Problem 125 - Calculus 3

Question

For the following problems, find the general solution. $$y^{\prime \prime}+9 y=0$$

Step-by-Step Solution

Answer

The general solution is \( y(x) = C_1 \cos(3x) + C_2 \sin(3x) \).

Step 1: Identify the Differential Equation Type

The given equation is a second-order linear homogeneous differential equation with constant coefficients. It has the form \( y'' + 9y = 0 \).

Step 2: Formulate the Characteristic Equation

To solve the differential equation, we first write its characteristic equation, which comes from assuming a solution of the form \( y = e^{rx} \). The characteristic equation is \( r^2 + 9 = 0 \).

Step 3: Solve the Characteristic Equation

The characteristic equation \( r^2 + 9 = 0 \) can be solved for \( r \) by moving the 9 to the right side and taking the square root, giving \( r^2 = -9 \), and thus \( r = \pm 3i \).

Step 4: Write the General Solution

Since the roots \( r = \pm 3i \) are purely imaginary, the general solution for the differential equation can be expressed in terms of sine and cosine functions: \( y(x) = C_1 \cos(3x) + C_2 \sin(3x) \), where \( C_1 \) and \( C_2 \) are arbitrary constants.

Key Concepts

Characteristic Equation

When solving a second-order linear homogeneous differential equation like \( y'' + 9y = 0 \), the process begins with finding the characteristic equation. This equation helps determine the behavior of possible solutions without directly solving the differential equation itself. The characteristic equation comes from assuming a solution of the form \( y = e^{rx} \), where \( r \) is a constant. By substituting \( y = e^{rx} \) into the differential equation \( y'' + 9y = 0 \), we get the characteristic equation: \[ r^2 + 9 = 0 \]Finding the roots of this characteristic equation is key to determining the form of the general solution of the differential equation. This step translates a complex differential problem into an algebraic one, making it significantly easier to solve.

Imaginary Roots

When solving the characteristic equation \( r^2 + 9 = 0 \), you find that the roots are imaginary. Imaginary roots occur when the equation involves a negative value under the square root, resulting in the need for imaginary numbers. In our case, the roots are found by solving:\[ r^2 = -9 \]This gives:\[ r = \pm 3i \]where \( i \) is the imaginary unit, defined by \( i^2 = -1 \). Imaginary roots indicate that the solutions of the differential equation will involve trigonometric functions, specifically sine and cosine. This happens because the exponential function with an imaginary exponent can be expressed in terms of sine and cosine, according to Euler's formula.

General Solution

With the roots of the characteristic equation \( r = \pm 3i \) determined as imaginary, we can now write the general solution of the differential equation. Imaginary roots lead to a solution in terms of trigonometric functions:\[ y(x) = C_1 \cos(3x) + C_2 \sin(3x) \]Here, \( C_1 \) and \( C_2 \) are arbitrary constants. These constants are crucial because they allow the solution to potentially match any initial conditions that may be given in particular problems.This form of the solution reflects the oscillatory nature of the functions sine and cosine, often appearing in systems with periodic behavior, like mechanical vibrations or alternating electric currents.