How to Solve This ODE with Substitution?

  • Context: MHB 
  • Thread starter Thread starter jasonmcc
  • Start date Start date
  • Tags Tags
    Ode Separable
Click For Summary
SUMMARY

The discussion focuses on solving the ordinary differential equation (ODE) given by the expression \(2xyy'=-x^2-y^2\). Participants suggest rewriting the equation as an exact differential, leading to the form \((x^{2} + y^{2})\ dx + 2\ x\ y\ d y =0\). Additionally, an alternative method involves using the substitution \(v=\frac{y}{x}\) to simplify the ODE to \(\frac{dy}{dx}=-\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)\). These techniques provide structured approaches to tackle the problem effectively.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with exact differentials
  • Knowledge of substitution methods in differential equations
  • Basic calculus concepts, including derivatives and integrals
NEXT STEPS
  • Study the method of exact differentials in ODEs
  • Learn about substitution techniques for solving ODEs
  • Explore polar coordinates and their applications in differential equations
  • Investigate the implications of using the substitution \(v=\frac{y}{x}\) in various contexts
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on differential equations, as well as educators seeking effective teaching methods for ODEs.

jasonmcc
Messages
10
Reaction score
0
I haven't done ODEs in a while nor have a book handing.

How do I tackle an equation of the form
\[
2xyy'=-x^2-y^2
\]
I tried polar but that didn't seem to work.
 
Physics news on Phys.org
Re: separable or not separable ODE

wmccunes said:
I haven't done ODEs in a while nor have a book handing.

How do I tackle an equation of the form
\[
2xyy'=-x^2-y^2
\]
I tried polar but that didn't seem to work.

With simple steps Yoy arrive to write...

$\displaystyle (x^{2} + y^{2})\ dx + 2\ x\ y\ d y =0\ (1)$ ... and the expression (1) is an 'exact differential'... Kind regards $\chi$ $\sigma$
 
Re: separable or not separable ODE

Another approach would be to write the ODE as:

$$\frac{dy}{dx}=-\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x} \right)$$

and use the substitution:

$$v=\frac{y}{x}$$
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K