91Ó°ÊÓ

Differential equations are equations that involve an unknown function and its derivatives. A first-order linear differential equation is a specific type of differential equation that can be written in the standard form:

• \( y' + P(x)y = Q(x) \)

Here, \( y' \) represents the first derivative of the unknown function \( y \) with respect to \( x \), and \( P(x) \) and \( Q(x) \) are continuous functions defined over a given interval. In the example provided, the differential equation is \((1+x) y' + y = \cos(x)\). To transform it into the standard form, we divide the whole equation by \(1+x\), resulting in:

• \( y' + \frac{1}{1+x}y = \frac{\cos(x)}{1+x} \)

This process aligns the equation with the form \( y' + P(x)y = Q(x) \), making it easier to solve.

An integrating factor is a specific function that, when multiplied with a differential equation, allows the equation to be written in a form that is easy to integrate. For first-order linear differential equations, the integrating factor \( \mu(x) \) is determined by:

• \( \mu(x) = e^{\int P(x) \, dx} \)

For the given equation, we identify \( P(x) = \frac{1}{1+x} \). Integrating \( P(x) \) yields:

• \( \int \frac{1}{1+x} \, dx = \ln|1+x| \)

Thus, the integrating factor becomes:

• \( \mu(x) = e^{\ln|1+x|} = |1+x| \)

Given the context where \( 1+x \) is positive, this simplifies to \( \mu(x) = 1+x \). Multiplying the entire differential equation by \( 1+x \) aligns terms for simple integration.

In solving differential equations, initial conditions play a crucial role in determining the specific solution from a family of possible solutions. An initial condition is often given as a value of the function \( y \) at a particular point \( x \). In this example, the initial condition is \( y(0) = 1 \). This means that when \( x = 0 \), the solution should satisfy \( y = 1 \).With the general solution form \((1+x)y = \sin(x) + C\), we can substitute \( x = 0 \) and \( y = 1 \):

• \( (1+0) \cdot 1 = \sin(0) + C \rightarrow 1 = 0 + C \)

From this, we determine that \( C = 1 \). Applying this back into the general solution completes the process, providing the specific solution. In this case, it simplifies to:

• \( y = \frac{\sin(x) + 1}{1+x} \)

The initial condition helps tailor the universal solution to satisfy specific requirements at a given point, ensuring accuracy and relevance to the problem at hand.