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The characteristic equation plays a crucial role in solving second-order linear differential equations. It comes from rewriting the differential equation in terms of its roots, or characteristic values. For the differential equation \( y'' - 4y' + 4y = 0 \), we assume a solution of the form \(y = e^{rt}\) and substitute it into the equation. This gives us the characteristic polynomial, as shown below: \( r^2 e^{rt} - 4r e^{rt} + 4 e^{rt} = 0 \) By factoring out the common term \( e^{rt} \), we get: \( r^2 - 4r + 4 = 0 \). Solving this quadratic equation by factoring, we find \( (r-2)^2 = 0 \) leading to a double root: \( r = 2 \). This sole root will help us form the general solution.

Initial conditions are given values that a solution to a differential equation must satisfy. They help determine the specific solution from the general solution of the differential equation. For instance, consider the initial conditions:

• \(\text{y(0) = 0}\)
• \(\text{y'(0) = 1}\)

These initial values are applied to find the constants, \( C_1 \) and \( C_2 \), in the general solution. Given the general solution \( y(t) = (C_1 + C_2 t)e^{2t} \), we substitute: For \(\text{y(0) = 0}\): \(0 = (C_1 + C_2 \times 0)e^{0} = C_1 \). Thus, \( C_1 = 0 \). Next, apply the initial condition \(\text{y'(0) = 1}\), but first find \( y'(t) \). By using the product rule: \( y'(t) = 2(C_1 + C_2 t)e^{2t} + C_2 e^{2t} \), Substituting \( C_1 = 0 \) and \( \text{t = 0} \), we get: \( 1 = 2(0)e^{0} + C_2 e^{0} = C_2\). Thus, \( C_2 = 1 \). Initial conditions effectively help tailor the particular solution to fit the given scenario.

The general solution of a second-order differential equation encompasses all possible solutions before we apply any initial conditions. For a differential equation with repeated roots, like \(r=2\), the general solution follows a specific form. Given our characteristic equation \((r-2)^2 = 0 \), with a double root at \(r = 2\), the general solution is: \(y(t) = (C_1 + C_2 t) e^{2t}\), Here, \(C_1\) and \(C_2\) are arbitrary constants. These constants will be determined later using the initial conditions. To break it down:

• The term \(e^{2t}\) corresponds to our characteristic root \(r=2\)
• \(C_1\) represents the part of the solution associated with \(e^{2t}\)
• \(C_2 t\) represents the part associated with the repeated root, which necessitates an additional factor of \(t\) to form a linearly independent solution.

This general solution encapsulates all potential behaviors of the differential equation prior to any specific constraints.

A particular solution satisfies both the differential equation and the initial conditions provided. In our example, we already determined \( C_1 = 0 \) and \( C_2 = 1 \) using the initial conditions.Here’s how the particular solution is formed: Starting with the general solution: \( y(t) = (C_1 + C_2 t) e^{2t} \), Substituting \( C_1 = 0 \) and \( C_2 = 1 \), we get: \( y(t) = (0 + 1 \times t) e^{2t} \) simplifying to: \( y(t) = t e^{2t} \). This is our particular solution. It satisfies the original differential equation and perfectly aligns with the given initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \). After solving with the initial conditions, you get a specific function that will satisfy the conditions at all points in the domain.