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Prove Theorem 6.22 by solving the initial-value problem dTdt=k(A−T)with T(0) = T0, where k and A are constants.

Short Answer

Expert verified

T(t)=A-(A-T0)e-kt

Step by step solution

01

Step 1. Given information

dTdt=k(A-T)

02

Step 2. Integrating both the sides

∫dT(A-T)=∫kdt

-ln(A-T)=Kt+C1

Therefore,

ln(A-T)=-Kt-C1

Since, e-C1=C

Therefore,

localid="1649156311528" A-T=Ce-kt

localid="1649156328670" T=A-Ce-kt

03

Step 3. Using initial conditions

Using initial conditions, we get,

C=A-T0

Substitute the value of C,

T(t)=A-(A-T0)e-kt

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