# Math Help - Calculus AB exam prep, dy/dt = my

1. ## 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

(A) 4e^(mty)
(B) 4e^(mt)
(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.

2. Originally Posted by EuptothepiI
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

(A) 4e^(mty)
(B) 4e^(mt)
(C) e^(mt) + 4
(D) mty + 4
(E) (m/2)y^2 +4
Differentiate A through E and see which give $my$ for the derivative.

CB

3. Originally Posted by EuptothepiI
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

(A) 4e^(mty)
(B) 4e^(mt)
(C) e^(mt) + 4
(D) mty + 4
(E) (m/2)y^2 +4
The type of equation this is rules out D and E already as e is involved.

You can solve this by separating the variables:

$\frac{dy}{y} = m\,dt$

$\int \frac{dy}{y} = \int m\,dt$

Spoiler:
$ln(y) = mt + k$

$Let k = ln(C)$

$y = e^{mt+ln(C)} = e^{mt}\cdot e^{ln(C)} = Ce^{mt}$