Problem 23 - Advanced Engineering Mathematics

Question

In Problems 13-28, use the procedures developed in this chapter to find the general solution of each differential equation. $$ y^{\prime \prime}-2 y^{\prime}+2 y=e^{x} \tan x $$

Step-by-Step Solution

Answer

The general solution is a combination of homogeneous solution and particular solution obtained via variation of parameters.

Step 1: Solve the Homogeneous Equation

The homogeneous part of the differential equation is \( y'' - 2y' + 2y = 0 \). To solve this, first find the characteristic equation: \( r^2 - 2r + 2 = 0 \). Solving this quadratic equation using the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we get \( r = 1 \pm i \). This complex root implies the solution to the homogeneous equation is \( y_h = e^x (C_1 \cos x + C_2 \sin x) \), where \( C_1 \) and \( C_2 \) are arbitrary constants.

Step 2: Identify the Particular Solution

For the particular solution, we need to account for the non-homogeneous part, \( e^x \tan x \). Note that \( \tan x \) suggests using a trial solution involving trigonometric functions multiplied by an exponential function. However, due to \( \tan x = \frac{\sin x}{\cos x} \), standard undetermined coefficients won't work, requiring variation of parameters instead.

Step 3: Apply Variation of Parameters

Using the variation of parameters method, assume a particular solution of the form \( y_p = u_1(x) e^x \cos x + u_2(x) e^x \sin x \). Here, \( u_1(x) \) and \( u_2(x) \) are functions to be determined. Differentiate and substitute into the original differential equation. Solve the resulting system of equations: \( u_1' e^x \cos x + u_2' e^x \sin x = 0 \) and \( u_1' e^x (-\sin x) + u_2' e^x \cos x = e^x \tan x \). These yield an integral form after substitution and simplification.

Step 4: Compute Integrals for Parameters

Integrate the equations derived from variation of parameters to obtain \( u_1(x) \) and \( u_2(x) \). Solve: \( u_1' = \int e^x \tan x \sin x \, dx \) and \( u_2' = \int e^x \tan x \cos x \, dx \). These may require integration by parts or substitution. Compute and simplify to get explicit functions for \( u_1(x) \) and \( u_2(x) \).

Step 5: Form the General Solution

Combine the homogeneous solution and the particular solution: \[ y(x) = y_h + y_p = e^x (C_1 \cos x + C_2 \sin x) + y_p \]Where \( y_p = u_1(x) e^x \cos x + u_2(x) e^x \sin x \) by solving and substituting back the integrals found previously.

Key Concepts

Homogeneous Equations

Differential equations can be categorized into two broad types: homogeneous and non-homogeneous. When dealing with a homogeneous differential equation, all the terms are dependent solely on the function and its derivatives. We start by examining the given differential equation:

• The homogeneous part of the equation is given by the expression: \( y'' - 2y' + 2y = 0 \).
• The goal here is to find the complementary function \( y_h \), which satisfies this homogeneous equation.
• To solve it, you typically convert it into a characteristic equation. This transforms a second order linear differential equation into a simpler algebraic problem.

First, we need to find the characteristic equation, as outlined in the solution steps. This helps us determine the roots, which in this case are complex. The characteristic equation corresponds to:\[ r^2 - 2r + 2 = 0 \]This leads us to complex roots given by \( r = 1 \pm i \). With a positive real part and imaginary part, the complementary solution takes the form:\[ y_h = e^x (C_1 \cos x + C_2 \sin x) \]This equation represents all possible solutions to the homogeneous differential equation. "Homogeneous" means there are no terms independent of the function \( y \). Each solution is based on constants \( C_1 \) and \( C_2 \), which indicate the generality of the solution for differing initial conditions.

Characteristic Equation

The characteristic equation is a crucial mathematical tool when solving linear homogeneous differential equations with constant coefficients. Using this, we can find the solutions to a differential equation more readily. Let's break this down further:

• The characteristic equation is formed by replacing the derivatives \( y', y''\) with powers of a variable like \( r \), resulting in a polynomial.
• For our differential equation, this gave us: \( r^2 - 2r + 2 = 0 \).
• By solving this polynomial equation, we identified the roots as \( r = 1 \pm i \), a complex conjugate pair.

These roots hint at an oscillatory solution when combined with the exponential function due to the complex nature. Each root of the characteristic equation reflects a component of the solution. For complex roots \( \alpha + \beta i \), the solutions include exponential and trigonometric functions:\[ \large y_h = e^x (C_1 \cos x + C_2 \sin x) \]This ensures the solution not only satisfies the differential equation but also adjusts to accommodate any imposed initial conditions or boundaries.

Variation of Parameters

In situations where the method of undetermined coefficients doesn't work, variation of parameters comes into play. This method allows us to find a particular solution for non-homogeneous differential equations. Here's how it unfolds:

• Unlike undetermined coefficients, variation of parameters doesn't presume the form of the solution.
• Instead, it systematically derives functions \( u_1(x) \) and \( u_2(x) \) that modify the linearly independent solutions of the homogeneous equation.
• These functions are unknown and must be calculated to account for the non-homogenous part \( e^x \tan x \).

The method proceeds as follows:

• Assume a solution form: \( y_p = u_1(x)e^x \cos x + u_2(x)e^x \sin x \).
• Find two new equations from this. Then, solve the equations to derive \( u_1'(x) \) and \( u_2'(x) \).
• Once these derivatives are obtained, integrating them provides the actual functions \( u_1(x) \) and \( u_2(x) \).

This approach effectively breaks down the complex behavior of non-homogeneous equations, allowing us to solve for and include the particular solution in the overall equation.

Particular Solution

In differential equations, finding the particular solution is essential for addressing the non-homogeneous parts of the equation. This means solving the inhomogeneous term separately. Let's elaborate:

• A particular solution \( y_p \) directly caters to any piece of the equation not solved by the homogeneous part.
• For our given problem, the term \( e^x \tan x \) requires particular attention.
• Given the nature of \( \tan x \), typical approaches like undetermined coefficients aren't feasible.

By using the variation of parameters, you derive a particular solution that complements the homogeneous solutions. This involves solving integrals for \( u_1(x) \) and \( u_2(x) \) after deducing their derivatives from earlier equations:

• The integrals solve to give expressions defining \( u_1(x) \) and \( u_2(x) \), specific for the inhomogeneous component.
• By solving, substituting back and integrating, we obtain explicit functions contributing to the overall solution.

In essence, the particular solution jointly with the homogeneous solution \( y_h \) completes the general solution for the differential equation. Together, they describe the system's entire behavior based on external forces, like \( e^x \tan x \) in this case, affecting it.