Problem 19 - Calculus Volume 3

Question

Find the general solution to the linear differential equation. $$ y^{\prime \prime}-121 y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(t) = C_1 e^{11t} + C_2 e^{-11t} \).

Step 1: Identify the Characteristic Equation

For a linear differential equation with constant coefficients such as \( y'' - 121y = 0 \), we start by writing the characteristic equation. Assume a solution of the form \( y = e^{rt} \), leading to \( r^2 - 121 = 0 \).

Step 2: Solve the Characteristic Equation

The characteristic equation \( r^2 - 121 = 0 \) can be solved by finding the roots using factorization or setting \( r^2 = 121 \). This gives \( r = \pm 11 \).

Step 3: Write the General Solution

For distinct real roots, the solution to the differential equation is of the form \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \), where \( r_1 \) and \( r_2 \) are the roots. Here, it becomes \( y(t) = C_1 e^{11t} + C_2 e^{-11t} \) as \( r_1 = 11 \) and \( r_2 = -11 \).

Key Concepts

Characteristic Equation

In the world of differential equations, especially when dealing with linear differential equations with constant coefficients, the characteristic equation plays a pivotal role. It helps us determine the behavior of solutions to the differential equation. To explain it simply, if you have a differential equation in the form:

• \( y'' - 121y = 0 \)

You would look for solutions of the form \( y = e^{rt} \). By substituting this assumed solution into the equation, you derive what is known as the characteristic equation. For our example, this substitution leads to the equation \( r^2 - 121 = 0 \). This characteristic equation becomes a polynomial equation that allows us to find the roots, which are essential for understanding the solution of the differential equation. These roots tell us about the exponential terms that will be part of the general solution.

Constant Coefficients

Constant coefficients refer to the numbers that multiply the derivatives in a linear differential equation which do not change over time. This makes solving such equations straightforward as they usually lead to characteristic equations with constant terms. For instance, in the example \( y'' - 121y = 0 \), the coefficient of \( y'' \) is 1 (implied) and that of \( y \) is -121. These constants allow more predictable behavior when finding the solution, making the process systematic.

• Since the coefficients are constant, you can employ algebraic methods like factoring to solve the corresponding characteristic equation \( r^2 - 121 = 0 \).
• This also means that the form of the solution depends directly on the nature of these coefficients, simplifying the determination of the solution behavior without needing to consider variable changes.

Working with constant coefficients thus provides a clear path to the general solution as the coefficients don’t vary, streamlining finding roots and eventually constructing the general solution.

General Solution

Once we determine the roots of the characteristic equation, crafting the general solution for the differential equation becomes the next step. The general solution embodies all possible solutions to a homogeneous linear differential equation with constant coefficients. For the given problem:\( y'' - 121y = 0 \)The characteristic equation is solved to give roots \( r = 11 \) and \( r = -11 \). These roots are distinct and real, which guides us to write the general solution in the form:

• \( y(t) = C_1 e^{11t} + C_2 e^{-11t} \)

Here, \( C_1 \) and \( C_2 \) are arbitrary constants that are determined by initial conditions, should they be provided. This general solution comprises two exponential functions representing growth and decay determined by the roots, capturing both possibilities for the differential equation's behavior across all time parameter values. Thus, it provides a complete picture of the potential system dynamics described by the differential equation.