Question

Finding a Particular Solution Using Separation of Variables In Exercises \(15-24\) , find the particular solution that satisfies the initial condition. $$ d T+k(T-70) d t=0 \quad T(0)=140 $$

Short answer

The particular solution that satisfies the given initial condition is \( T = 70(1+e^{-kt}) \)

Step-by-step solution

Separating Variables

The given differential equation is \( dT + k(T-70)dt = 0 \). We can rewrite this equation into a form where all terms involving T are on one side and those involving t are on the other. We have: \(\frac{1}{T-70}dT = -kdt \)

Integration

Now both sides of the equation can be integrated to obtain: \[ \int \frac{1}{T-70} dT = -k \int dt \] The left side of the equation can be integrated using the simple rule for the natural logarithm of a function, while the right side integration results in -kt. Simplifying, this leads to: \[ ln|T-70| = -kt + C \] where C is the constant of integration.

Applying the initial condition

The given initial condition is T(0) = 140. Substituting t = 0 and T = 140 into the equation from Step 2, it leads to: \[ ln|140-70| = -k(0) + C \] Solving for C gives: C = ln|70|

Substituting the value of C and solving for T

Now we substitute the calculated value of C from step 3 back into the equation from step 2 to obtain: \[ ln|T-70| = -kt + ln|70| \] Exponentiating both sides to remove the logarithmic function gives: \[ T-70 = 70e^{-kt} \] Solving for T results in: \[ T = 70(1+e^{-kt}) \] This is the required particular solution.

Key concepts

Differential Equations

Differential equations are a type of mathematical equation that involve functions and their derivatives. They are fundamental in expressing phenomena where there is a constant rate of change, such as physics, engineering, and other sciences. The equation provided in the exercise,
\( dT + k(T-70) dt = 0 \),
is a first-order linear differential equation, and it dictates how a temperature, T, changes over time, t. Solving differential equations often requires finding an unknown function that satisfies the given relationship between variables and their rates of change. Separation of variables is a technique used to solve such equations by isolating each variable with its corresponding differential on opposite sides of the equation.

Integration

Integration is the process of finding the integral of a function, which is essentially the reverse operation of differentiation. When we integrate both sides of the separated differential equation, as shown in Step 2 of the original solution,
\[ \int \frac{1}{T-70} dT = -k \int dt \],
we're summing up the small changes to find the overall quantity. In this case, we're finding the total change in temperature over time. The results of simple integrations are often known functions, like the natural logarithm in this exercise, and the indefinite integration introduces a constant of integration, C, which is later determined using initial conditions.

Initial Condition

Initial conditions are specific values provided for the variables of a differential equation at the start of the observation. These are used to find the particular solution of a differential equation that not only satisfies the differential equation but also fits a specific scenario. For our exercise, the initial condition is
\( T(0) = 140 \),
which means when \( t = 0 \), the temperature was 140 units. This condition enables us to find the precise value of the constant of integration, C, as shown in Step 3. Initial conditions are crucial as they ground the abstract mathematical solution in reality, aligning the solution to an observable quantity at a given time.

Particular Solution

A particular solution to a differential equation is a solution that not only satisfies the general form of the equation but also meets the criteria set by the initial conditions. After applying the initial condition, we find a specific value for the constant C, which leads us to a unique solution, tailored for our given scenario. As illustrated in Step 4, with the value of C determined, we can express the particular solution without ambiguity:
\( T = 70(1+e^{-kt}) \).
The particular solution is crucial for practical applications where specific outcomes are predicted or expected, such as determining the temperature at a given time during a process. Understanding the difference between a general solution—which incorporates an arbitrary constant—and a particular solution that precisely caters to given conditions is important in the application of differential equations.