Interesting Differential equation problem $xfracdydx + 3fracdxdy = y^2 $.

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Solve the differential equation to get $y $ as a function of $x$.

$$xfracdydx + 3fracdxdy = y^2$$

I tried taking $dfracdydx$ to be $p$ . Then proceed to factorise the expression but nothing happened. Please help me in this question as I don't know how to proceed.

calculus algebra-precalculus differential-equations

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edited Jul 29 at 12:38

Abcd

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asked Jul 28 at 14:53

Mastermind

656

add a comment | 

up vote
1
down vote

favorite

2

Solve the differential equation to get $y $ as a function of $x$.

$$xfracdydx + 3fracdxdy = y^2$$

I tried taking $dfracdydx$ to be $p$ . Then proceed to factorise the expression but nothing happened. Please help me in this question as I don't know how to proceed.

calculus algebra-precalculus differential-equations

share|cite|improve this question

edited Jul 29 at 12:38

Abcd

2,3151624

asked Jul 28 at 14:53

Mastermind

656

add a comment | 

up vote
1
down vote

favorite

2

up vote
1
down vote

favorite

2

2

Solve the differential equation to get $y $ as a function of $x$.

$$xfracdydx + 3fracdxdy = y^2$$

I tried taking $dfracdydx$ to be $p$ . Then proceed to factorise the expression but nothing happened. Please help me in this question as I don't know how to proceed.

calculus algebra-precalculus differential-equations

share|cite|improve this question

edited Jul 29 at 12:38

Abcd

2,3151624

asked Jul 28 at 14:53

Mastermind

656

Solve the differential equation to get $y $ as a function of $x$.

$$xfracdydx + 3fracdxdy = y^2$$

I tried taking $dfracdydx$ to be $p$ . Then proceed to factorise the expression but nothing happened. Please help me in this question as I don't know how to proceed.

calculus algebra-precalculus differential-equations

share|cite|improve this question

edited Jul 29 at 12:38

Abcd

2,3151624

asked Jul 28 at 14:53

Mastermind

656

share|cite|improve this question

share|cite|improve this question

edited Jul 29 at 12:38

Abcd

2,3151624

edited Jul 29 at 12:38

Abcd

2,3151624

edited Jul 29 at 12:38

Abcd

2,3151624

2,3151624

asked Jul 28 at 14:53

Mastermind

656

asked Jul 28 at 14:53

Mastermind

656

asked Jul 28 at 14:53

Mastermind

656

656

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

We change the role of $x$ and $y$, therfore
$$ydfracdxdy+3dfracdydx=x^2$$
gives the equation
$$x^2y'-3y'^2=y$$
which has two following roots
$$y'=dfrac-x^2pmsqrtx^4-12y-6$$
now let $z^2=x^4-12y$ which leads us to DE
$$(2x^3-x^2+z)dx-zdz=0$$
that could be convert to an exact equation with $dfracpartial Mpartial z=1$ and $dfracpartial Npartial x=0$ (with standard notation). After solve that and substitution, exchange the previous role of $x$ and $y$.

share|cite|improve this answer

answered Jul 28 at 15:47

user 108128

19k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much sir for the method. Nice approach.
  – Mastermind
  Jul 29 at 3:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry. I couldn't solve it completely!
  – user 108128
  Jul 29 at 12:30
  

  
  
  
  

add a comment | 

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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



accepted

We change the role of $x$ and $y$, therfore
$$ydfracdxdy+3dfracdydx=x^2$$
gives the equation
$$x^2y'-3y'^2=y$$
which has two following roots
$$y'=dfrac-x^2pmsqrtx^4-12y-6$$
now let $z^2=x^4-12y$ which leads us to DE
$$(2x^3-x^2+z)dx-zdz=0$$
that could be convert to an exact equation with $dfracpartial Mpartial z=1$ and $dfracpartial Npartial x=0$ (with standard notation). After solve that and substitution, exchange the previous role of $x$ and $y$.

share|cite|improve this answer

answered Jul 28 at 15:47

user 108128

19k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much sir for the method. Nice approach.
  – Mastermind
  Jul 29 at 3:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry. I couldn't solve it completely!
  – user 108128
  Jul 29 at 12:30
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

We change the role of $x$ and $y$, therfore
$$ydfracdxdy+3dfracdydx=x^2$$
gives the equation
$$x^2y'-3y'^2=y$$
which has two following roots
$$y'=dfrac-x^2pmsqrtx^4-12y-6$$
now let $z^2=x^4-12y$ which leads us to DE
$$(2x^3-x^2+z)dx-zdz=0$$
that could be convert to an exact equation with $dfracpartial Mpartial z=1$ and $dfracpartial Npartial x=0$ (with standard notation). After solve that and substitution, exchange the previous role of $x$ and $y$.

share|cite|improve this answer

answered Jul 28 at 15:47

user 108128

19k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much sir for the method. Nice approach.
  – Mastermind
  Jul 29 at 3:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry. I couldn't solve it completely!
  – user 108128
  Jul 29 at 12:30
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

We change the role of $x$ and $y$, therfore
$$ydfracdxdy+3dfracdydx=x^2$$
gives the equation
$$x^2y'-3y'^2=y$$
which has two following roots
$$y'=dfrac-x^2pmsqrtx^4-12y-6$$
now let $z^2=x^4-12y$ which leads us to DE
$$(2x^3-x^2+z)dx-zdz=0$$
that could be convert to an exact equation with $dfracpartial Mpartial z=1$ and $dfracpartial Npartial x=0$ (with standard notation). After solve that and substitution, exchange the previous role of $x$ and $y$.

share|cite|improve this answer

answered Jul 28 at 15:47

user 108128

19k41544

We change the role of $x$ and $y$, therfore
$$ydfracdxdy+3dfracdydx=x^2$$
gives the equation
$$x^2y'-3y'^2=y$$
which has two following roots
$$y'=dfrac-x^2pmsqrtx^4-12y-6$$
now let $z^2=x^4-12y$ which leads us to DE
$$(2x^3-x^2+z)dx-zdz=0$$
that could be convert to an exact equation with $dfracpartial Mpartial z=1$ and $dfracpartial Npartial x=0$ (with standard notation). After solve that and substitution, exchange the previous role of $x$ and $y$.

share|cite|improve this answer

answered Jul 28 at 15:47

user 108128

19k41544

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 15:47

user 108128

19k41544

answered Jul 28 at 15:47

user 108128

19k41544

answered Jul 28 at 15:47

user 108128

19k41544

19k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much sir for the method. Nice approach.
  – Mastermind
  Jul 29 at 3:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry. I couldn't solve it completely!
  – user 108128
  Jul 29 at 12:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much sir for the method. Nice approach.
  – Mastermind
  Jul 29 at 3:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry. I couldn't solve it completely!
  – user 108128
  Jul 29 at 12:30
  

  
  
  
  

Thank you very much sir for the method. Nice approach.
– Mastermind
Jul 29 at 3:03

Thank you very much sir for the method. Nice approach.
– Mastermind
Jul 29 at 3:03

Sorry. I couldn't solve it completely!
– user 108128
Jul 29 at 12:30

Sorry. I couldn't solve it completely!
– user 108128
Jul 29 at 12:30

add a comment | 

 

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