2 answers:
You might be interested in
Please see attached document for the graphs:
The functions with the given graphs can't be solutions of
the differential equation dy dt = et(y − 1)2.
(a) won’t work, since it has negative slope on portions of its solutions. The
slope, e t (y − 1)2 , is a nonnegative function.
(b) won’t work, since it has positive slope at y = 1, where the slope must be flat.
