Let g be a function everywhere continuous and not identically zero.
Show that if f'(t) = g(t)f(t) for all real t, then either f is identically zero or f does not take on the value zero.
Thanks for the help.
note we have the 1st order differential eqn (y = f(t))
dy/dt = g(t)y
Then y = 0 is an equillibrium solution as dy/dt = 0 = g*0= 0
At this point we could invoke the existence and uniqueness thm
to conclude any other solution cannot pass through 0 or we can proceed
directly:
We can separate the variables
dy/y = g(t)dt
Integrating we obtain ln|y| = int(g(t)dt) + C
Therfore its obvious y cannot take on the value 0 as ln|y| is not defined
if y = 0