First order ODE problem

1. May 4, 2008

mhill

Let be the first order ODE's

$$y'(x)g(x)=0$$ and $$y'(x)g(x)=\delta (x-a)$$

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

$$y(x)=C$$ and $$y(x)= C+ \int dx \frac{\delta (x-a)}{g(x)}$$

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

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.

2. May 4, 2008

HallsofIvy

Staff Emeritus
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.