Problem 134 - Calculus 3

Question

For the following problems, find the solution to the initial-value problem, if possible. $$y^{\prime \prime}=3 y-\cos (x), \quad y(0)=\frac{9}{4}, \quad y^{\prime}(0)=0$$

Step-by-Step Solution

Answer

The solution is \( y = e^{\sqrt{3} x} + e^{-\sqrt{3} x} + \frac{1}{4}\cos(x) \).

Step 1: Understand the Problem

We need to solve the second-order differential equation \( y'' = 3y - \cos(x) \) with the initial conditions \( y(0) = \frac{9}{4} \) and \( y'(0) = 0 \). This problem involves an inhomogeneous linear differential equation.

Step 2: Solve the Homogeneous Equation

The homogeneous part of the equation is \( y'' = 3y \). The characteristic equation is \( r^2 = 3 \), which gives us roots \( r = \sqrt{3} \) and \( r = -\sqrt{3}\). Therefore, the general solution of the homogeneous equation is \( y_h = C_1 e^{\sqrt{3} x} + C_2 e^{-\sqrt{3} x} \).

Step 3: Solve the Particular Solution

To solve the particular solution for the non-homogeneous part \( -\cos(x) \), assume \( y_p = A\cos(x) + B\sin(x) \). Differentiating twice, \( y_p'' = -A\cos(x) - B\sin(x) \). Insert into the differential equation to get \( -A\cos(x) - B\sin(x) = 3(A\cos(x) + B\sin(x)) - \cos(x) \). Simplifying, this equation leads to \( -4A\cos(x) - 4B\sin(x) = -\cos(x)\). Thus, \( A = \frac{1}{4} \) and \( B = 0 \). Therefore, the particular solution is \( y_p = \frac{1}{4}\cos(x) \).

Step 4: Form the General Solution

The general solution is the sum of the homogeneous solution and the particular solution: \( y = C_1 e^{\sqrt{3} x} + C_2 e^{-\sqrt{3} x} + \frac{1}{4}\cos(x) \).

Step 5: Apply Initial Conditions

Use the initial condition \( y(0) = \frac{9}{4} \) in the general solution: \( \frac{9}{4} = C_1 + C_2 + \frac{1}{4} \). This implies \( C_1 + C_2 = 2 \). For the derivative, \( y' = C_1 \sqrt{3} e^{\sqrt{3} x} - C_2 \sqrt{3} e^{-\sqrt{3} x} - \frac{1}{4}\sin(x) \). Use \( y'(0) = 0 \): \( 0 = C_1 \sqrt{3} - C_2 \sqrt{3} \), which implies \( C_1 = C_2 \). Solve these two conditions to find \( C_1 = C_2 = 1 \).

Step 6: Write the Solution with Known Constants

Include the constants \( C_1 \) and \( C_2 \) into the general solution: \( y = e^{\sqrt{3} x} + e^{-\sqrt{3} x} + \frac{1}{4}\cos(x) \).

Key Concepts

Second-Order Differential Equation

A second-order differential equation is a type of differential equation that involves the second derivative of a function. These equations often arise in physics and engineering, where they can represent phenomena like oscillations and waves. For example, the given equation \( y'' = 3y - \cos(x) \) is a second-order differential equation because it involves the second derivative \( y'' \). Here, \( y'' \) represents the acceleration if the function \( y \) models motion. Second-order differential equations can be classified as either homogeneous or non-homogeneous, depending on whether they include a term that is not a function of the dependent variable and its derivatives. In our example, the presence of \( -\cos(x) \) makes the equation non-homogeneous.

Homogeneous Solution

For a second-order differential equation like \( y'' = 3y \), the homogeneous solution addresses only the portion of the equation without any external forcing function, such as \( -\cos(x) \). The homogeneous part is solved by setting \( y'' = 3y \) and solving for the general form of the solution. This results in the characteristic equation which will be explored in a later section. The homogeneous solution, represented as \( y_h = C_1 e^{\sqrt{3} x} + C_2 e^{-\sqrt{3} x} \), captures how the system behaves naturally, without any external influences.

Particular Solution

The particular solution of a second-order differential equation is specific to the non-homogeneous part, in this case, \( -\cos(x) \). This solution captures the effects of any external forces or inputs that vary with respect to the independent variable, here \( x \). We hypothesize a form for the particular solution, typically involving functions similar to the non-homogeneous term itself. For example, guessing \( y_p = A\cos(x) + B\sin(x) \), and solving the resulting equation, helps in obtaining the particular solution. For our equation, it yields \( y_p = \frac{1}{4}\cos(x) \), showing how the solution specifically addresses the non-homogeneous input.

Characteristic Equation

The characteristic equation is a crucial component in solving homogeneous second-order differential equations. For the given problem, simplifying the homogeneous equation \( y'' = 3y \) leads to the characteristic equation \( r^2 = 3 \). This algebraic equation helps determine the form of the general solution to the homogeneous part by providing the roots, which identify the solution's exponential terms. Solving \( r^2 = 3 \) results in the roots \( r = \sqrt{3} \) and \( r = -\sqrt{3} \). These roots symbolize the rates at which the natural responses to the differential equation grow or decay. Using these roots, the homogeneous solution \( y_h = C_1 e^{\sqrt{3} x} + C_2 e^{-\sqrt{3} x} \) is constructed.