# Some calc 2

1. Feb 26, 2007

### cgward

I have to solve the differential equation and let C represent an arbitrary constant.

(dz/dt)+e^(t+z)=0

I cant seem to figure it out i wind up getting z=-2e^(t^2)+e^(C)

2. Feb 26, 2007

### Matthew Rodman

Ok, divide by $$e^z$$ on both sides to get

$$e^{-z} \frac{dz}{dt} = -e^t$$

then integrate with respect to t,

$$-e^{-z} = k - e^t$$

re-arrange to get

$$z(t) = - \ln{(k + e^t)}$$.

(k is a constant).

Last edited: Feb 26, 2007