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PDE : $x^2 z_x + y^2 z_y = z(x+y)$

First order quasi-linear PDE $zz_x+z_y=1$Solving for PDE $z=z(x,y)$ and explain the need to consider separate segments of initial curveSolving : $x^2 z_x + y^2 z_y = 2xy$I.V.P. : $ x^2 z_x + y^2 z_y = z^2 ; ; ; z=1 $ on the initial curve $C : y=2x $IVP $x(y-z) z_x + y(z-x) z_y = z(x-y) ; ; ; z=t $ over the initial Curve : $C : x=t, y = 2t/(t^2-1) ; 0 < t < 1$Existence and uniqueness of PDE IVP : $z z_x + y z_y = x$, $C : x=t, y=t ; ; ; t >0$Existence and uniqueness of PDE IVP : $z z_x + y z_y = x$, $C : x=t, y=t ; ; ; t >0$ and $z=t$ over $C$.Solving the problem : $z z_x + z_y = 0, quad z(x,0) = x^2$PDE IVP : $zz_x + z_y = 0, ; ; z(x,0) = -3x$

4

1

$begingroup$

Solving the PDE $$x^2 z_x + y^2 z_y = z(x+y)$$

I came across an error in my calculations which I cannot find :

$$frac{mathrm{d}x}{x^2}=frac{mathrm{d}y}{y^2}=frac{mathrm{d}z}{z(x+y)}$$

The first integral curve is given as :

$$frac{mathrm{d}x}{x^2}=frac{mathrm{d}y}{y^2}implies z_1 = frac{1}{x} - frac{1}{y}$$

Now, for the second integral curve, I use the following differential subtraction trick to get rid of the $(x+y)$ term :

$$frac{mathrm{d}x-mathrm{d}y}{x^2-y^2}=frac{mathrm{d}z}{z(x+y)}$$
$$Rightarrow$$
$$frac{mathrm{d}(x-y)}{x-y} = frac{mathrm{d}z}{z}$$
$$implies$$
$$z_2 = frac{z}{x-y}$$

And thus the general solution is :
$$z_2=F(z_1)Rightarrow z(x,y)=(x-y)Fbigg(frac{1}{x}-frac{1}{y}bigg)$$
where $F$ is a $C^1$ function of $x$ and $y$.

Wolfram Alpha though, states that the solution is $z(x,y)=xyF(1/x-1/y$), which means that I have found the integral curve $z_2$ wrong. I cannot find any fault in my calculations though. The correct integral curve should be :
$$z_2 = frac{z}{xy}$$
How would one come to this calculation though ?

I would really appreciate your help

integration ordinary-differential-equations multivariable-calculus pde

share|cite|improve this question

asked Jun 1 '18 at 10:48

RebellosRebellos

15.4k31250

$endgroup$

add a comment | 

4

1

$begingroup$

Solving the PDE $$x^2 z_x + y^2 z_y = z(x+y)$$

I came across an error in my calculations which I cannot find :

$$frac{mathrm{d}x}{x^2}=frac{mathrm{d}y}{y^2}=frac{mathrm{d}z}{z(x+y)}$$

The first integral curve is given as :

$$frac{mathrm{d}x}{x^2}=frac{mathrm{d}y}{y^2}implies z_1 = frac{1}{x} - frac{1}{y}$$

Now, for the second integral curve, I use the following differential subtraction trick to get rid of the $(x+y)$ term :

$$frac{mathrm{d}x-mathrm{d}y}{x^2-y^2}=frac{mathrm{d}z}{z(x+y)}$$
$$Rightarrow$$
$$frac{mathrm{d}(x-y)}{x-y} = frac{mathrm{d}z}{z}$$
$$implies$$
$$z_2 = frac{z}{x-y}$$

And thus the general solution is :
$$z_2=F(z_1)Rightarrow z(x,y)=(x-y)Fbigg(frac{1}{x}-frac{1}{y}bigg)$$
where $F$ is a $C^1$ function of $x$ and $y$.

Wolfram Alpha though, states that the solution is $z(x,y)=xyF(1/x-1/y$), which means that I have found the integral curve $z_2$ wrong. I cannot find any fault in my calculations though. The correct integral curve should be :
$$z_2 = frac{z}{xy}$$
How would one come to this calculation though ?

I would really appreciate your help

integration ordinary-differential-equations multivariable-calculus pde

share|cite|improve this question

asked Jun 1 '18 at 10:48

RebellosRebellos

15.4k31250

$endgroup$

add a comment | 

$begingroup$

Solving the PDE $$x^2 z_x + y^2 z_y = z(x+y)$$

I came across an error in my calculations which I cannot find :

$$frac{mathrm{d}x}{x^2}=frac{mathrm{d}y}{y^2}=frac{mathrm{d}z}{z(x+y)}$$

The first integral curve is given as :

$$frac{mathrm{d}x}{x^2}=frac{mathrm{d}y}{y^2}implies z_1 = frac{1}{x} - frac{1}{y}$$

Now, for the second integral curve, I use the following differential subtraction trick to get rid of the $(x+y)$ term :

$$frac{mathrm{d}x-mathrm{d}y}{x^2-y^2}=frac{mathrm{d}z}{z(x+y)}$$
$$Rightarrow$$
$$frac{mathrm{d}(x-y)}{x-y} = frac{mathrm{d}z}{z}$$
$$implies$$
$$z_2 = frac{z}{x-y}$$

And thus the general solution is :
$$z_2=F(z_1)Rightarrow z(x,y)=(x-y)Fbigg(frac{1}{x}-frac{1}{y}bigg)$$
where $F$ is a $C^1$ function of $x$ and $y$.

Wolfram Alpha though, states that the solution is $z(x,y)=xyF(1/x-1/y$), which means that I have found the integral curve $z_2$ wrong. I cannot find any fault in my calculations though. The correct integral curve should be :
$$z_2 = frac{z}{xy}$$
How would one come to this calculation though ?

I would really appreciate your help

integration ordinary-differential-equations multivariable-calculus pde

share|cite|improve this question

asked Jun 1 '18 at 10:48

RebellosRebellos

15.4k31250

$endgroup$

share|cite|improve this question

asked Jun 1 '18 at 10:48

RebellosRebellos

15.4k31250

share|cite|improve this question

asked Jun 1 '18 at 10:48

RebellosRebellos

15.4k31250

asked Jun 1 '18 at 10:48

RebellosRebellos

15.4k31250

asked Jun 1 '18 at 10:48

RebellosRebellos

15.4k31250

3 Answers
3

active

oldest

votes

2

$begingroup$

From

$$
frac{1}{x}-frac{1}{y}=C_1 Rightarrow y-x=C_1 x y
$$

From

$$
frac{dy-dx}{y-x}=frac{dz}{z}Rightarrow y-x = C_2 z
$$

hence

$$
C_2 z = C_1 x y Rightarrow C_4 = frac{z}{x y}
$$

share|cite|improve this answer

answered Jun 1 '18 at 11:13

CesareoCesareo

9,4713517

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So essentially both solutions are correct ?
  $endgroup$
  – Rebellos
  Jun 1 '18 at 11:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. They are equivalent.
  $endgroup$
  – Cesareo
  Jun 1 '18 at 11:16
  

  
  
  
  

add a comment | 

1

$begingroup$

Some algebra helps too:

$(x-y)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyleft(dfrac{x-y}{xy}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyleft(dfrac{1}{y}-dfrac{1}{x}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyGleft(dfrac{1}{x}-dfrac{1}{y}right)$

with $Gleft(dfrac{1}{x}-dfrac{1}{y}right)=-left(dfrac{1}{x}-dfrac{1}{y}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)$

So, they are essentialy the same.

share|cite|improve this answer

edited Jun 2 '18 at 12:10

answered Jun 2 '18 at 12:05

Rafa BudríaRafa Budría

5,9101825

$endgroup$

add a comment | 

0

$begingroup$

$$dt =frac{mathrm{d}x}{x^2}=frac{mathrm{d}y}{y^2}=frac{mathrm{d}z}{z(x+y)}$$
is equivalent to
$$ dt =frac{zy dx + zx dy - xy dz }{0 },$$

hence the numerator is constant, which only depends on $C_1 = frac{ 1}{x } - frac{ 1}{ y}. $

This implies by integrating each side
$$F(C_1) = zyx + zxy - xyz = zxy,$$
Thus,
$$z = frac{F(frac{ 1}{x } - frac{ 1}{ y}) }{ xy} .$$

share|cite|improve this answer

answered Mar 13 at 15:15

onurcanbektasonurcanbektas

3,48911037

$endgroup$

add a comment | 

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2

$begingroup$

From

$$
frac{1}{x}-frac{1}{y}=C_1 Rightarrow y-x=C_1 x y
$$

From

$$
frac{dy-dx}{y-x}=frac{dz}{z}Rightarrow y-x = C_2 z
$$

hence

$$
C_2 z = C_1 x y Rightarrow C_4 = frac{z}{x y}
$$

share|cite|improve this answer

answered Jun 1 '18 at 11:13

CesareoCesareo

9,4713517

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So essentially both solutions are correct ?
  $endgroup$
  – Rebellos
  Jun 1 '18 at 11:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. They are equivalent.
  $endgroup$
  – Cesareo
  Jun 1 '18 at 11:16
  

  
  
  
  

add a comment | 

2

$begingroup$

From

$$
frac{1}{x}-frac{1}{y}=C_1 Rightarrow y-x=C_1 x y
$$

From

$$
frac{dy-dx}{y-x}=frac{dz}{z}Rightarrow y-x = C_2 z
$$

hence

$$
C_2 z = C_1 x y Rightarrow C_4 = frac{z}{x y}
$$

share|cite|improve this answer

answered Jun 1 '18 at 11:13

CesareoCesareo

9,4713517

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So essentially both solutions are correct ?
  $endgroup$
  – Rebellos
  Jun 1 '18 at 11:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. They are equivalent.
  $endgroup$
  – Cesareo
  Jun 1 '18 at 11:16
  

  
  
  
  

add a comment | 

$begingroup$

From

$$
frac{1}{x}-frac{1}{y}=C_1 Rightarrow y-x=C_1 x y
$$

From

$$
frac{dy-dx}{y-x}=frac{dz}{z}Rightarrow y-x = C_2 z
$$

hence

$$
C_2 z = C_1 x y Rightarrow C_4 = frac{z}{x y}
$$

share|cite|improve this answer

answered Jun 1 '18 at 11:13

CesareoCesareo

9,4713517

$endgroup$

share|cite|improve this answer

answered Jun 1 '18 at 11:13

CesareoCesareo

9,4713517

answered Jun 1 '18 at 11:13

CesareoCesareo

9,4713517

answered Jun 1 '18 at 11:13

CesareoCesareo

9,4713517

1

$begingroup$

Some algebra helps too:

$(x-y)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyleft(dfrac{x-y}{xy}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyleft(dfrac{1}{y}-dfrac{1}{x}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyGleft(dfrac{1}{x}-dfrac{1}{y}right)$

with $Gleft(dfrac{1}{x}-dfrac{1}{y}right)=-left(dfrac{1}{x}-dfrac{1}{y}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)$

So, they are essentialy the same.

share|cite|improve this answer

edited Jun 2 '18 at 12:10

answered Jun 2 '18 at 12:05

Rafa BudríaRafa Budría

5,9101825

$endgroup$

add a comment | 

1

$begingroup$

Some algebra helps too:

$(x-y)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyleft(dfrac{x-y}{xy}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyleft(dfrac{1}{y}-dfrac{1}{x}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyGleft(dfrac{1}{x}-dfrac{1}{y}right)$

with $Gleft(dfrac{1}{x}-dfrac{1}{y}right)=-left(dfrac{1}{x}-dfrac{1}{y}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)$

So, they are essentialy the same.

share|cite|improve this answer

edited Jun 2 '18 at 12:10

answered Jun 2 '18 at 12:05

Rafa BudríaRafa Budría

5,9101825

$endgroup$

add a comment | 

$begingroup$

Some algebra helps too:

$(x-y)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyleft(dfrac{x-y}{xy}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyleft(dfrac{1}{y}-dfrac{1}{x}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)=xyGleft(dfrac{1}{x}-dfrac{1}{y}right)$

with $Gleft(dfrac{1}{x}-dfrac{1}{y}right)=-left(dfrac{1}{x}-dfrac{1}{y}right)Fbigg(dfrac{1}{x}-dfrac{1}{y}bigg)$

So, they are essentialy the same.

share|cite|improve this answer

edited Jun 2 '18 at 12:10

answered Jun 2 '18 at 12:05

Rafa BudríaRafa Budría

5,9101825

$endgroup$

share|cite|improve this answer

edited Jun 2 '18 at 12:10

answered Jun 2 '18 at 12:05

Rafa BudríaRafa Budría

5,9101825

answered Jun 2 '18 at 12:05

Rafa BudríaRafa Budría

5,9101825

answered Jun 2 '18 at 12:05

Rafa BudríaRafa Budría

5,9101825

0

$begingroup$

$$dt =frac{mathrm{d}x}{x^2}=frac{mathrm{d}y}{y^2}=frac{mathrm{d}z}{z(x+y)}$$
is equivalent to
$$ dt =frac{zy dx + zx dy - xy dz }{0 },$$

hence the numerator is constant, which only depends on $C_1 = frac{ 1}{x } - frac{ 1}{ y}. $

This implies by integrating each side
$$F(C_1) = zyx + zxy - xyz = zxy,$$
Thus,
$$z = frac{F(frac{ 1}{x } - frac{ 1}{ y}) }{ xy} .$$

share|cite|improve this answer

answered Mar 13 at 15:15

onurcanbektasonurcanbektas

3,48911037

$endgroup$

add a comment | 

0

$begingroup$

$$dt =frac{mathrm{d}x}{x^2}=frac{mathrm{d}y}{y^2}=frac{mathrm{d}z}{z(x+y)}$$
is equivalent to
$$ dt =frac{zy dx + zx dy - xy dz }{0 },$$

hence the numerator is constant, which only depends on $C_1 = frac{ 1}{x } - frac{ 1}{ y}. $

This implies by integrating each side
$$F(C_1) = zyx + zxy - xyz = zxy,$$
Thus,
$$z = frac{F(frac{ 1}{x } - frac{ 1}{ y}) }{ xy} .$$

share|cite|improve this answer

answered Mar 13 at 15:15

onurcanbektasonurcanbektas

3,48911037

$endgroup$

add a comment | 

$begingroup$

$$dt =frac{mathrm{d}x}{x^2}=frac{mathrm{d}y}{y^2}=frac{mathrm{d}z}{z(x+y)}$$
is equivalent to
$$ dt =frac{zy dx + zx dy - xy dz }{0 },$$

hence the numerator is constant, which only depends on $C_1 = frac{ 1}{x } - frac{ 1}{ y}. $

This implies by integrating each side
$$F(C_1) = zyx + zxy - xyz = zxy,$$
Thus,
$$z = frac{F(frac{ 1}{x } - frac{ 1}{ y}) }{ xy} .$$

share|cite|improve this answer

answered Mar 13 at 15:15

onurcanbektasonurcanbektas

3,48911037

$endgroup$

share|cite|improve this answer

answered Mar 13 at 15:15

onurcanbektasonurcanbektas

3,48911037

answered Mar 13 at 15:15

onurcanbektasonurcanbektas

3,48911037

answered Mar 13 at 15:15

onurcanbektasonurcanbektas

3,48911037