Solve Differential Eq.: dy/dx= 1/(xy+x^2y^3)

  • Context: Undergrad 
  • Thread starter Thread starter estalniath
  • Start date Start date
  • Tags Tags
    Method
Click For Summary
SUMMARY

The differential equation dy/dx = 1/(xy + x^2y^3) can be solved using the substitution x = 1/u, transforming it into a non-homogeneous linear ordinary differential equation (ODE). The new ODE becomes du/dy = -yu - y^3, which is separable. The discussion highlights the importance of recognizing the form of the equation and applying appropriate substitution methods to simplify the problem. Additionally, participants shared tips on using LaTeX for formatting mathematical equations in forum posts.

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with substitution methods in ODEs
  • Knowledge of linear and non-homogeneous equations
  • Basic skills in LaTeX for mathematical formatting
NEXT STEPS
  • Study methods for solving non-homogeneous linear ODEs
  • Learn about substitution techniques in differential equations
  • Explore LaTeX documentation for mathematical typesetting
  • Practice solving separable differential equations
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations, as well as anyone interested in enhancing their LaTeX skills for mathematical documentation.

estalniath
Messages
17
Reaction score
0
Hello guys,

Which method can we use to solve the differential equation below?
dy/dx= 1/(xy+x^2y^3)

It doesn't seem to be of any form which I had studied before(linear differential equations,bernoulli,exact differential,homogeneous,seperable equations) for first order differential equations yet it appeared in one of out past examination questions :cry:
 
Physics news on Phys.org
It's easy.For me

\frac{dy}{dx}=\frac{1}{xy+x^{2}y^{3}}

Therefore

\frac{dx}{dy}=yx+y^{3}x^{2}

Make the substitution

x=\frac{1}{u}

,under which

\frac{dx}{dy}=-\frac{1}{u^{2}}\frac{du}{dy}

Therefore the new ODE is

-\frac{1}{u^{2}}\frac{du}{dy}=\frac{y}{u}+y^{3}\frac{1}{u^{2}}

equivalently

\frac{du}{dy}=-yu-y^{3}

I trust you can take it from here.It's an nonhomegenous linear ODE...(the homogenous eq is separable).


Daniel.
 
Wow! That was a fast reply. =) Thank you for replying and how do you type the mathematical equations in between? I don't seem to see any functions in this thread which allows us to type mathematical functions?
 
Write formulas in latex code and use preview option for checking it b4 clicking submit.

Daniel.
 
I see. Thank you for the advice =)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
9K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K