91Ó°ÊÓ

Linear first-order differential equations are a class of differential equations characterized by their specific form, which is given as \( y' + Py = Q \). Here, \( y' \) represents the derivative of \( y \) with respect to \( t \), \( P \) is a constant or function of \( t \), and \( Q \) is a constant or a function of \( t \). These equations are termed "first-order" because they involve the first derivative of the unknown function \( y \). They are called "linear" because both \( y' \) and \( y \) appear to the first power and are not multiplied by each other.

In the context of our problem, we have the equation \( y'(t) = 3y - 6 \). We can rewrite it as \( y'(t) - 3y = -6 \) to more clearly identify the terms, where \( P = -3 \) and \( Q = -6 \). Understanding this form is crucial because it guides us to use the right method for solving, namely the integrating factor method.

The method of solving linear first-order differential equations often involves finding an integrating factor. An integrating factor is a function, typically denoted by \( \mu(t) \), which, when multiplied through the differential equation, transforms it into an easily integrable form.

For the equation \( y'(t) - 3y = -6 \), the integrating factor \( \mu(t) \) is based on the term \( P \), which is \(-3\) in this case. The integrating factor is obtained by taking the exponential of the integral of \( P \), expressed as:
\[\mu(t) = e^{\int -3 \, dt} = e^{-3t}\]

This integrating factor converts our original equation into:
\( (e^{-3t} y(t))' = -6e^{-3t} \)

The left side is now a derivative of the product of the integrating factor and \( y \). This makes it straightforward to integrate both sides to solve for \( y \).

The concept of an integrating factor is powerful because it transforms the differential equation into a form that is much simpler to solve, leveraging the properties of derivatives.

Initial conditions allow us to find the specific solution to a differential equation, in contrast to the general solution. They are individual values given for the function itself or its derivatives at certain points, which help determine the constant of integration that arises when solving differential equations.

In the equation \( y(t) = 2 + Ce^{3t} \), the general solution includes an arbitrary constant \( C \). To find the particular solution, we apply the initial condition \( y(0) = 9 \). This means at \( t = 0 \), \( y \) equals 9.

By substituting \( y(0) = 9 \) into \( y(t) = 2 + Ce^{3t} \), we set up the equation:
\( 9 = 2 + Ce^{0} \).

This simplifies to \( 9 = 2 + C \), and solving this gives \( C = 7 \). Therefore, the particular solution that satisfies the initial condition is \( y(t) = 2 + 7e^{3t} \).

Initial conditions are crucial when modeling real-world scenarios, as they ensure that the solution fits the observed or initial data of a problem.