91Ó°ÊÓ

To solve second-order linear homogeneous differential equations like the one in the exercise, a key step is finding the characteristic equation. The differential equation given is \( y'' - 10y' + 25y = 0 \). To form the characteristic equation, you transform each derivative to powers of \( r \): replace \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1.
This process leads to the quadratic characteristic equation: \( r^2 - 10r + 25 = 0 \). This equation is fundamental because its roots help in determining the general solution of the differential equation.

• Replacement Rule: \( y'' \to r^2, \; y' \to r, \; y \to 1 \)
• Characteristic Equation: \( r^2 - 10r + 25 = 0 \)

The nature of the roots obtained from this equation (real and distinct, real and repeated, or complex) influences the form of the solutions.

Boundary conditions are crucial in determining the specific solution to a differential equation that meets particular criteria. In our example, the boundary conditions are given as \( y(0) = 1 \) and \( y(1) = 0 \).
These conditions impose restrictions on the constants \( C_1 \) and \( C_2 \) in the general solution. By applying each boundary condition to the general solution, \( y(t) = (C_1 + C_2 t) e^{5t} \), you can solve for these constants.

• First Condition: Substituting \( t = 0 \) and \( y = 1 \) results in finding \( C_1 = 1 \).
• Second Condition: With \( t = 1 \) and \( y = 0 \), it resolves to finding \( C_2 = -\frac{1}{e^5} \).

Using boundary conditions ensures the solution is tailored to specific scenarios, providing the particular solution needed.

The general solution of a second-order linear homogeneous differential equation depends on the characteristic equation's roots. For the obtained characteristic equation \( r^2 - 10r + 25 = 0 \), we found a repeated root, \( r = 5 \).
A repeated real root affects the form of the general solution which takes on the form \( y(t) = (C_1 + C_2 t) e^{5t} \), where \( C_1 \) and \( C_2 \) are constants that need further definition through boundary conditions.

• General Form with Repeated Roots: \( y(t) = (C_1 + C_2 t) e^{rt} \)
• In our case: \( r = 5 \) leading to \( y(t) = (C_1 + C_2 t) e^{5t} \)

This formulation ensures flexibility through constants \( C_1 \) and \( C_2 \), shaping the solution to match initial or boundary conditions.