How do you find the particular solution to #dT+k(T-70)dt=0# that satisfies #T(0)=140#?

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1 Answer
Steve M
Nov 8, 2016

# T(t) = 70 + 70e^(-kt) #

Explanation:

We have # dT + k(T-70)dt = 0 # which is a First Order separable DE

We can rearrange as follows:

# dT =- k(T-70)dt => (dT)/(k(T-70)) = -dt #
# :. 1/kint 1/(T-70) dT= - int dt #

Integrating gives:
# :. 1/kln (T-70) = -t+C #

I always prefer to aptly the initial conditions as soon as possible to minimise the change of an algebraic slip:

#T=140# when #t=0#
# :. 1/kln (140-70) = C => 1/kln (70) #

Substituting back into our DE solution gives us:
# 1/kln (T-70) = -t+1/kln (70) #
# :. ln (T-70) = -kt+ln (70) #
# :. T-70 = e^(-kt+ln (70)) #
# :. T-70 = e^(-kt)e^(ln (70)) #
# :. T-70 = e^(-kt)(70) #
# :. T = 70 + 70e^(-kt) #

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