Transform an inexact differential into an exact differential using an integrating factor

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There're answered questions here about what I'm asking; however, I didn't study differential equations yet, and those answers use technics to solve differential equations. But my teacher, knowing that, asked me to do this:

Determine a function $u(x,y)$, such that, the differential $(x³ + x + y)u(x,y)dx - xu(x,y)dy$ be exact, and specify a potential for that differential. ($u$ is called an integrating factor for that differential)

I tried to use the assumption that

$$begincases dfracpartial varphipartial x = (x³ + x + y)u(x,y) \ dfracpartial varphipartial y = - xu(x,y) endcases$$

but I came to this differential equation:

$$dfracpartial u(x,y)partial y (x³ + x + y) + dfracpartial u(x,y)partial x x = -2u(x,y)$$

and I don't know how to proceed.

differential-equations

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edited Jul 16 at 0:12

asked Jul 16 at 0:01

Almir M. da Cunha

33

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up vote
0
down vote

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There're answered questions here about what I'm asking; however, I didn't study differential equations yet, and those answers use technics to solve differential equations. But my teacher, knowing that, asked me to do this:

Determine a function $u(x,y)$, such that, the differential $(x³ + x + y)u(x,y)dx - xu(x,y)dy$ be exact, and specify a potential for that differential. ($u$ is called an integrating factor for that differential)

I tried to use the assumption that

$$begincases dfracpartial varphipartial x = (x³ + x + y)u(x,y) \ dfracpartial varphipartial y = - xu(x,y) endcases$$

but I came to this differential equation:

$$dfracpartial u(x,y)partial y (x³ + x + y) + dfracpartial u(x,y)partial x x = -2u(x,y)$$

and I don't know how to proceed.

differential-equations

share|cite|improve this question

edited Jul 16 at 0:12

asked Jul 16 at 0:01

Almir M. da Cunha

33

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

There're answered questions here about what I'm asking; however, I didn't study differential equations yet, and those answers use technics to solve differential equations. But my teacher, knowing that, asked me to do this:

Determine a function $u(x,y)$, such that, the differential $(x³ + x + y)u(x,y)dx - xu(x,y)dy$ be exact, and specify a potential for that differential. ($u$ is called an integrating factor for that differential)

I tried to use the assumption that

$$begincases dfracpartial varphipartial x = (x³ + x + y)u(x,y) \ dfracpartial varphipartial y = - xu(x,y) endcases$$

but I came to this differential equation:

$$dfracpartial u(x,y)partial y (x³ + x + y) + dfracpartial u(x,y)partial x x = -2u(x,y)$$

and I don't know how to proceed.

differential-equations

share|cite|improve this question

edited Jul 16 at 0:12

asked Jul 16 at 0:01

Almir M. da Cunha

33

There're answered questions here about what I'm asking; however, I didn't study differential equations yet, and those answers use technics to solve differential equations. But my teacher, knowing that, asked me to do this:

Determine a function $u(x,y)$, such that, the differential $(x³ + x + y)u(x,y)dx - xu(x,y)dy$ be exact, and specify a potential for that differential. ($u$ is called an integrating factor for that differential)

I tried to use the assumption that

$$begincases dfracpartial varphipartial x = (x³ + x + y)u(x,y) \ dfracpartial varphipartial y = - xu(x,y) endcases$$

but I came to this differential equation:

$$dfracpartial u(x,y)partial y (x³ + x + y) + dfracpartial u(x,y)partial x x = -2u(x,y)$$

and I don't know how to proceed.

differential-equations

share|cite|improve this question

edited Jul 16 at 0:12

asked Jul 16 at 0:01

Almir M. da Cunha

33

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 0:12

edited Jul 16 at 0:12

edited Jul 16 at 0:12

asked Jul 16 at 0:01

Almir M. da Cunha

33

asked Jul 16 at 0:01

Almir M. da Cunha

33

asked Jul 16 at 0:01

Almir M. da Cunha

33

33

add a comment | 

add a comment | 

2 Answers
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I tried $u(x,y)=u(x)$
Then got this

$$u(x)=-(u+xu') $$
$$int frac duu=-2int frac dxx$$
$$implies u(x)=frac 1 x^2$$

$$(x³ + x + y)u(x)dx - xu(x)dy$$
You have that :
$$partial_y left (frac x³ + x + yx^2 right )= partial_x left (-frac 1x right )$$

share|cite|improve this answer

edited Jul 16 at 0:22

answered Jul 16 at 0:16

Isham

10.6k3829

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Just staring onto the equation I noticed that it would be simpler if $u(x,y)$ was constant in $y$.

That assumption will lead you to the ODE of Euler type.

share|cite|improve this answer

answered Jul 16 at 0:15

user254873

112

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

I tried $u(x,y)=u(x)$
Then got this

$$u(x)=-(u+xu') $$
$$int frac duu=-2int frac dxx$$
$$implies u(x)=frac 1 x^2$$

$$(x³ + x + y)u(x)dx - xu(x)dy$$
You have that :
$$partial_y left (frac x³ + x + yx^2 right )= partial_x left (-frac 1x right )$$

share|cite|improve this answer

edited Jul 16 at 0:22

answered Jul 16 at 0:16

Isham

10.6k3829

add a comment | 

up vote
0
down vote



accepted

I tried $u(x,y)=u(x)$
Then got this

$$u(x)=-(u+xu') $$
$$int frac duu=-2int frac dxx$$
$$implies u(x)=frac 1 x^2$$

$$(x³ + x + y)u(x)dx - xu(x)dy$$
You have that :
$$partial_y left (frac x³ + x + yx^2 right )= partial_x left (-frac 1x right )$$

share|cite|improve this answer

edited Jul 16 at 0:22

answered Jul 16 at 0:16

Isham

10.6k3829

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

I tried $u(x,y)=u(x)$
Then got this

$$u(x)=-(u+xu') $$
$$int frac duu=-2int frac dxx$$
$$implies u(x)=frac 1 x^2$$

$$(x³ + x + y)u(x)dx - xu(x)dy$$
You have that :
$$partial_y left (frac x³ + x + yx^2 right )= partial_x left (-frac 1x right )$$

share|cite|improve this answer

edited Jul 16 at 0:22

answered Jul 16 at 0:16

Isham

10.6k3829

I tried $u(x,y)=u(x)$
Then got this

$$u(x)=-(u+xu') $$
$$int frac duu=-2int frac dxx$$
$$implies u(x)=frac 1 x^2$$

$$(x³ + x + y)u(x)dx - xu(x)dy$$
You have that :
$$partial_y left (frac x³ + x + yx^2 right )= partial_x left (-frac 1x right )$$

share|cite|improve this answer

edited Jul 16 at 0:22

answered Jul 16 at 0:16

Isham

10.6k3829

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 0:22

edited Jul 16 at 0:22

edited Jul 16 at 0:22

answered Jul 16 at 0:16

Isham

10.6k3829

answered Jul 16 at 0:16

Isham

10.6k3829

answered Jul 16 at 0:16

Isham

10.6k3829

10.6k3829

add a comment | 

add a comment | 

up vote
0
down vote

Just staring onto the equation I noticed that it would be simpler if $u(x,y)$ was constant in $y$.

That assumption will lead you to the ODE of Euler type.

share|cite|improve this answer

answered Jul 16 at 0:15

user254873

112

add a comment | 

up vote
0
down vote

Just staring onto the equation I noticed that it would be simpler if $u(x,y)$ was constant in $y$.

That assumption will lead you to the ODE of Euler type.

share|cite|improve this answer

answered Jul 16 at 0:15

user254873

112

add a comment | 

up vote
0
down vote

up vote
0
down vote

Just staring onto the equation I noticed that it would be simpler if $u(x,y)$ was constant in $y$.

That assumption will lead you to the ODE of Euler type.

share|cite|improve this answer

answered Jul 16 at 0:15

user254873

112

Just staring onto the equation I noticed that it would be simpler if $u(x,y)$ was constant in $y$.

That assumption will lead you to the ODE of Euler type.

share|cite|improve this answer

answered Jul 16 at 0:15

user254873

112

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 0:15

user254873

112

answered Jul 16 at 0:15

user254873

112

answered Jul 16 at 0:15

user254873

112

112

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add a comment | 

 

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