91Ó°ÊÓ

In the realm of differential equations, the characteristic equation plays a pivotal role. To find this, we first express the differential equation in its standard form:

• Consider a solution of the form \( y = e^{rt} \) where \( r \) represents a constant.
• Substitute this expression into the given differential equation.

The resulting equation \( y'' + 12y' + 36y = 0 \) becomes \( r^2e^{rt} + 12re^{rt} + 36e^{rt} = 0 \). Next, we factor out \( e^{rt} \) since it is never zero, simplifying to: \[ r^2 + 12r + 36 = 0 \] This simplified quadratic equation is what we call the characteristic equation. Solving it provides the roots that are essential for forming the general solution of the differential equation.

When solving the characteristic equation \( r^2 + 12r + 36 = 0 \), we find \( r = -6 \) as a repeated root. Repeated roots arise when the discriminant in the quadratic formula \( b^2 - 4ac \) equals zero. In this equation, it simplifies to:\( r = \frac{-12 \pm \sqrt{0}}{2} = -6 \). Repeated roots occur when the quadratic formula yields the same value for both roots. The repeated nature indicates a special form for the general solution:

• For repeated roots \( r \), the general solution is \( y(t) = (C_1 + C_2 t)e^{rt} \).
• This is different from distinct roots, where the solution would be \( C_1 e^{r_1t} + C_2 e^{r_2t} \).

Understanding repeated roots helps in tailoring the solution of the differential equation to match the specific characteristics of the problem.

Initial conditions are given values that help determine specific constants in the general solution. These values are often given as \( y(t_0) = y_0 \) and \( y'(t_0) = y_0' \). For our exercise, they are \( y(1) = 0 \) and \( y'(1) = -1 \). Using initial conditions:

• Set \( y(1) = 0 \) in the derived general solution \( (C_1 + C_2 t)e^{-6t} \) and solve for the relationships between constants.
• Differentiate the general solution to get \( y'(t) \), then apply \( y'(1) = -1 \).

This process requires substituting these initial conditions into the solution to determine \( C_1 \) and \( C_2 \): - Essentially solving a system of equations to find the precise function that satisfies the differential equation and, simultaneously, these initial conditions.

The general solution of a differential equation provides an essential framework for understanding potential functions that solve the equation. Based on the characteristic equation and its roots, we can form a general solution. In this exercise,

• Since the root repeated at \( -6 \), this influences the structure of our solution.
• The general solution thus takes the form \( y(t) = (C_1 + C_2 t)e^{-6t} \).

This solution encompasses a range of possible functions depending on the constants \( C_1 \) and \( C_2 \).

Specific Solution with Initial Conditions

After integrating the initial conditions into our general solution, we determine specific values for the constants:

• From the initial conditions, calculate \( C_1 = \frac{e^6}{5} \) and \( C_2 = -\frac{e^6}{5} \).

This allows us to write the particular solution:\( y(t) = \left( \frac{e^6}{5} - \frac{e^6}{5} t \right)e^{-6t} \). This unique version of the general solution satisfies both the differential equation and the specified initial conditions, completing our task.