Chapter 6: Q. 78 (page 573)

Prove Theorem 6.22 by solving the initial-value problem $\frac{d T}{d t} = k (A - T)$ with T(0) = T0, where k and A are constants.

Short Answer

Expert verified

$T (t) = A - (A - T_{0}) e^{- k t}$

Step by step solution

Step 1. Given information

$\frac{d T}{d t} = k (A - T)$

Step 2. Integrating both the sides

$\int \frac{d T}{(A - T)} = \int k d t$

$- \ln (A - T) = K t + C 1$

Therefore,

$\ln (A - T) = - K t - C 1$

Since, $e^{- C 1} = C$

Therefore,

localid="1649156311528" $A - T = C e^{- k t}$

localid="1649156328670" $T = A - C e^{- k t}$

Step 3. Using initial conditions

Using initial conditions, we get,

$C = A - T_{0}$

Substitute the value of C,

$T (t) = A - (A - T_{0}) e^{- k t}$

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Prove Theorem 6.22 by solving the initial-value problem $\frac{d T}{d t} = k (A - T)$ with T(0) = T0, where k and A are constants.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Integrating both the sides

Step 3. Using initial conditions

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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

Step by step solution

Step 1. Given information

Step 2. Integrating both the sides

Step 3. Using initial conditions

One App. One Place for Learning.

Most popular questions from this chapter

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Prove Theorem 6.22 by solving the initial-value problem $\frac{d T}{d t} = k (A - T)$ with T(0) = T0, where k and A are constants.