Problem 26 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d w}{d \theta}=\theta w^{2} \sin \theta^{2}, \quad w(0)=1$$

Step-by-Step Solution

Answer

The solution is \( w = \frac{2}{\cos(\theta^2) + 1} \).

Step 1: Identify the Differential Equation Form

The differential equation given is \( \frac{d w}{d \theta} = \theta w^2 \sin \theta^2 \). We notice that this equation allows us to use the method of separation of variables.

Step 2: Separate Variables

We rearrange the equation to separate the variables \( w \) and \( \theta \). Divide both sides by \( w^2 \), to get: \[ \frac{1}{w^2} \frac{d w}{d \theta} = \theta \sin \theta^2 \] Now express it as \[ \frac{1}{w^2} dw = \theta \sin \theta^2 d\theta \] so that each side contains only one variable.

Step 3: Integrate Both Sides

Integrate both sides of the equation: \[ \int \frac{1}{w^2} dw = \int \theta \sin \theta^2 d\theta \] the left side integrates to \[ -\frac{1}{w} + C_1, \] and the integral on the right will require substitution \( u = \theta^2 \), \( du = 2\theta d\theta \), thus: \[ \frac{1}{2} \int \sin u \, du = -\frac{1}{2} \cos u = -\frac{1}{2} \cos(\theta^2) + C_2 \].

Step 4: Combine and Solve for \( w \)

Set the integrated expressions equal to express \[ -\frac{1}{w} = -\frac{1}{2}\cos(\theta^2) + C. \] Solving for \( w \), we have: \[ \frac{1}{w} = \frac{1}{2} \cos(\theta^2) - C \] Thus, \[ w = \frac{1}{1/2 \cos(\theta^2) - C} \].

Step 5: Use the Initial Condition

Apply the initial condition \( w(0) = 1 \) to determine \( C \). Set \( \theta = 0 \) in the equation \[ w = \frac{1}{1/2 \cos(\theta^2) - C} \], so \[ 1 = \frac{1}{1/2 \cos(0) - C} \] which gives \[ 1/2 - C = 1 \], thus \[ C = -1/2 \].

Step 6: Write the Final Solution

Substitute \( C = -1/2 \) back into the equation: \[ w = \frac{1}{1/2 \cos(\theta^2) + 1/2} \] which simplifies to \[ w = \frac{2}{\cos(\theta^2) + 1} \].

Key Concepts

Differential Equations

Differential equations are mathematical equations involving an unknown function and its derivatives. This type of equation can describe a variety of real-world phenomena, such as population growth, heat distribution, and mechanical motion.
When working with differential equations, particularly first-order equations like the one presented in the problem, it's crucial to recognize the structure. First-order differential equations involve the first derivative of an unknown function.
In the given exercise, we have a differential equation in the form \( \frac{d w}{d \theta} = \theta w^2 \sin \theta^2 \). The goal is to find an explicit function for \( w \) in terms of \( \theta \), satisfying some initial conditions. This kind of problem is common in calculus and often requires specific techniques like separation of variables to solve.

Initial Conditions in Calculus

Initial conditions are specific values that a solution to a differential equation must satisfy. These conditions are crucial because they help determine the unique solution among many possibilities. Without them, differential equations may have infinitely many solutions.
In the current exercise, the initial condition given is \( w(0) = 1 \). This means when \( \theta = 0 \), the value of \( w \) should be 1. We use this information later in our calculations to solve for any constants introduced during integration.
Applying initial conditions is a step in verifying and finalizing the solution; it helps pinpoint the exact function among the family of all potential solutions for the differential equation.

Integration Techniques

Integration is a fundamental technique in solving differential equations. In this problem, after separating variables, we have two integrals to solve: \( \int \frac{1}{w^2} \, dw \) and \( \int \theta \sin \theta^2 \, d\theta \).

• On the left side, \( \int \frac{1}{w^2} \, dw \) simplifies using a basic power rule for integration, resulting in \( -\frac{1}{w} + C_1 \).
• The right-side integral, \( \int \theta \sin \theta^2 \, d\theta \), is a bit more complex and requires substitution. By letting \( u = \theta^2 \), \( du = 2\theta \, d\theta \), the integral becomes \( \frac{1}{2} \int \sin u \, du \), leading us to \( -\frac{1}{2} \cos u = -\frac{1}{2} \cos(\theta^2) + C_2 \).

Integration techniques such as substitution are powerful tools in tackling more complex integrals. After integrating both sides, we solve for the constant using the initial condition before finding the explicit form of the solution.