Proof of Solutions for y' = xg(x,y) Equation

  • Thread starter Thread starter Fibonacci88
  • Start date Start date
  • Tags Tags
    Existence Ode
Click For Summary
SUMMARY

The discussion centers on the differential equation y' = xg(x,y), where g and its partial derivative dg/dy are continuous for all (x,y). It is established that y(x) = 0 is not a valid solution, as demonstrated by the counterexample g(x,y) = 1, leading to the conclusion that y' = x does not satisfy y(x) = 0. Furthermore, it is confirmed that if y = y(x) is a solution in the interval (a,b) and y(x0) > 0 for some x0 in (a,b), then y(x) remains positive for all x in that interval.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with continuity and differentiability concepts
  • Knowledge of the existence and uniqueness theorem for ODEs
  • Basic calculus, particularly integration techniques
NEXT STEPS
  • Study the existence and uniqueness theorem for ordinary differential equations
  • Explore the implications of continuity in solutions of differential equations
  • Investigate the behavior of solutions to nonlinear differential equations
  • Learn about the method of separation of variables in solving ODEs
USEFUL FOR

Mathematicians, students studying differential equations, and anyone interested in the analysis of nonlinear dynamic systems.

Fibonacci88
Messages
2
Reaction score
0
Given the equation y'= xg(x,y) , suppose that g and (partial) dg/dy are defined and continuous for all (x,y). Show the following:

1) y(x)=0 is a solution

2)if y=y(x), x in (a,b) is a solution and if y(x0)>0, x0 in (a,b), then y(x)>0 for all x in (a,b)

Please i need your help.
 
Physics news on Phys.org
Fibonacci88 said:
Given the equation y'= xg(x,y) , suppose that g and (partial) dg/dy are defined and continuous for all (x,y). Show the following:

1) y(x)=0 is a solution
You can't- it's not true. For example, supose g(x,y)= 1 so the equation is y'= x. Then [itex]y(x)= (1/2)x^2+ C[/itex]. y(x)= 0 is not a solution since then y'= 0 and so [itex]y'= 0\ne x[/itex]. there may be some other condition that you have left out.

2)if y=y(x), x in (a,b) is a solution and if y(x0)>0, x0 in (a,b), then y(x)>0 for all x in (a,b)

Please i need your help.
 
sorry for mistake. I am sending the correct version of the problem now.

i need answer for (ii) and (iii)
 

Attachments

  • odee.jpg
    odee.jpg
    31.4 KB · Views: 376

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
8
Views
2K
Replies
7
Views
2K
Replies
2
Views
1K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K