve: $(3xy-2ay^2)dx+(x^2-2axy)dy=0$
The given differential equation is not exact.
The given differential equation is homogeneous.
Therefore,
$I.F=frac1Mx+Ny$
$=frac1(3xy-2ay^2)x+(x^2-2axy)y$
$=frac1xy(4x-4ay)$
With Integrating Factor of this form (ugly value), it is difficult to compute further integration, that is involved in solving the equation.
So, can we not solve the equation using the rule:
If in the differential eqn $Mdx+Ndy=0$, provided $ Mx+Ny ne 0$, andM
and N ar eboth homogeneous then
$I.F=frac1Mx+Ny$
differential-equations
edited Aug 8 at 9:19
asked Aug 8 at 9:13
Soumee
626213
626213
put $y=vx,$ .............
â Magneto
Aug 8 at 9:44
add a comment |Â
put $y=vx,$ .............
â Magneto
Aug 8 at 9:44
put $y=vx,$ .............
â Magneto
Aug 8 at 9:44
put $y=vx,$ .............
â Magneto
Aug 8 at 9:44
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Hint: $$(3xy-2ay^2)dx+(x^2-2axy)dy=0$$
With
$$M=3xy-2ay^2~~~~~~,~~~~~~N=x^2-2axy$$
then
$$M_y=3x-4ay~~~~~~,~~~~~~N_x=2x-2ay$$
so the equation is not exact. For integrating factor
$$mu=dfracM_y-N_xN=dfracx-2ayx(x-2ay)=dfrac1x$$
therefore $I=e^int mu dx=x$ is integrating factor.
Can you proceed?
Yes, I can proceed like this. But, can we not solve it using the homogeneous method?
â Soumee
Aug 8 at 12:28
other answer is wity homogeneous method
â Nosrati
Aug 8 at 12:51
add a comment |Â
up vote
1
down vote
Put y=vx
$fracdydx = frac2ay^2-3xyx^2-2axy \ v+xfracdvdx = frac2av^2-3v1-2av\xfracdvdx =frac4v(av-1)1-2av \ frac1-2av4v(av-1)dv=fracdxx \(frac-14v-fraca4(av-1))dv=fracdxx\v(av-1)x^4=C\x^2y(ay-x)=C, space some space constant$
Other simpler way would be multiply with x
Given differential equation becomes:
$d(x^3 y)-ad(x^2y^2)=0$
U can directly integrate from here
Sir, can you please help me with this question: math.stackexchange.com/q/2881062/394202
â Soumee
Aug 13 at 7:00
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: $$(3xy-2ay^2)dx+(x^2-2axy)dy=0$$
With
$$M=3xy-2ay^2~~~~~~,~~~~~~N=x^2-2axy$$
then
$$M_y=3x-4ay~~~~~~,~~~~~~N_x=2x-2ay$$
so the equation is not exact. For integrating factor
$$mu=dfracM_y-N_xN=dfracx-2ayx(x-2ay)=dfrac1x$$
therefore $I=e^int mu dx=x$ is integrating factor.
Can you proceed?
Yes, I can proceed like this. But, can we not solve it using the homogeneous method?
â Soumee
Aug 8 at 12:28
other answer is wity homogeneous method
â Nosrati
Aug 8 at 12:51
add a comment |Â
up vote
1
down vote
accepted
Hint: $$(3xy-2ay^2)dx+(x^2-2axy)dy=0$$
With
$$M=3xy-2ay^2~~~~~~,~~~~~~N=x^2-2axy$$
then
$$M_y=3x-4ay~~~~~~,~~~~~~N_x=2x-2ay$$
so the equation is not exact. For integrating factor
$$mu=dfracM_y-N_xN=dfracx-2ayx(x-2ay)=dfrac1x$$
therefore $I=e^int mu dx=x$ is integrating factor.
Can you proceed?
Yes, I can proceed like this. But, can we not solve it using the homogeneous method?
â Soumee
Aug 8 at 12:28
other answer is wity homogeneous method
â Nosrati
Aug 8 at 12:51
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: $$(3xy-2ay^2)dx+(x^2-2axy)dy=0$$
With
$$M=3xy-2ay^2~~~~~~,~~~~~~N=x^2-2axy$$
then
$$M_y=3x-4ay~~~~~~,~~~~~~N_x=2x-2ay$$
so the equation is not exact. For integrating factor
$$mu=dfracM_y-N_xN=dfracx-2ayx(x-2ay)=dfrac1x$$
therefore $I=e^int mu dx=x$ is integrating factor.
Can you proceed?
Hint: $$(3xy-2ay^2)dx+(x^2-2axy)dy=0$$
With
$$M=3xy-2ay^2~~~~~~,~~~~~~N=x^2-2axy$$
then
$$M_y=3x-4ay~~~~~~,~~~~~~N_x=2x-2ay$$
so the equation is not exact. For integrating factor
$$mu=dfracM_y-N_xN=dfracx-2ayx(x-2ay)=dfrac1x$$
therefore $I=e^int mu dx=x$ is integrating factor.
Can you proceed?
answered Aug 8 at 11:05
Nosrati
20.2k41644
20.2k41644
Yes, I can proceed like this. But, can we not solve it using the homogeneous method?
â Soumee
Aug 8 at 12:28
other answer is wity homogeneous method
â Nosrati
Aug 8 at 12:51
add a comment |Â
Yes, I can proceed like this. But, can we not solve it using the homogeneous method?
â Soumee
Aug 8 at 12:28
other answer is wity homogeneous method
â Nosrati
Aug 8 at 12:51
Yes, I can proceed like this. But, can we not solve it using the homogeneous method?
â Soumee
Aug 8 at 12:28
Yes, I can proceed like this. But, can we not solve it using the homogeneous method?
â Soumee
Aug 8 at 12:28
other answer is wity homogeneous method
â Nosrati
Aug 8 at 12:51
other answer is wity homogeneous method
â Nosrati
Aug 8 at 12:51
add a comment |Â
up vote
1
down vote
Put y=vx
$fracdydx = frac2ay^2-3xyx^2-2axy \ v+xfracdvdx = frac2av^2-3v1-2av\xfracdvdx =frac4v(av-1)1-2av \ frac1-2av4v(av-1)dv=fracdxx \(frac-14v-fraca4(av-1))dv=fracdxx\v(av-1)x^4=C\x^2y(ay-x)=C, space some space constant$
Other simpler way would be multiply with x
Given differential equation becomes:
$d(x^3 y)-ad(x^2y^2)=0$
U can directly integrate from here
Sir, can you please help me with this question: math.stackexchange.com/q/2881062/394202
â Soumee
Aug 13 at 7:00
add a comment |Â
up vote
1
down vote
Put y=vx
$fracdydx = frac2ay^2-3xyx^2-2axy \ v+xfracdvdx = frac2av^2-3v1-2av\xfracdvdx =frac4v(av-1)1-2av \ frac1-2av4v(av-1)dv=fracdxx \(frac-14v-fraca4(av-1))dv=fracdxx\v(av-1)x^4=C\x^2y(ay-x)=C, space some space constant$
Other simpler way would be multiply with x
Given differential equation becomes:
$d(x^3 y)-ad(x^2y^2)=0$
U can directly integrate from here
Sir, can you please help me with this question: math.stackexchange.com/q/2881062/394202
â Soumee
Aug 13 at 7:00
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Put y=vx
$fracdydx = frac2ay^2-3xyx^2-2axy \ v+xfracdvdx = frac2av^2-3v1-2av\xfracdvdx =frac4v(av-1)1-2av \ frac1-2av4v(av-1)dv=fracdxx \(frac-14v-fraca4(av-1))dv=fracdxx\v(av-1)x^4=C\x^2y(ay-x)=C, space some space constant$
Other simpler way would be multiply with x
Given differential equation becomes:
$d(x^3 y)-ad(x^2y^2)=0$
U can directly integrate from here
Put y=vx
$fracdydx = frac2ay^2-3xyx^2-2axy \ v+xfracdvdx = frac2av^2-3v1-2av\xfracdvdx =frac4v(av-1)1-2av \ frac1-2av4v(av-1)dv=fracdxx \(frac-14v-fraca4(av-1))dv=fracdxx\v(av-1)x^4=C\x^2y(ay-x)=C, space some space constant$
Other simpler way would be multiply with x
Given differential equation becomes:
$d(x^3 y)-ad(x^2y^2)=0$
U can directly integrate from here
answered Aug 8 at 9:47
Magneto
789213
789213
Sir, can you please help me with this question: math.stackexchange.com/q/2881062/394202
â Soumee
Aug 13 at 7:00
add a comment |Â
Sir, can you please help me with this question: math.stackexchange.com/q/2881062/394202
â Soumee
Aug 13 at 7:00
Sir, can you please help me with this question: math.stackexchange.com/q/2881062/394202
â Soumee
Aug 13 at 7:00
Sir, can you please help me with this question: math.stackexchange.com/q/2881062/394202
â Soumee
Aug 13 at 7:00
add a comment |Â
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put $y=vx,$ .............
â Magneto
Aug 8 at 9:44