Problem 32 - Calculus and its applications

Question

(a) find the particular solution of each differential equation as determined by the initial condition, and (b) check the solution by substituting into the differential equation. \(\frac{d P}{d t}=0.024 P,\) where \(P=32\) when \(t=0\)

Step-by-Step Solution

Answer

The particular solution is \( P(t) = 32 e^{0.024t} \). Verified by substitution.

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{dP}{dt} = 0.024P \). This is a separable first-order linear differential equation.

Step 2: Separate the Variables

Separate the variables by dividing both sides by \( P \) and multiplying by \( dt \): \[ \frac{1}{P}dP = 0.024\,dt \]

Step 3: Integrate Both Sides

Integrate both sides of the equation: \[ \int \frac{1}{P}dP = \int 0.024\,dt \] This gives: \[ \ln |P| = 0.024t + C \] where \( C \) is the integration constant.

Step 4: Solve for P

To solve for \( P \), exponentiate both sides to eliminate the natural logarithm: \[ |P| = e^{0.024t + C} \] Let \( C_1 = e^C \), which is a positive constant, hence: \[ P(t) = C_1 e^{0.024t} \]

Step 5: Apply the Initial Condition

Use the initial condition \( P(0) = 32 \):\[ 32 = C_1 e^{0.024 \times 0} \]Simplifying gives \( 32 = C_1 \). Therefore, the particular solution is: \[ P(t) = 32 e^{0.024t} \]

Step 6: Verify the Solution

Substitute \( P(t) = 32 e^{0.024t} \) back into the original differential equation:\[ \frac{dP}{dt} = \frac{d}{dt}(32 e^{0.024t}) = 32 \cdot 0.024 e^{0.024t} \]This simplifies to \( 0.024 \cdot 32 e^{0.024t} \), which confirms \( \frac{dP}{dt} = 0.024P \), verifying the solution.

Key Concepts

Separable Differential Equations

Separable differential equations are a class of differential equations that can be rewritten such that all terms containing one variable are on one side, and all terms containing the other variable are on the other side. This process allows us to integrate each side with respect to its own variable.

In our given equation, \( \frac{dP}{dt} = 0.024P \), it's clear that the equation is already expressed as a product of a function of \( t \) and a function of \( P \).

**How to Separate Variables**:

• Begin by rearranging the terms: divide both sides by \( P \), and multiply by \( dt \). This gives us \( \frac{1}{P} dP = 0.024 dt \).
• Notice that the left side is a function of \( P \) and the right side is a function of \( t \). This separation of variables allows for integration on both sides.

Integration Techniques

Integration is a key step in solving separable differential equations. Once the variables are separated, each side is integrated with respect to its respective variable. Here’s how integration plays a role:

**Integrating Both Sides**:

• On the left side, you integrate \( \frac{1}{P} \) with respect to \( P \), resulting in \( \ln |P| \).
• On the right side, integrate \( 0.024 \) with respect to \( t \), giving \( 0.024t \).
• Combine these results with the integration constant \( C \) to get: \( \ln |P| = 0.024t + C \).

**Solving for \( P \)**:

• To remove the natural log, exponentiate both sides leading to \( |P| = e^{0.024t + C} \).
• Let \( C_1 = e^C \), transforming the equation to \( P(t) = C_1 e^{0.024t} \).

The integral constants in differential equations are crucial in ensuring any specific initial conditions are met.

Initial Conditions

Initial conditions allow us to find a particular solution from the general solution of a differential equation. In the problem, the initial condition provided is \( P = 32 \) when \( t = 0 \). This condition will help us solve for \( C_1 \) in the function.

**Applying the Initial Condition**:

• Substitute \( t = 0 \) and \( P = 32 \) into the equation \( P(t) = C_1 e^{0.024t} \).
• Simplify to find \( C_1 \): \( 32 = C_1 \, e^{0.024 \times 0} \).
• Since \( e^0 = 1 \), it follows that \( C_1 = 32 \).

We now have the particular solution \( P(t) = 32 e^{0.024t} \). Initial conditions are essential because they tailor the solution to pass through a specific point, ensuring it fits the context of the problem being solved.