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From system of coupled ODEs to separable ODE

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3

$begingroup$

How does one go from

beginalign
dotx&=y\
doty&=-x^3
endalign

to the following ODE?

$$fracdydx = -fracx^3y$$

ordinary-differential-equations

share|cite|improve this question

edited 8 hours ago

Rodrigo de Azevedo

13.4k42065

asked 8 hours ago

ParsevalParseval

3,2061720

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $fracdydt = fracdydxfracdxdt Rightarrow -x^3 = fracdydx y$
  $endgroup$
  – Cesareo
  8 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

How does one go from

beginalign
dotx&=y\
doty&=-x^3
endalign

to the following ODE?

$$fracdydx = -fracx^3y$$

ordinary-differential-equations

share|cite|improve this question

edited 8 hours ago

Rodrigo de Azevedo

13.4k42065

asked 8 hours ago

ParsevalParseval

3,2061720

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $fracdydt = fracdydxfracdxdt Rightarrow -x^3 = fracdydx y$
  $endgroup$
  – Cesareo
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

How does one go from

beginalign
dotx&=y\
doty&=-x^3
endalign

to the following ODE?

$$fracdydx = -fracx^3y$$

ordinary-differential-equations

share|cite|improve this question

edited 8 hours ago

Rodrigo de Azevedo

13.4k42065

asked 8 hours ago

ParsevalParseval

3,2061720

$endgroup$

share|cite|improve this question

edited 8 hours ago

Rodrigo de Azevedo

13.4k42065

asked 8 hours ago

ParsevalParseval

3,2061720

share|cite|improve this question

edited 8 hours ago

Rodrigo de Azevedo

13.4k42065

asked 8 hours ago

ParsevalParseval

3,2061720

edited 8 hours ago

Rodrigo de Azevedo

13.4k42065

edited 8 hours ago

Rodrigo de Azevedo

13.4k42065

asked 8 hours ago

ParsevalParseval

3,2061720

asked 8 hours ago

ParsevalParseval

3,2061720

3 Answers
3

active

oldest

votes

5

$begingroup$

Notice that:

$$dotx = y Rightarrow dx = ydt,$$

and

$$doty = -x^3 Rightarrow dy = -x^3 dt.$$

Therefore:

$$fracdydx = frac-x^3dtydt = -fracx^3y$$.

share|cite|improve this answer

answered 8 hours ago

the_candymanthe_candyman

9,41732147

$endgroup$

add a comment | 

4

$begingroup$

Hint:
$$fracfracdydtfracdxdt=fracdydx$$

share|cite|improve this answer

answered 8 hours ago

E.H.EE.H.E

18.2k12070

$endgroup$

add a comment | 

2

$begingroup$

We know that $doty=-x^3$. Divide this by $dotx=y$ and we get $fracdotydotx=frac-x^3y$. But $fracdotydotx$ is just $fracdydx$ so we're done.

share|cite|improve this answer

answered 8 hours ago

auscryptauscrypt

4,326212

$endgroup$

add a comment | 

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5

$begingroup$

Notice that:

$$dotx = y Rightarrow dx = ydt,$$

and

$$doty = -x^3 Rightarrow dy = -x^3 dt.$$

Therefore:

$$fracdydx = frac-x^3dtydt = -fracx^3y$$.

share|cite|improve this answer

answered 8 hours ago

the_candymanthe_candyman

9,41732147

$endgroup$

add a comment | 

5

$begingroup$

Notice that:

$$dotx = y Rightarrow dx = ydt,$$

and

$$doty = -x^3 Rightarrow dy = -x^3 dt.$$

Therefore:

$$fracdydx = frac-x^3dtydt = -fracx^3y$$.

share|cite|improve this answer

answered 8 hours ago

the_candymanthe_candyman

9,41732147

$endgroup$

add a comment | 

$begingroup$

Notice that:

$$dotx = y Rightarrow dx = ydt,$$

and

$$doty = -x^3 Rightarrow dy = -x^3 dt.$$

Therefore:

$$fracdydx = frac-x^3dtydt = -fracx^3y$$.

share|cite|improve this answer

answered 8 hours ago

the_candymanthe_candyman

9,41732147

$endgroup$

share|cite|improve this answer

answered 8 hours ago

the_candymanthe_candyman

9,41732147

answered 8 hours ago

the_candymanthe_candyman

9,41732147

answered 8 hours ago

the_candymanthe_candyman

9,41732147

4

$begingroup$

Hint:
$$fracfracdydtfracdxdt=fracdydx$$

share|cite|improve this answer

answered 8 hours ago

E.H.EE.H.E

18.2k12070

$endgroup$

add a comment | 

4

$begingroup$

Hint:
$$fracfracdydtfracdxdt=fracdydx$$

share|cite|improve this answer

answered 8 hours ago

E.H.EE.H.E

18.2k12070

$endgroup$

add a comment | 

$begingroup$

Hint:
$$fracfracdydtfracdxdt=fracdydx$$

share|cite|improve this answer

answered 8 hours ago

E.H.EE.H.E

18.2k12070

$endgroup$

share|cite|improve this answer

answered 8 hours ago

E.H.EE.H.E

18.2k12070

answered 8 hours ago

E.H.EE.H.E

18.2k12070

answered 8 hours ago

E.H.EE.H.E

18.2k12070

2

$begingroup$

We know that $doty=-x^3$. Divide this by $dotx=y$ and we get $fracdotydotx=frac-x^3y$. But $fracdotydotx$ is just $fracdydx$ so we're done.

share|cite|improve this answer

answered 8 hours ago

auscryptauscrypt

4,326212

$endgroup$

add a comment | 

2

$begingroup$

We know that $doty=-x^3$. Divide this by $dotx=y$ and we get $fracdotydotx=frac-x^3y$. But $fracdotydotx$ is just $fracdydx$ so we're done.

share|cite|improve this answer

answered 8 hours ago

auscryptauscrypt

4,326212

$endgroup$

add a comment | 

$begingroup$

We know that $doty=-x^3$. Divide this by $dotx=y$ and we get $fracdotydotx=frac-x^3y$. But $fracdotydotx$ is just $fracdydx$ so we're done.

share|cite|improve this answer

answered 8 hours ago

auscryptauscrypt

4,326212

$endgroup$

share|cite|improve this answer

answered 8 hours ago

auscryptauscrypt

4,326212

answered 8 hours ago

auscryptauscrypt

4,326212

answered 8 hours ago

auscryptauscrypt

4,326212