Problem 19 - Advanced Engineering Mathematics

Question

Solve the given differential equation by undetermined coefficients. \(y^{\prime \prime}+2 y^{\prime}+y=\sin x+3 \cos 2 x\)

Step-by-Step Solution

Answer

The solution is \( y = C_1 e^{-x} + C_2 xe^{-x} - \frac{1}{2} \sin x + \frac{3}{10} \cos 2x \).

Step 1: Solve the Homogeneous Equation

Begin by solving the associated homogeneous differential equation: \( y'' + 2y' + y = 0 \). Use the characteristic equation method. The characteristic equation is \( r^2 + 2r + 1 = 0 \), which factors to \( (r+1)^2 = 0 \). Thus, \( r = -1 \) is a double root. The general solution of the homogeneous equation is \( y_h = C_1 e^{-x} + C_2 xe^{-x} \).

Step 2: Form Particular Solution for Non-Homogeneous Part

The non-homogeneous equation is \( y'' + 2y' + y = \sin x + 3\cos 2x \). We look for particular solutions for each term separately. Assume \( y_1 = A\sin x + B\cos x \) for the \( \sin x \) part, and \( y_2 = C\cos 2x + D\sin 2x \) for the \( 3\cos 2x \) part.

Step 3: Particular Solution for \( \sin x \)

Substitute \( y_1 = A\sin x + B\cos x \) into the left side of equation. Calculate derivatives, \( y_1' = A\cos x - B\sin x \) and \( y_1'' = -A\sin x - B\cos x \). Substitute into the differential equation and solve for \( A \) and \( B \). You find that \( A = -1/2 \) and \( B = 0 \). Thus, \( y_1 = -\frac{1}{2} \sin x \).

Step 4: Particular Solution for \( 3\cos 2x \)

Substitute \( y_2 = C\cos 2x + D\sin 2x \) into the left side of equation. Find derivatives \( y_2' = -2C\sin 2x + 2D\cos 2x \) and \( y_2'' = -4C\cos 2x - 4D\sin 2x \). Substitute into the differential equation and solve for \( C \) and \( D \). You find \( C = \frac{3}{10} \) and \( D = 0 \). Thus, \( y_2 = \frac{3}{10} \cos 2x \) as the particular solution.

Step 5: Combine Solutions

The complete solution to the original differential equation is the sum of the homogeneous and particular solutions: \( y = y_h + y_p = C_1 e^{-x} + C_2 xe^{-x} - \frac{1}{2} \sin x + \frac{3}{10} \cos 2x \). This is the final solution.

Key Concepts

Undetermined Coefficients

Undetermined coefficients is a powerful method to find the particular solution of linear non-homogeneous differential equations. In this approach, we assume a particular solution with unknown coefficients and then adjust them until the equation is satisfied. This method shines when the non-homogeneous part of the differential equation is a combination of polynomial, exponential, sine, and cosine terms.
The main idea is to "guess" the form of a particular solution based on the function outside the derivative, such as in \( \sin x + 3\cos 2x \). For instance, if you see a \( \sin x \) term, you start with an assumed solution of \( A\sin x + B\cos x \) where \( A \) and \( B \) are undetermined coefficients.

• This assumed solution gets plugged into the differential equation.
• Then, by manipulating the equation and matching coefficients, we find the values of \( A \) and \( B \).

It's systematic and straightforward for many types of non-homogeneous equations.

Characteristic Equation

The characteristic equation is crucial for solving linear homogeneous differential equations, especially with constant coefficients. It helps find the homogeneous solution by transforming the differential equation into an algebraic one.
By replacing derivatives with powers of \( r \) (representing roots), the differential equation \( y'' + 2y' + y = 0 \) becomes the characteristic equation \( r^2 + 2r + 1 = 0 \). Solving this provides the roots, which then help construct the general solution of the homogeneous equation.

• The characteristic equation helps determine the type of exponential functions—whether real or complex—needed for the homogeneous solution.
• Each root gives rise to a fundamental solution, such as an exponential function \( e^{rx} \) for a real root \( r \).

This forms the foundation for finding solutions to differential equations with constant coefficients.

Homogeneous Solution

The homogeneous solution is a set of solutions to the differential equation when the equation is set to zero (no external forces or inputs). It represents the natural response of the system described by the differential equation.
In the given equation, when solving \( y'' + 2y' + y = 0 \), we found that \( r = -1 \) was a double root of the characteristic equation. Thus, the form for the homogeneous solution, \( y_h \), combines terms like \( C_1 e^{-x} \) and \( C_2 xe^{-x} \) to account for the repeated root.

• Where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.
• The inclusion of \( xe^{-x} \) handles the repeat in the root \( r = -1 \).

Ultimately, the homogeneous solution characterizes the intrinsic behavior of the system disregarding external inputs.

Particular Solution

A particular solution directly addresses the non-homogeneous part of a differential equation. It represents the forced response due to external inputs or driving functions like \( \sin x \) or \( 3\cos 2x \).
The process involves assuming a form that mirrors these functions, leading to terms such as \( y_1 = A\sin x + B\cos x \) and \( y_2 = C\cos 2x + D\sin 2x \). These are then differentiated and substituted back into the non-homogeneous differential equation to find the coefficients \( A, B, C, \) and \( D \).

• For \( \sin x \), we determined \( A = -1/2 \) and \( B = 0 \).
• For \( 3\cos 2x \), \( C = 3/10 \) and \( D = 0 \).

These values are solved through equating and solving systems of equations originating from plugging the guessed solution into the differential equation.

Non-Homogeneous Differential Equation

A non-homogeneous differential equation is one which includes terms without derivatives. Unlike homogeneous equations, these include an "external force" or "driver," such as \( \sin x \) or \( 3\cos 2x \), that forces the system away from its natural state.
The solution process involves two main components:

• The homogeneous solution (intrinsic characteristics of the system).
• The particular solution (response to the external input).

Working through the equation \( y'' + 2y' + y = \sin x + 3\cos 2x \), involves finding both solutions separately and combining them. In essence, the complete solution is a sum \( y = y_h + y_p \), well illustrated step-by-step:
The homogeneous part addresses how the system naturally behaves, whereas the particular part shows how it reacts to external influences.