QUESTION 6 [TOTAL MARKS: 9 MARKS]
The cooling of an engine is modelled by , where is the temperature (in degree
Celsius) and is the time (in minutes). The initial temperature is 100 degrees Celsius. Using
integrating factor, find the function of temperature .
[9 marks]

Ask by Gough Summers. in Brunei

Mar 22,2025

Question

QUESTION 6 [TOTAL MARKS: 9 MARKS]
The cooling of an engine is modelled by , where is the temperature (in degree
Celsius) and is the time (in minutes). The initial temperature is 100 degrees Celsius. Using
integrating factor, find the function of temperature .
[9 marks]

Ask by Gough Summers. in Brunei

Mar 22,2025

UpStudy AI · Accepted Answer

We are given the differential equation

$$
\frac{dT}{dt} + 5T = 200,
$$

with the initial condition

$$
T(0) = 100.
$$

**Step 1: Determine the integrating factor**

The integrating factor is given by

$$
\mu(t) = e^{\int 5\, dt} = e^{5t}.
$$

**Step 2: Multiply the differential equation by the integrating factor**

Multiplying both sides by $$ e^{5t} $$ gives

$$
e^{5t} \frac{dT}{dt} + 5e^{5t}T = 200e^{5t}.
$$

Notice that the left-hand side is the derivative of $$e^{5t}T$$:

$$
\frac{d}{dt}(e^{5t}T) = 200e^{5t}.
$$

**Step 3: Integrate both sides**

Integrate with respect to $$ t $$:

$$
\int \frac{d}{dt}(e^{5t}T)\, dt = \int 200e^{5t}\, dt.
$$

This simplifies to

$$
e^{5t}T = 200 \cdot \frac{1}{5} e^{5t} + C,
$$

or

$$
e^{5t}T = 40e^{5t} + C.
$$

**Step 4: Solve for $$ T $$**

Divide by $$ e^{5t} $$:

$$
T = 40 + Ce^{-5t}.
$$

**Step 5: Apply the initial condition**

Substitute $$ t = 0 $$ and $$ T(0) = 100 $$:

$$
100 = 40 + C \cdot e^{-5 \cdot 0} = 40 + C.
$$

Hence,

$$
C = 100 - 40 = 60.
$$

**Final Answer:**

$$
T(t) = 40 + 60e^{-5t}.
$$
The temperature function is $$ T(t) = 40 + 60e^{-5t} $$.