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In the realm of first-order linear differential equations, the integrating factor is a crucial tool used for finding solutions. It transforms a non-exact differential equation into an exact one, making it easier to integrate. To find an integrating factor, you first need to express the differential equation in its standard form: \(\frac{d y}{d \theta} + P(\theta) y = Q(\theta)\).
The integrating factor \(\mu(\theta)\) is derived using the formula \(\mu(\theta) = e^{\int P(\theta) \, d\theta}\). Here, the function \(P(\theta)\) is obtained from the coefficient next to \(y\) in the standard form of the differential equation. For the problem at hand, \(P(\theta) = -\frac{2}{\theta}\).
Hence, the integrating factor is calculated as:

• \(\mu(\theta) = e^{-\int \frac{2}{\theta} \, d\theta} = \theta^{-2}\)

Multiplying the entire differential equation by this integrating factor transforms it into a form where the left-hand side becomes the derivative of \((\theta^{-2} y)\). This simplification allows us to integrate easily to find the general solution.

An initial value problem adds more structure to a differential equation by specifying a value of the solution at a particular point. This particular specification helps determine a unique solution out of the infinite possible solutions to a differential equation.
In this exercise, the initial value is given as \(y(\pi/3)=2\). This means that when \(\theta\) is \(\pi/3\), the function \(y(\theta)\) takes the value 2. Integrating this initial condition into the solution allows us to solve for the constant \(C\) in the general solution.
When applying the initial condition, substitute \(\theta = \pi/3\) and \(y = 2\) into the general solution \(y(\theta) = \theta^2 \sec \theta + C \theta^2\). This results in an equation with \(C\) as the only unknown, which can be algebraically solved to find the precise value of \(C\).

• Substitute: \(2 = (\pi/3)^2 \cdot 2 + C (\pi/3)^2\)
• Solve for \(C\)

This forms the completed specific solution, distinguishing it from the general solution by locking it to the initial condition.

The general solution to a differential equation represents a family of solutions that include an arbitrary constant, often denoted as \(C\). This constant arises from the integration process, reflecting that there are infinitely many functions that satisfy the differential equation.
For first-order linear differential equations, finding the general solution involves integrating over the expression transformed using the integrating factor. This leads to the form \(y = g(\theta) + C\), where \(g(\theta)\) is a particular solution found by the integration, and \(C\) encompasses all possible vertical shifts of this function.
In the context of the exercise, after multiplying the differential equation by the integrating factor and integrating both sides, the general solution was found:

• \(\theta^{-2} y = \sec \theta + C\)

Solving this for \(y\) gives \(y(\theta) = \theta^2 \sec \theta + C \theta^2\). This expression contains \(C\), representing the set of all functions that solve the differential equation before applying any specific conditions. To find a unique solution, such as for an initial value problem, \(C\) is determined through additional conditions provided by the problem.