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