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A linear first-order differential equation is a type of differential equation that involves the first derivative of the unknown function (usually denoted as y or y'). The general form of a linear first-order differential equation is: \[ y' + P(t)y = Q(t) \] In this equation, \( P(t) \) and \( Q(t) \) are functions of t (the independent variable), and y is the dependent variable we are trying to solve for. For example, in the given exercise: (1+t)y' + y = -1 We need to rewrite it in the standard form by dividing every term by (1+t), giving the equation: \[ y' + \frac{1}{1+t}y = \frac{-1}{1+t} \] This form is crucial because it helps us identify the necessary steps to solve the differential equation using techniques such as the integrating factor method.

An integrating factor is a special function used to solve linear first-order differential equations. It simplifies the equation, making it easier to integrate. The integrating factor \( \mu(t) \) is given by: \[ \mu(t) = e^{\int P(t) dt} \] In the given example: \[ P(t) = \frac{1}{1+t} \] Therefore, the integral of \( P(t) \) is: \[ \int \frac{1}{1+t} dt = \ln|1+t| \] Thus, the integrating factor becomes: \[ \mu(t) = e^{\ln|1+t|} = |1+t| = 1+t \] This integrating factor \( (1+t) \) will be used to simplify the original differential equation.

To solve a linear first-order differential equation using the integrating factor, follow these steps:

• Identify \( P(t) \) and \( Q(t) \) in the standard form \( y' + P(t)y = Q(t) \)
• Calculate the integrating factor \( \mu(t) = e^{\int P(t) dt} \)
• Multiply the entire differential equation by the integrating factor
• Simplify the equation to recognize the derivative of a product
• Integrate both sides with respect to the independent variable
• Solve for the dependent variable y

In our given example, after finding \( \mu(t) = (1+t) \), we can multiply the entire differential equation: \[ y' + \frac{1}{1+t}y = \frac{-1}{1+t} \] by \( 1+t \) to get: \[ y' (1+t) + y = -1 \] Then, recognize that the left-hand side is the derivative of the product \( y(1+t) \): \[ \frac{d}{dt}[y(1+t)] = -1 \] Finally, integrate both sides with respect to t: \[ \int \frac{d}{dt}[y(1+t)] dt = \int -1 dt \] resulting in: \[ y(1+t) = -t + C \]

A key part of solving differential equations is being comfortable with various integration techniques. The most relevant technique for this problem is the integration of simple rational functions. In the given exercise, we computed the integral: \[ \int \frac{1}{1+t} dt \] Using the natural logarithm: \[ \int \frac{1}{1+t} dt = \ln|1+t| \] Additionally, when integrating both sides of the equation in Step 7: \[ \int \frac{d}{dt}[y(1+t)] dt = \int -1 dt \] we use the fact that: \[ \int -1 dt = -t + C \] Understanding these techniques is crucial for solving differential equations effectively.

When integrating, we always include a constant of integration (typically denoted by C). This is because the process of integrating introduces an unknown constant value. In the final steps of solving the given differential equation, after integrating both sides: \[ y(1+t) = -t + C \] We observe that C represents an arbitrary constant. This constant accounts for the general solution to the differential equation, fulfilling any initial or boundary conditions that may be applied later. After rearranging and solving for y, the final solution incorporates this constant: \[ y = \frac{-t + C}{1+t} \] The constant of integration ensures that we have the most general form of the solution, adaptable to various potential constraints or additional conditions we may encounter.