Problem 77 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(67-86\). $$ \frac{d v}{d t}=\frac{1}{2} \sec t \tan t, \quad v(0)=1 $$

Step-by-Step Solution

Answer

The solution is \( v(t) = \frac{1}{2} \sec t + \frac{1}{2} \).

Step 1: Identify the Differential Equation

The given problem is a first-order differential equation: \( \frac{dv}{dt} = \frac{1}{2} \sec t \tan t \). We also have an initial condition: \( v(0)=1 \).

Step 2: Separate Variables

Separate the variables by moving terms involving \(v\) on one side and terms involving \(t\) on the other. However, for this problem let's directly consider that the given function just involves \(t\). The equation can be integrated directly: \( dv = \frac{1}{2} \sec t \tan t \, dt \).

Step 3: Integrate Both Sides

Integrate both sides with respect to \(t\). The left side is simply \( \int dv = v \). The right side can be integrated as \( \int \frac{1}{2} \sec t \tan t \, dt = \frac{1}{2} \int \sec t \tan t \, dt = \frac{1}{2} \sec t + C \).

Step 4: Apply the Initial Condition

Apply the initial condition \( v(0) = 1 \) to find \( C \). At \( t = 0 \), \( \sec 0 = 1 \), so: \( 1 = \frac{1}{2} \cdot 1 + C \). Solve for \( C \) to find \( C = \frac{1}{2} \).

Step 5: Write the Solution

Substitute \( C \) back into the integrated equation to find \( v(t) = \frac{1}{2} \sec t + \frac{1}{2} \).

Key Concepts

Initial Value Problem

An initial value problem, commonly abbreviated as IVP, involves solving a differential equation along with a given initial condition. The goal is to find a function that satisfies not only the differential equation but also the initial condition, providing a specific solution rather than a general family of solutions.
This type of problem is crucial in many real-world applications as it allows for modeling processes that begin at a certain point in time with a known state.

• The given differential equation is \ \( \frac{dv}{dt} = \frac{1}{2} \sec t \tan t \) \.
• The initial condition given is \ \( v(0) = 1 \) \.

These pieces of information together formulate the IVP, guiding us in finding a unique solution for \( v(t) \).

Separate Variables

Separating variables is a technique often used to solve differential equations where all terms involving one variable are moved to one side of the equation, and all terms involving another variable are moved to the other side.
This method simplifies the process of integration, as it often reduces the problem to simpler integrals that can be solved individually.
In the given problem, we notice that terms can readily be prepared for integration:

• Write the differential equation as \ \( dv = \frac{1}{2} \sec t \tan t \, dt \).

This equation is already separated in terms of \( v \) and \( t \), allowing direct integration to proceed.

Integration

Integration is the process of finding the antiderivative or the integral of a function. In solving differential equations, this step is crucial as it helps in deriving the function from the rate of change.

• The left side integrates simply: \ \( \int dv = v \) \.
• For the right side, integrate as: \ \( \int \frac{1}{2} \sec t \tan t \, dt = \frac{1}{2} \int \sec t \tan t \, dt \) \, which results in \ \( \frac{1}{2} \sec t + C \) \.

Here, we used the known integral of \( \sec t \tan t \) to find the particular antiderivative.
This integration gives us the general solution of the differential equation before applying any initial conditions.

Applying Initial Conditions

Applying initial conditions allows us to find a specific constant in the general solution of a differential equation, resulting in the unique solution to the initial value problem.
Here, the general solution after integration is:
\( v(t) = \frac{1}{2} \sec t + C \).Apply the initial condition \( v(0) = 1 \) to solve for \( C \):

• Substitute \( t = 0 \) in the solution, knowing \( \sec 0 = 1 \).
• This gives \( 1 = \frac{1}{2} \cdot 1 + C \).
• Solve for \( C \) to find \( C = \frac{1}{2} \).

With \( C \) determined, substitute back to get the specific solution: \( v(t) = \frac{1}{2} \sec t + \frac{1}{2} \).
Thus, the initial condition personalizes the solution to the particular scenario described by the initial value problem.