Maths show that question calculus

Formula:
dT/dt = -k (T-Ts) + P

The temperature inside the engine of John's car is 65°C at 6pm, when he turns the engine off, and the outside temp, Ts = 20°C, from 6pm to 6am the next morning. At 9pm the temp inside the engine dropped to 33°C. Assume Ts is constant from 6pm to 6am the following morning.

Show that temperature is T = 20 + 45e^-0.4t

I am not sure what the P represents but it does not appear in Newton's Law of cooling
dT/dt = -k(T-Ts)
dT/(T-Ts) = -k*dt now integrate
ln(T-Ts) = -kt + c
when t = 0
ln(65-20) = c
c = ln(45)
ln(T-Ts) = -kt + ln(45)
T-Ts = e^(-kt+ln(45))
T-Ts = e^(-kt)*e^(ln(45)
T-Ts = e^(-kt)*45 = 45e^(-kt)
At 9.00 pm t =3
33-20 = 45e^(-3k)
13/45 = e^(-3k)
ln(13/45) = -3k
k=0.41
T-Ts = 45e^(-0.4t)
Ts = 20
T = 20 + 45e^(-0.4t) ...QED