Problem 64 - Calculus for Biology and Medicine

Question

Solve the initial-value problem. $$ \frac{d N}{d t}=\frac{t+2}{t}, \text { for } t \geq 1 \text { with } N(1)=2 $$

Step-by-Step Solution

Answer

The solution is \( N(t) = t + 2\ln|t| + 1 \).

Step 1: Separate Variables

To solve the differential equation \( \frac{d N}{d t} = \frac{t+2}{t} \), first observe that it is already in a form where the right-hand side depends only on \( t \). Therefore, no separation is necessary in this case. We will proceed by integrating both sides with respect to \( t \).

Step 2: Integrate Both Sides

Integrate both sides with respect to \( t \). The left side becomes \( \int dN \) which is \( N(t) \) and the right side becomes \( \int \left( 1 + \frac{2}{t} \right) dt \).

Step 3: Compute the Integrals

Calculate the integral on the right side: \( \int \left( 1 + \frac{2}{t} \right) dt = \int 1 \, dt + \int \frac{2}{t} \, dt \). This evaluates to \( t + 2\ln|t| + C \), where \( C \) is the constant of integration.

Step 4: Write the General Solution

The general solution to the differential equation is given by \( N(t) = t + 2\ln|t| + C \).

Step 5: Apply the Initial Condition

Use the initial condition \( N(1) = 2 \) to find \( C \). Substitute \( t = 1 \) and \( N = 2 \) into the equation: \( 2 = 1 + 2\ln|1| + C \). Since \( \ln(1) = 0 \), this simplifies to \( 2 = 1 + C \).

Step 6: Solve for C

Solving \( 2 = 1 + C \) gives \( C = 1 \).

Step 7: Write the Specific Solution

Substitute \( C = 1 \) into the general solution to get the specific solution: \( N(t) = t + 2\ln|t| + 1 \).

Key Concepts

Initial Value Problem

An initial value problem in differential equations specifies a differential equation along with a specific value, known as the initial condition, at a particular point. This condition helps in finding a unique solution to the equation. In our problem, the differential equation is \( \frac{d N}{d t} = \frac{t+2}{t} \) and the initial condition given is \( N(1) = 2 \). This means that when the variable \( t \) is equal to 1, the function \( N(t) \) takes the value 2.
Understanding initial value problems is crucial because, unlike general solutions that include a constant term \( C \), an initial value problem allows us to pinpoint the exact value of \( C \), thus offering a specific solution for the problem. Initial value problems are particularly useful in modeling real-world situations where specific starting conditions are known.

Integration

Integration is a key technique in solving differential equations. It involves finding the anti-derivative of a function. In our situation, to solve \( \frac{d N}{d t} = \frac{t+2}{t} \), we need to integrate the right-hand side with respect to \( t \). This step transforms the differential equation into a more manageable form.
For the integral \( \int (1 + \frac{2}{t}) \, dt \), this can be split into two simpler integrals: \( \int 1 \, dt \) and \( \int \frac{2}{t} \, dt \).

• \( \int 1 \, dt = t \)
• \( \int \frac{2}{t} \, dt = 2 \ln |t| \)

Thus, the integration process gives us \( t + 2 \ln |t| \), which includes an integration constant \( C \). This constant is crucial for developing the general solution.

General Solution

The general solution of a differential equation refers to a solution that includes one or more arbitrary constants (usually denoted by \( C \)). It encompasses a family of functions that can be adjusted to satisfy various initial conditions. For the differential equation \( \frac{d N}{d t} = \frac{t+2}{t} \), the general solution is derived as \( N(t) = t + 2\ln|t| + C \).
This form is quite adaptable: by changing \( C \), we can potentially develop an infinite number of specific solutions, each corresponding to a different initial condition. The presence of \( C \) reflects the general nature of the solution, hinting that further specific information is needed, like an initial condition, to find the exact function representing the situation at hand.

Specific Solution

The specific solution of a differential equation is acquired when a general solution is tailored to correspond with a given initial condition. Adopting our general solution \( N(t) = t + 2\ln|t| + C \), we apply the initial condition \( N(1) = 2 \).
Substituting these values leads us to \( 2 = 1 + 0 + C \), which simplifies to \( C = 1 \) since \( \ln(1) = 0 \). We then plug \( C = 1 \) back into the general solution equation, giving \( N(t) = t + 2\ln|t| + 1 \).
This specific solution defines the function that precisely fits the given situation, indicating how \( N(t) \) behaves based on the initial condition provided. It's essential to understand that solving for the specific solution ensures the differential equation truly models the scenario with the required starting values.