91Ó°ÊÓ

In the context of solving a first-order linear differential equation, an integrating factor is a crucial tool. The idea is to transform the differential equation into an easier form that is straightforward to integrate. For our specific differential equation, we began by identifying it as

• \( y' + P(t)y = 0 \)
• where \( P(t) = \frac{1}{1+t^2} \)

To find the integrating factor, denoted as \( \mu(t) \), we calculate\[ \mu(t) = e^{\int P(t)\, dt} = e^{\int \frac{1}{1+t^2} \, dt}. \]When integrating \( \frac{1}{1+t^2} \), we recognize the result as the inverse tangent function, yielding \( \arctan(t) \). This gives the integrating factor:\[ \mu(t) = e^{\arctan(t)}. \]Using an integrating factor allows us to rewrite the differential equation in such a way that the left-hand side becomes the derivative of a product of functions, thus simplifying the integration process.

A general solution provides a complete expression for all possible solutions of the differential equation. By employing an integrating factor, the given differential equation \( y' + \frac{y}{1+t^2} = 0 \) is transformed through multiplication:\[ e^{\arctan(t)}(y' + \frac{y}{1+t^2}) = 0. \]This simplifies to the derivative of the product \( e^{\arctan(t)}y \) as\[ \frac{d}{dt}(e^{\arctan(t)}y) = 0. \]The equation above tells us that the derivative of \( e^{\arctan(t)}y \) with respect to \( t \) is zero. Therefore, \( e^{\arctan(t)}y \) must be a constant, leading to\[ e^{\arctan(t)}y = C, \]where \( C \) represents the constant of integration determined by initial conditions if specified. Solving for \( y \), the general solution is:\[ y = \frac{C}{e^{\arctan(t)}}. \]This expression describes how \( y \) behaves for all values of \( t \), contingent on the arbitrary constant \( C \).

The integration of rational functions often involves recognizing standard forms. In our problem, integrating \( \frac{1}{1+t^2} \) leads directly to the function \( \arctan(t) \). This arises because

• The integral \( \int \frac{1}{1+t^2} \, dt \) is a known antiderivative, \( \tan^{-1}(t) \) or \( \arctan(t) \), forming part of basic calculus results.

The inverse tangent function, \( \arctan(t) \), is central in recognizing this integration for computational use. Understanding this standard integral is key in many contexts, especially in differential equations where it appears often. Utilizing \( \arctan(t) \) in calculating the integrating factor is a perfect example, showing both its importance and frequent recurrence in solving mathematical problems. This makes fluency in these standard forms very beneficial when working through mathematics problems that require integration as a tool.