First order ODE problem

  • #1
188
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
  • #2
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.
 

Suggested for: First order ODE problem

Replies
28
Views
2K
Replies
1
Views
1K
Replies
8
Views
1K
Replies
13
Views
1K
Replies
1
Views
1K
Back
Top