Calculus AB exam prep, dy/dt = my
Howdy everyone. I was wondering if someone here could help me out with this test prep question:
If dy/dt = my and m is a nonzero constant, then y could be
(C) e^(mt) + 4
(D) mty + 4
(E) (m/2)y^2 +4
I thought that the first logical step would be antidifferentiation, which, in my mind, produced:
y = tm * (y^2/2)
I then multiplied out and got :
y = (tmy^2)/2
multiplying both sides by two yields :
2y = tmy^2
dividing both sides by y yields:
2 = tmy
Finally, dividing by tm has me believing that y= 2/tm
However, that is not an answer choice, lol.
All help is appreciated!
Have a good one.