Problem 16 - Advanced Engineering Mathematics

Question

In Problems 1-22, solve the given differential equation by separation of variables. $$ \frac{d Q}{d t}=k(Q-70) $$

Step-by-Step Solution

Answer

The solution is \( Q = 70 + A e^{kt} \).

Step 1: Recognize the Equation Type

The given equation \(\frac{dQ}{dt} = k(Q-70)\) is a separable differential equation, which means we can separate the variables to solve it.

Step 2: Separate Variables

Rewrite the differential equation by moving all terms involving \(Q\) to one side and the terms involving \(t\) to the other side:\[ \frac{1}{Q-70} \, dQ = k \, dt \]

Step 3: Integrate Both Sides

Integrate both sides of the equation with respect to their respective variables:\[ \int \frac{1}{Q-70} \, dQ = \int k \, dt \]

Step 4: Solve the Integrals

Compute the integrals:\[ \ln|Q-70| = kt + C \]where \(C\) is the constant of integration.

Step 5: Solve for Q

Exponentiate both sides to solve for \(Q\):\[ |Q-70| = e^{kt+C} = e^C \cdot e^{kt} \]

Step 6: Express the Solution

Let \(A = e^C\), a new constant, hence:\[ Q - 70 = A e^{kt} \]Finally, write the solution as:\[ Q = 70 + A e^{kt} \]

Key Concepts

Variable Separation

In solving differential equations, a common technique is variable separation. This method is particularly useful for certain types of differential equations known as separable equations. The main idea is to manipulate the equation so that all terms involving one variable appear on one side of the equation, and all terms involving the other variable appear on the other side. In the context of the exercise, this means adjusting the equation \( \frac{dQ}{dt} = k(Q-70) \) such that all the \( Q \) terms are on one side and the \( t \) terms are on the other.

• Mathematically, this involves using algebraic operations to isolate \( Q \) and \( dQ \) on one side of the equation.
• On the opposite side, isolate \( t \) and \( dt \) along with any functions of \( t \). This gives a complete set up for the next step, integration.

Once you're able to separate the variables clearly, the pathway to solving the differential equation becomes much clearer, and this setup is crucial for progress.

Integration of Differential Equations

After separating the variables in a differential equation, the next step is to integrate each side. This process involves treating the differential components \(dQ\) and \(dt\) almost like "extra small" elements that express how changes in \( Q \) relate to changes in \( t \). For the given problem \( \frac{1}{Q-70} \, dQ = k \, dt \):

• The left side, \( \int \frac{1}{Q-70} \, dQ \), can be solved by recognizing it as a logarithmic integral and hence integrates to \( \ln|Q-70| \).
• The right side, \( \int k \, dt \), is a simple integral that results in \( kt \), assuming \( k \) is a constant.

Understanding each integration part is critical. It involves recognizing the type of function you are integrating, such as exponential, polynomial, or logarithmic, and applying the correct integration rules.

Solving Differential Equations Using Separation

To solve a differential equation using the separation of variables, once you have integrated both sides, it often involves additional algebraic manipulation to express the dependent variable as a function of the independent variable. For our differential equation, after integrating, we derive the expression: \[ \ln|Q-70| = kt + C \] To further solve, we must exponentiate both sides to remove the natural logarithm:

• This transforms \( |Q-70| = e^{kt+C} \) into \( |Q-70| = e^C \cdot e^{kt} \), simplifying our result substantially.
• It's crucial to understand how exponentiation cancels logarithms, effectively reversing the logarithmic operation applied in previous integration.
• Adjust \( e^C \) to be a single constant \( A \), allowing us to express the equation in a neat form.

Constant of Integration

When integrating, we introduce an arbitrary constant known as the constant of integration, often denoted as \( C \). This constant arises because the process of integration determines an indefinite integral, which is a general form expressing all potential solutions of the differential equation. In our equation after integration, the constant \( C \) appears so:

• This constant accounts for any initial conditions or specific solutions to the differential equation problem.
• By exponentiating the result, \( C \) becomes part of the overall multiplicative constant, represented as \( e^C \) or a single constant \( A \).
• Practical application or determination of "initial conditions" can specify this constant, thereby tailoring the general solution to a particular scenario.

Including \( C \) in your solution ensures that you respect the need for generality, and it later facilitates incorporating any specific initial conditions provided in the problem.