# Calc problem

1. Jan 9, 2005

We have a set of problems for hw. I am stuck on 1 where I know the answer but cant seem to get it.

If dy/dt=ky and k is a nonzero constant then y could be
a. 2e^kty b. 2e^kt c. e^kt d. kty+5 e. 1/2ky^2 +1/2

I know the answer is b but i cant get that answer
Here is my work
S=integral sign

dy/dt=ky
dy/y=kdt
Sdy/y=kSdt
lny=kt
e^lny=e^kt
y=e^kt

How do u get a 2 in there for choice b

Last edited: Jan 9, 2005
2. Jan 9, 2005

### jamesrc

You forgot about the constant of integration:
$$\frac{dy}{dt}=ky$$
$$\frac{dy}y=kdt$$
$$\int{\frac{dy}y} = \int{kdt}$$
$$\ln y = kt +C$$
$$e^{\ln y} = e^{kt + C}$$
$$y = e^{kt}\cdot e^C$$
eC is also a constant, so it can be written as C1 if you like.The value of C1 will depend on the initial conditions. Unless there's a typo in your answer list, I can see two answers that are of this form:
b. $$y=2e^{kt}$$
and c. $$y = e^{kt}$$

I hope that helps.

3. Jan 9, 2005