dT + k(T-70)dt = 0 , T(0) = 140 Find the particular solution that satisfies the initial condition

Given the differential equation : dT+K(T-70)dt=0, T(0)=140
We have to find a particular solution that satisfies the initial condition.
 
We can write,
dT=-K(T-70)dt
\frac{dT}{T-70}=-Kdt
\int \frac{dT}{T-70}=\int -Kdt
ln(T-70)=-Kt+C  where C is a constant.
Now,
T-70=e^{-Kt+C}
         =e^{-Kt}.e^{C}
         =C'e^{-Kt}    where e^C=C' is again a constant.
Hence we have,
T=70+C'e^{-Kt}
Applying the initial condition we get,
140=70+C' implies C'=70
Therefore we have the solution:
T=70(1+e^{-Kt})