91Ó°ÊÓ

The integrating factor is a crucial tool for solving first-order linear differential equations like the one presented in the exercise. When dealing with differential equations of the form \( \frac{dy}{dt} + P(t)y = Q(t) \), the integrating factor \( \mu(t) \) is used to simplify the equation into a form that is easier to integrate.
The integrating factor is calculated as \( \mu(t) = e^{\int P(t) \, dt} \). In this example, because \( P(t) = 2 \), the integrating factor becomes \( e^{2t} \).
This factor is then multiplied throughout the differential equation, transforming it into \( \frac{d}{dt}(e^{2t}y) = 3e^{2t} \). This transformation is possible because the derivative of the product \( e^{2t}y \) respects the structure of the equation, making it integrated straightforwardly.

An initial value problem not only involves solving a differential equation but also demands finding a specific solution that meets certain initial conditions. In this exercise, the initial condition given is \( y(0) = 1 \).
The purpose of the initial condition is to provide specific values at a particular instant, ensuring the solution you find is unique to the problem's requirements. Once a general solution of the differential equation is found, as in \( y(t) = \frac{3}{2} + Ce^{-2t} \), the initial condition is used to find the specific value of the constant \( C \). By applying \( y(0) = 1 \), we substitute into the equation and solve for \( C \). This completes the initial value problem by ensuring the solution \( y(t) \) adheres to the conditions provided.

Solving a first-order linear differential equation involves a step-by-step procedure that eventually derives a solution in terms of the dependent variable, in this case, \( y(t) \). Once the equation is multiplied by the integrating factor, it takes a form that can be easily integrated. The key steps include:

• Integration leads to an expression like \( e^{2t} y = \frac{3}{2} e^{2t} + C \).
• Isolating \( y(t) \) by dividing both sides by the integrating factor \( e^{2t} \).
  

The final differential equation solution expresses \( y(t) \) in terms of constants determined via initial conditions or other provided settings.
This process not only solves the equation but elucidates the behavior of the solution over time.

The constant of integration \( C \) arises when you integrate an expression, as integration can yield infinitely many solutions differing by a constant. In differential equations, particularly with an initial value problem, \( C \) is essential in tailoring the generic solution to meet specific criteria.
In this case, after integrating, we find the general solution \( y(t) = \frac{3}{2} + Ce^{-2t} \), where \( C \) is initially unknown. Using the initial condition \( y(0) = 1 \), you substitute \( t = 0 \) into the equation, solve for \( C \), and obtain a specific value: \( C = -\frac{1}{2} \).
Hence, the solution becomes a specific one: \( y(t) = \frac{3}{2} - \frac{1}{2}e^{-2t} \), uniquely meeting the initial conditions set out in the problem.