Are the solutions to first order ODE's the same or different for x=a?

  • Context: Graduate 
  • Thread starter Thread starter mhill
  • Start date Start date
  • Tags Tags
    First order Ode
Click For Summary
SUMMARY

The discussion centers on the solutions to two first-order ordinary differential equations (ODEs): y'(x)g(x)=0 and y'(x)g(x)=δ(x-a). While both equations yield the same form except at x=a, their solutions differ significantly. The first equation results in y(x)=C, while the second yields y(x)=C+∫dx (δ(x-a)/g(x)), simplifying to y(x)=C+1/g(a). The distinction lies in the dependence of the second solution on the function g(x), highlighting that the constants C and C' are not necessarily equal.

PREREQUISITES
  • Understanding of first-order ordinary differential equations (ODEs)
  • Familiarity with the Dirac delta function and its properties
  • Knowledge of integration techniques involving distributions
  • Basic concepts of differential equations and their solutions
NEXT STEPS
  • Study the properties of the Dirac delta function in detail
  • Learn about the method of integrating ODEs with discontinuous functions
  • Explore the implications of different forms of g(x) on ODE solutions
  • Investigate the uniqueness of solutions in the context of initial value problems
USEFUL FOR

Mathematicians, physics students, and engineers dealing with differential equations, particularly those interested in the behavior of solutions near singularities or discontinuities.

mhill
Messages
180
Reaction score
1
Let be the first order ODE's

[tex]y'(x)g(x)=0[/tex] and [tex]y'(x)g(x)=\delta (x-a)[/tex]

except when x=a the two equations are equal , however the solutions are very different

[tex]y(x)=C[/tex] and [tex]y(x)= C+ \int dx \frac{\delta (x-a)}{g(x)}[/tex]

or using the properties of Dirac delta [tex]y(x)=C+\frac{1}{g(a)}[/tex]

the second equation depends on the form of g(x) whereas the first does not, however except at the point x=a the 2 ODE's are completely equal.
 
Physics news on Phys.org
Why are you saying the two solutions are different? You should be writing C for one and, say, C' for the other- the two constants are not necessarily the same. In fact, all you are saying is that C= C'+ 1/g(a). Which is perfectly reasonable since 1/g(a) is itself a constant.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K