Problem 15 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Problem Type

The given differential equation is \( y^{\prime \prime} + y = 2x \sin x \). This is a second-order linear differential equation with constant coefficients and a non-homogeneous term \( 2x \sin x \). We will solve it using the method of undetermined coefficients.

Step 2: Solve the Homogeneous Equation

First, solve the homogeneous differential equation \( y^{\prime \prime} + y = 0 \). The auxiliary equation is \( r^2 + 1 = 0 \), which has roots \( r = i \) and \( r = -i \). Therefore, the general solution of the homogeneous equation is \( y_h = C_1 \cos x + C_2 \sin x \).

Step 3: Formulate the Particular Solution

Since the non-homogeneous term is \( 2x \sin x \), for a particular solution, we assume a solution of the form \( y_p = x(A \cos x + B \sin x) \). This form accounts for the polynomial term \( x \) and the trigonometric term \( \sin x \).

Step 4: Derive the Particular Solution

Calculate the derivatives: \( y_p' = A\cos x - Ax \sin x + B \sin x + Bx \cos x \) and \( y_p'' = -A\sin x - A\cos x - Ax \cos x - B\cos x + B\sin x - Bx \sin x \). Substitute \( y_p, y_p', \) and \( y_p'' \) into the original differential equation.

Step 5: Equate and Solve for Undetermined Coefficients

After substitution, we get an equation: \(-2A x \sin x = 2x \sin x\). Since the terms must match, equate coefficients: \(-2A = 2\), thus solve for \( A = -1 \) and \( B = 0 \). So the particular solution is \( y_p = -x \cos x \).

Step 6: Combine Solutions

Combine the solutions: the general solution is \( y = y_h + y_p = C_1 \cos x + C_2 \sin x - x \cos x \).

Step 7: Verify the Solution

Verify by substituting \( y \) back into the original equation to ensure all terms satisfy \( y^{\prime \prime}+y=2x\sin x \). This confirms the calculated solution is correct.

Key Concepts

Understanding Second-Order Linear Differential Equations

A second-order linear differential equation is a type of equation that involves the second derivative of a function. These equations often appear in various fields, such as physics and engineering, where they describe phenomena like motion or heat transfer. The term "linear" means that the function and its derivatives appear in a linear form, without powers or products of the function or its derivatives.

Here's a basic structure for understanding these equations:

• The general form is expressed as: \[ a(x)y'' + b(x)y' + c(x)y = g(x) \], where \(y''\) is the second derivative, \(y'\) is the first derivative, and \(y\) is the function itself.
• Coefficients \(a(x), b(x),\) and \(c(x)\) can be constant or variable, whereas \(g(x)\) represents an external influence or forcing term.
• Solutions to these equations help describe how systems evolve over time or space.

Identifying the nature of these coefficients and the function \(g(x)\) is crucial for choosing the appropriate method for finding the solution.

Exploring Non-Homogeneous Differential Equations

Non-homogeneous differential equations include an additional term, \(g(x)\), which makes their solutions more complex since they don't solely depend on the function or its derivatives. The term "non-homogeneous" highlights the presence of this forcing term.

Key features of non-homogeneous equations include:

• The structure: \[ a(x)y'' + b(x)y' + c(x)y = g(x) \]
• Studying such equations involves finding a general solution, which is the sum of the solution to the associated homogeneous equation and a particular solution to the non-homogeneous part.
• The homogeneous equation is derived by setting \(g(x) = 0\).

For the exercise at hand, the non-homogeneous term is \(2x \sin x\), which significantly influences the form of the particular solution. Finding the particular solution requires skillfully guessing a solution form, a technique known as the method of undetermined coefficients.

The Role of Constant Coefficients

When dealing with constant coefficients, it means that the coefficients \(a, b,\) and \(c\) in the second-order differential equation are constants, not functions of \(x\). This characteristic greatly simplifies solving differential equations.

Here’s how constant coefficients affect the solution process:

• The equation can be written simply as \[ ay'' + by' + cy = g(x) \].
• Such equations are easier to solve since the characteristic equation derived from the homogeneous part yields real or complex roots.
• The roots are used to construct the general homogeneous solution, often involving exponential functions like \(e^{rx}\) or trigonometric functions like \(\cos x\) and \(\sin x\).

In our specific problem, the constant coefficient is \(1\) for both \(y''\) and \(y\), simplifying calculations and allowing for easy solution by recognition of the form of the solution.

Determining the Homogeneous Solution

Finding the homogeneous solution is the first critical step in solving a non-homogeneous differential equation. The homogeneous solution solves the equation when the non-homogeneous part, \(g(x)\), equals zero.

The process involves:

• Formulating the homogeneous equation from the original equation by dropping the non-homogeneous term: \( y'' + y = 0 \).
• Finding the characteristic or auxiliary equation, which, in this problem, is \( r^2 + 1 = 0 \).
• Solving this equation yields complex roots \( r = i \) and \( r = -i \).
• These roots give rise to trigonometric functions in the solution: \( y_h = C_1 \cos x + C_2 \sin x \).

This solution forms the basis for building the overall solution to the original non-homogeneous differential equation, in conjunction with a particular solution capturing the effects of \(g(x)\).