Math
posted by Lucy .
on a day when the air temperature is 34 degrees celsius, a can of soft drink is taken from the fridge at a temperature of 4 degrees celsius at 2pm. Thereafter the temperature of the can satisfies the differential equation; dT/dt=0.1(34T), where t is the number of minutes after 2pm. What is the temperature of the can at 2:15pm? At what time does the temperature reach 30 degrees?

dT/dt=0.1(34T)
dT/(34T) = 0.1dt
Integrate both sides
ln(34T)=0.1t+c
34T=Ce^(0.1t)
T=34Ce^(0.1t)
Initial conditions:
t=0, T=4 => 4=34C => C=30
so
T=3430e^(0.1t)
A) Substitute t=15 to solve for T.
B) Substitute T=30 to solve for t. 
wow thank you so much

You're welcome!
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