91Ó°ÊÓ

A homogeneous linear differential equation is a type of differential equation where every term is dependent on the unknown function and its derivatives. In simpler terms, there is no "free-standing" constant term, everything is multiplied by some function of the dependent variable. The general form of a second-order homogeneous linear differential equation is

• \( ay^{\prime\prime} + by^{\prime} + cy = 0 \)

Here, \( a \), \( b \), and \( c \) are constants, and \( y \) is the function we need to find. The presence of zero on the right side means there is no external force or term affecting the system. For instance, in our exercise, the given equation is:

• \( 4 y^{\prime \prime} - 4 y^{\prime} + y = 0 \)

This equation is homogeneous because everything is expressed in terms of \( y \) and its derivatives, with nothing else added or subtracted from zero.
Homogeneous linear differential equations are important because they do not change the form of the solutions significantly when the coefficients \( a \), \( b \), and \( c \) are varied. Solutions to these equations can provide insights into systems that have no external inputs or constant terms interfering with their natural behavior.

The characteristic equation is a crucial step in solving linear differential equations, especially when they are constant coefficient equations. When solving a homogeneous linear differential equation, we usually assume a solution of the form

• \( y = e^{rt} \)

where \( r \) is a constant that we need to determine. Plugging this trial solution into the homogeneous differential equation transforms it into an algebraic equation in terms of \( r \). This new equation is called the characteristic equation.
For the exercise given, after assuming \( y = e^{rt} \) and substituting into the differential equation:

• \( 4(r^2 e^{rt}) - 4(r e^{rt}) + e^{rt} = 0 \)

we simplify to obtain:

• \( 4r^2 - 4r + 1 = 0 \)

This quadratic equation can be solved using methods like factoring or the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The roots determine the type of solution to the differential equation. In our problem, the roots are repeated, both being \( r = \frac{1}{2} \). This repetition influences the general solution form of our differential equation, emphasizing its importance in the solution process.

An initial value problem (IVP) in differential equations includes finding a specific solution derived from a general solution by fitting given initial conditions. Initial conditions describe the state of the system at the start (usually time \( t = 0 \)) and are essential to pinning down one unique solution among the infinite possible ones.
In our exercise, the initial value problem is defined as follows:

• \( y(0) = 4 \)
• \( y^{\prime}(0) = 4 \)

These conditions imply that the function \( y(t) \) has to pass through \( y = 4 \) when \( t = 0 \), and its first derivative needs to be 4 at the same point.
To apply these conditions:1. Start with the general solution obtained: - \( y(t) = (C_1 + C_2t)e^{\frac{t}{2}} \)2. Use \( y(0) = 4 \), leading to \( C_1 = 4 \) because \( e^0 = 1 \).3. Differentiate \( y(t) \) and apply \( y^{\prime}(0) = 4 \) to solve for \( C_2 \).
Thus, initial value problems allow us to identify a precise curve that not only satisfies the differential equation but also adheres to specific conditions given at \( t = 0 \). In this way, an IVP ensures we find the exact mathematical function describing the behavior of the modeled system.