Problem 17 - Advanced Engineering Mathematics

Question

Solve the given differential equation by undetermined coefficients. \(y^{\prime \prime}-2 y^{\prime}+5 y=e^{x} \cos 2 x\)

Step-by-Step Solution

Answer

The solution is \( y = e^x(C_1 \cos 2x + C_2 \sin 2x) + e^x(A \cos 2x + B \sin 2x) \).

Step 1: Identify the Homogeneous Equation

The given differential equation is \( y'' - 2y' + 5y = e^x \cos(2x) \). Start by solving the corresponding homogeneous equation: \( y'' - 2y' + 5y = 0 \).

Step 2: Characteristic Equation

Write the characteristic equation for the homogeneous equation: \( r^2 - 2r + 5 = 0 \). Solve for \( r \) using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1, b = -2, c = 5 \).

Step 3: Solve Characteristic Equation

\( r = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(5)}}{2(1)} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} \). This simplifies to \( r = 1 \pm 2i \).

Step 4: Write General Solution of Homogeneous Equation

The solution of the homogeneous equation is \( y_h = e^x(C_1 \cos 2x + C_2 \sin 2x) \).

Step 5: Guess a Particular Solution Form

For the non-homogeneous term \( e^x \cos 2x \), we guess the particular solution form as \( y_p = e^x(A \cos 2x + B \sin 2x) \).

Step 6: Substitute Particular Solution and Derivatives

Calculate the derivatives: \( y_p' = e^x(A \cos 2x + B \sin 2x) + e^x(-2A \sin 2x + 2B \cos 2x) \) and \( y_p'' = e^x(-4A \cos 2x - 4B \sin 2x) + 2e^x(-2A \sin 2x + 2B \cos 2x) \).

Step 7: Plug into Differential Equation

Substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the differential equation. Simplify and combine like terms to find coefficients for \( A \) and \( B \).

Step 8: Solve for Coefficients

After substituting the derivatives and simplifying, set the coefficients of like terms from both sides of the equation equal to each other. This gives two equations that allow you to solve for \( A \) and \( B \).

Step 9: Write the Full Solution

Combine the particular solution obtained after determining \( A \) and \( B \) with the homogeneous solution: \( y(x) = y_h + y_p \).

Step 10: Final Answer Formulation

The final solution is the sum of the homogeneous solution \( y_h = e^x(C_1 \cos 2x + C_2 \sin 2x) \) and the particular solution \( y_p \). Write the complete solution.

Key Concepts

Homogeneous Differential Equation

In the context of differential equations, a homogeneous differential equation is one where all terms depend solely on the function and its derivatives, with no external forcing component (i.e., the non-homogeneous part is zero). For example, consider the equation:

• \( y'' - 2y' + 5y = 0 \)

This equation is homogeneous because it's set to zero, containing only terms involving the unknown function \( y \) and its derivatives \( y'' \) and \( y' \). To find the solution, we focus first on this homogeneous equation. This approach removes any non-homogeneous terms and helps in determining the foundational behavior of the system.
The homogeneous solution provides us with a basic form that can be used as a scaffold or structure, which we then build upon to find the complete solution for the non-homogeneous differential equation.

Characteristic Equation

The characteristic equation is a crucial step in solving homogeneous differential equations. It arises from the assumption that the solution to the differential equation can be expressed as an exponential function, such as \( e^{rx} \). Substituting \( y = e^{rx} \) into the differential equation leads to an algebraic equation in terms of \( r \), known as the characteristic equation.For the homogeneous equation \( y'' - 2y' + 5y = 0 \), the characteristic equation is:

• \( r^2 - 2r + 5 = 0 \)

Solve this quadratic equation to find the roots \( r \). Using the quadratic formula, we find that the roots are complex: \( r = 1 \pm 2i \). These roots determine the form of the solution to the homogeneous differential equation.
Complex roots typically lead to oscillatory solutions, represented using exponential terms and sine or cosine functions, capturing both the growth or decay and periodicity of the system.

General Solution

The general solution of a homogeneous linear differential equation is a linear combination of the fundamental solutions derived from the roots of the characteristic equation. For an equation with complex roots like \( r = 1 \pm 2i \), the general solution is expressed as a combination involving both exponential and trigonometric functions:

• \( y_h = e^x(C_1 \cos 2x + C_2 \sin 2x) \)

Here, \( C_1 \) and \( C_2 \) are constants determined by initial or boundary conditions, if provided.
This expression captures the essence of both exponential growth (or decay) given by \( e^x \) and oscillations due to \( \cos 2x \) and \( \sin 2x \). The general solution is integral to forming the complete solution of a differential equation, combining with the particular solution to encapsulate both the natural dynamics and any external influences on the system.

Particular Solution

To solve a non-homogeneous differential equation, like \( y'' - 2y' + 5y = e^x \cos 2x \), we seek a particular solution that specifically addresses the non-homogeneous part. The method of undetermined coefficients is often used in such cases, where we make an educated guess about the form of the particular solution and solve for unknown coefficients. For the given non-homogeneous term, we guess:

• \( y_p = e^x(A \cos 2x + B \sin 2x) \)

This guess reflects the structure of the non-homogeneous term \( e^x \cos 2x \). We calculate the derivatives of \( y_p \), substitute them back into the differential equation, and solve for \( A \) and \( B \) by equating coefficients. This provides a particular solution that mirrors the external forcing applied to the system.
Once found, the particular solution is combined with the general solution of the homogeneous equation to form the complete solution, representing the total response of the system to both inherent dynamics and external forces.