91Ó°ÊÓ

Differential equations are mathematical equations that relate a function with its derivatives. In the context of this exercise, we are given a first-order linear differential equation. The general form of a first-order linear differential equation is \( y' + p(t)y = q(t) \), where \( y' \) is the derivative of \( y \) with respect to \( t \), and \( p(t) \) and \( q(t) \) are functions of \( t \).

The goal when solving such differential equations is to find the function \( y(t) \) that satisfies the equation. It involves a systematic method that often requires integrating factors or other techniques to simplify and solve.

• Differential equations are fundamental in modeling real-world phenomena.
• Solving them involves finding the function \( y(t) \) that fulfills the equation.

The example equation we have is \( y'(t) = 3y - 6 \). Our task is to find the particular solution given the initial condition.

The integrating factor is a pivotal technique used to solve first-order linear differential equations. It helps in turning a non-exact differential equation into an exact one, making it easier to solve. The integrating factor \( \mu(t) \) is typically of the form \( e^{\int p(t) \, dt} \), where \( p(t) \) is derived from the differential equation.

For the exercise at hand, the equation is rearranged to \( y'(t) - 3y = -6 \), making it resemble the standard form. The function \( p(t) \) here is \(-3\), and the integrating factor then becomes \( e^{-3t} \). Multiplying the entire differential equation by this factor transforms the left side into the derivative of a product \((y(t) \cdot e^{-3t})\).

• Using the integrating factor converts the equation into an easily integrable form.
• It facilitates finding the product of the solution and the integrating factor.

This method simplifies the solving process significantly, making integration straightforward.

Initial conditions are vital in solving differential equations as they help determine the specific solution from a family of solutions. In our example, the initial condition provided is \( y(0) = 9 \). This condition specifies the value of the function \( y(t) \) at \( t = 0 \), which is crucial for finding the constant that arises during the integration process.

After integrating the equation, we obtain a general solution that includes an arbitrary constant \( C \). Applying the initial condition helps us solve for this constant:

• Substitute the initial condition values into the general solution.
• Solve for the constant \( C \) using basic algebra.

In our scenario, substituting \( y(0) = 9 \) into the equation determined \( C = 7 \). With \( C \) known, the specific solution of the differential equation is fully defined.

Exponential functions play a critical role in solving differential equations, especially when using integrating factors. These functions take the form \( e^{kt} \), where \( k \) is a constant, and they appear naturally in the solutions of linear differential equations. In this task, they are employed in the integrating factor method.

In practice, exponential functions help 'undo' the effects of differential operations or rates of change in the equation. For instance, after multiplying by the integrating factor \( e^{-3t} \) and integrating, we end up with terms like \( 2e^{-3t} \) and \( 7 \). Finally, to express \( y(t) \) alone, we use the exponential rule \( e^{-x} = \frac{1}{e^{x}} \), rearranging to make \( y(t) \) the subject.

• Exponential functions are essential in the solutions of differential equations.
• They help express solutions in a form that satisfies the original equation.

Thus, the final solution is \( y(t) = 2 + 7e^{3t} \), showcasing the elegant nature of exponential functions in mathematics.