Differential equation for a circle at the origin with an arbitrary radius C

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I my mathematical journey I've stumbled upon an a way to represent any circle of a radius $R$ given that the original equation is of the form: $$y=pm sqrtR^2 - x^2,$$
in terms of a differential equation, we get: $$yfracdydx + x = 0$$

Now i was thinking on deriving a form made by myself, and was wondering how to best tackle the problem...I was thinking along the lines of an exact differential equation of the first order using the properties of the total differential. Considering that a circle can be thought of the level cuve of a cylindrical surface that can be expressed as: $$f(x,y) = C$$

Am I on the right path though? Please, no one show how to explicitly do it...a few tips should suffice.

Thank you!

differential-equations differential-forms

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asked Jul 14 at 13:57

SSBASE

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up vote
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down vote

favorite

I my mathematical journey I've stumbled upon an a way to represent any circle of a radius $R$ given that the original equation is of the form: $$y=pm sqrtR^2 - x^2,$$
in terms of a differential equation, we get: $$yfracdydx + x = 0$$

Now i was thinking on deriving a form made by myself, and was wondering how to best tackle the problem...I was thinking along the lines of an exact differential equation of the first order using the properties of the total differential. Considering that a circle can be thought of the level cuve of a cylindrical surface that can be expressed as: $$f(x,y) = C$$

Am I on the right path though? Please, no one show how to explicitly do it...a few tips should suffice.

Thank you!

differential-equations differential-forms

share|cite|improve this question

asked Jul 14 at 13:57

SSBASE

644

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I my mathematical journey I've stumbled upon an a way to represent any circle of a radius $R$ given that the original equation is of the form: $$y=pm sqrtR^2 - x^2,$$
in terms of a differential equation, we get: $$yfracdydx + x = 0$$

Now i was thinking on deriving a form made by myself, and was wondering how to best tackle the problem...I was thinking along the lines of an exact differential equation of the first order using the properties of the total differential. Considering that a circle can be thought of the level cuve of a cylindrical surface that can be expressed as: $$f(x,y) = C$$

Am I on the right path though? Please, no one show how to explicitly do it...a few tips should suffice.

Thank you!

differential-equations differential-forms

share|cite|improve this question

asked Jul 14 at 13:57

SSBASE

644

I my mathematical journey I've stumbled upon an a way to represent any circle of a radius $R$ given that the original equation is of the form: $$y=pm sqrtR^2 - x^2,$$
in terms of a differential equation, we get: $$yfracdydx + x = 0$$

Now i was thinking on deriving a form made by myself, and was wondering how to best tackle the problem...I was thinking along the lines of an exact differential equation of the first order using the properties of the total differential. Considering that a circle can be thought of the level cuve of a cylindrical surface that can be expressed as: $$f(x,y) = C$$

Am I on the right path though? Please, no one show how to explicitly do it...a few tips should suffice.

Thank you!

differential-equations differential-forms

share|cite|improve this question

asked Jul 14 at 13:57

SSBASE

644

share|cite|improve this question

share|cite|improve this question

asked Jul 14 at 13:57

SSBASE

644

asked Jul 14 at 13:57

SSBASE

644

asked Jul 14 at 13:57

SSBASE

644

644

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote

Well you can start with $$x^2 + y^2 = R^2$$ and differentiate implicitly, to get $$2x+2yy'=0$$

Solve for $y'$ to get your differential equation.

share|cite|improve this answer

answered Jul 14 at 14:07

Mohammad Riazi-Kermani

27.7k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 more simple with implicit differentiation
  – Isham
  Jul 14 at 14:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MohammadRiazi-Kermani Right just now i came up with df = 2x dx + 2y dy = 0, but I am not quite sure about this...
  – SSBASE
  Jul 14 at 14:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  wait, i suspect this to be wrong...
  – SSBASE
  Jul 14 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SSBASE You can divide by $2$ or solve for $frac dydx$
  – Mohammad Riazi-Kermani
  Jul 14 at 14:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MommadRiazi-Kermani If I divide by 2, then wouldn't I get df/2?
  – SSBASE
  Jul 14 at 14:59
  

  
  
  
  

 | 
show 4 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Well you can start with $$x^2 + y^2 = R^2$$ and differentiate implicitly, to get $$2x+2yy'=0$$

Solve for $y'$ to get your differential equation.

share|cite|improve this answer

answered Jul 14 at 14:07

Mohammad Riazi-Kermani

27.7k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 more simple with implicit differentiation
  – Isham
  Jul 14 at 14:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MohammadRiazi-Kermani Right just now i came up with df = 2x dx + 2y dy = 0, but I am not quite sure about this...
  – SSBASE
  Jul 14 at 14:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  wait, i suspect this to be wrong...
  – SSBASE
  Jul 14 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SSBASE You can divide by $2$ or solve for $frac dydx$
  – Mohammad Riazi-Kermani
  Jul 14 at 14:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MommadRiazi-Kermani If I divide by 2, then wouldn't I get df/2?
  – SSBASE
  Jul 14 at 14:59
  

  
  
  
  

 | 
show 4 more comments

up vote
2
down vote

Well you can start with $$x^2 + y^2 = R^2$$ and differentiate implicitly, to get $$2x+2yy'=0$$

Solve for $y'$ to get your differential equation.

share|cite|improve this answer

answered Jul 14 at 14:07

Mohammad Riazi-Kermani

27.7k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 more simple with implicit differentiation
  – Isham
  Jul 14 at 14:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MohammadRiazi-Kermani Right just now i came up with df = 2x dx + 2y dy = 0, but I am not quite sure about this...
  – SSBASE
  Jul 14 at 14:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  wait, i suspect this to be wrong...
  – SSBASE
  Jul 14 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SSBASE You can divide by $2$ or solve for $frac dydx$
  – Mohammad Riazi-Kermani
  Jul 14 at 14:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MommadRiazi-Kermani If I divide by 2, then wouldn't I get df/2?
  – SSBASE
  Jul 14 at 14:59
  

  
  
  
  

 | 
show 4 more comments

up vote
2
down vote

up vote
2
down vote

Well you can start with $$x^2 + y^2 = R^2$$ and differentiate implicitly, to get $$2x+2yy'=0$$

Solve for $y'$ to get your differential equation.

share|cite|improve this answer

answered Jul 14 at 14:07

Mohammad Riazi-Kermani

27.7k41852

Well you can start with $$x^2 + y^2 = R^2$$ and differentiate implicitly, to get $$2x+2yy'=0$$

Solve for $y'$ to get your differential equation.

share|cite|improve this answer

answered Jul 14 at 14:07

Mohammad Riazi-Kermani

27.7k41852

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 14:07

Mohammad Riazi-Kermani

27.7k41852

answered Jul 14 at 14:07

Mohammad Riazi-Kermani

27.7k41852

answered Jul 14 at 14:07

Mohammad Riazi-Kermani

27.7k41852

27.7k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 more simple with implicit differentiation
  – Isham
  Jul 14 at 14:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MohammadRiazi-Kermani Right just now i came up with df = 2x dx + 2y dy = 0, but I am not quite sure about this...
  – SSBASE
  Jul 14 at 14:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  wait, i suspect this to be wrong...
  – SSBASE
  Jul 14 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SSBASE You can divide by $2$ or solve for $frac dydx$
  – Mohammad Riazi-Kermani
  Jul 14 at 14:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MommadRiazi-Kermani If I divide by 2, then wouldn't I get df/2?
  – SSBASE
  Jul 14 at 14:59
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 more simple with implicit differentiation
  – Isham
  Jul 14 at 14:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MohammadRiazi-Kermani Right just now i came up with df = 2x dx + 2y dy = 0, but I am not quite sure about this...
  – SSBASE
  Jul 14 at 14:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  wait, i suspect this to be wrong...
  – SSBASE
  Jul 14 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SSBASE You can divide by $2$ or solve for $frac dydx$
  – Mohammad Riazi-Kermani
  Jul 14 at 14:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MommadRiazi-Kermani If I divide by 2, then wouldn't I get df/2?
  – SSBASE
  Jul 14 at 14:59
  

  
  
  
  

+1 more simple with implicit differentiation
– Isham
Jul 14 at 14:26

+1 more simple with implicit differentiation
– Isham
Jul 14 at 14:26

@MohammadRiazi-Kermani Right just now i came up with df = 2x dx + 2y dy = 0, but I am not quite sure about this...
– SSBASE
Jul 14 at 14:35

@MohammadRiazi-Kermani Right just now i came up with df = 2x dx + 2y dy = 0, but I am not quite sure about this...
– SSBASE
Jul 14 at 14:35

wait, i suspect this to be wrong...
– SSBASE
Jul 14 at 14:44

wait, i suspect this to be wrong...
– SSBASE
Jul 14 at 14:44

@SSBASE You can divide by $2$ or solve for $frac dydx$
– Mohammad Riazi-Kermani
Jul 14 at 14:51

@SSBASE You can divide by $2$ or solve for $frac dydx$
– Mohammad Riazi-Kermani
Jul 14 at 14:51

@MommadRiazi-Kermani If I divide by 2, then wouldn't I get df/2?
– SSBASE
Jul 14 at 14:59

@MommadRiazi-Kermani If I divide by 2, then wouldn't I get df/2?
– SSBASE
Jul 14 at 14:59

 | 
show 4 more comments

 

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