91Ó°ÊÓ

A differential equation is a type of equation that involves an unknown function and its derivatives. In the provided exercise, we are working with a first-order linear differential equation of the form \( y' - p(t)y = q(t) \).

Here, \( y' \) represents the first derivative of \( y \) with respect to \( t \). The functions \( p(t) \) and \( q(t) \) are known functions of \( t \).

Differential equations are essential in describing various physical phenomena, such as motion, heat flow, and electrical circuits. Solving these equations often requires finding an unknown function that satisfies the given relationship between the function and its derivatives.

In our exercise, we identified that \( p(t) = -\frac{1}{10+t} \) and \( q(t) = 2 \). This step is crucial as it sets up the problem for finding the integrating factor, which simplifies solving the equation.

The integral is a mathematical concept that finds the area under a curve. In the context of our exercise, we use integration to find the integrating factor.

The integrating factor \( \mu(t) \) is essential to transform a non-exact differential equation into an exact one, which can be solved more easily. The general formula for the integrating factor is: \[ \mu(t) = e^{\int p(t) \; dt} \].

In our exercise, \( p(t) = -\frac{1}{10+t} \), so we compute the integral: \[ \int p(t) \; dt = \int -\frac{1}{10+t} \; dt \] which simplifies to \( -\ln|10 + t| \).

Simplified further, we get the integrating factor:
\( \mu(t) = e^{-\text{ln}(10+t)} = \frac{1}{10+t} \). This factor, when multiplied with the original differential equation, simplifies the task of solving for \( y \).

Logarithms are mathematical operations that help simplify expressions involving exponents. A logarithm provides the exponent to which a base must be raised to produce a given number.

In our exercise, we encounter the natural logarithm because of the integration of \( p(t) = -\frac{1}{10+t} \). The integral result is \( -\text{ln}|10 + t| \).

Properties of logarithms allow us to simplify further:
\( e^{-\text{ln}(10+t)} \)
Using the relationship \( e^{\text{ln}(x)} = x \), we simplify \( e^{-\text{ln}(10+t)} \) to \( \frac{1}{10+t} \).

Understanding logarithms is crucial for handling exponential functions, which frequently appear in differential equations and their solutions.