Problem 14 - Advanced Engineering Mathematics

Question

Solve the given differential equation by undetermined coefficients. \(y^{\prime \prime}-4 y=\left(x^{2}-3\right) \sin 2 x\)

Step-by-Step Solution

Answer

Solve the homogeneous solution first and then find the particular solution using guessed coefficients.

Step 1: Solve the Homogeneous Equation

The given differential equation is \(y'' - 4y = \left(x^2 - 3\right) \sin 2x\). Start by solving the associated homogeneous equation \(y'' - 4y = 0\).The characteristic equation for this is \(r^2 - 4 = 0\), which factors to \((r-2)(r+2) = 0\).Thus, the roots are \(r = 2\) and \(r = -2\). So, the general solution to the homogeneous equation is:\[ y_h = C_1e^{2x} + C_2e^{-2x} \].

Step 2: Form a Particular Solution Guess

The non-homogeneous part is \((x^2 - 3) \sin 2x\). As \(\sin 2x\) is a trigonometric function and the polynomial \(x^2 - 3\) is of degree 2, we consider guessing a particular solution of the form:\[ y_p = (Ax^2 + Bx + C) \cos 2x + (Dx^2 + Ex + F) \sin 2x \].We include both cosine and sine terms because the differential equation involves sinusoidal functions.

Step 3: Compute the Derivatives of the Particular Solution

Calculate the first and second derivatives of the guessed particular solution:\( y_p = (Ax^2 + Bx + C) \cos 2x + (Dx^2 + Ex + F) \sin 2x \).First derivative: \[ y_p' = [2Ax + B]\cos 2x - 2(Ax^2 + Bx + C)\sin 2x + [2Dx + E]\sin 2x + 2(Dx^2 + Ex + F)\cos 2x \].Second derivative: \[ y_p'' = [2A - 4(2Ax + B) \sin 2x - 4(Ax^2 + Bx + C) \cos 2x]\cos 2x + [-4(2Dx + E)\cos 2x + 4(Dx^2 + Ex + F)\sin 2x]\sin 2x \].This step will require some simplification to plug back into the original equation.

Step 4: Equate and Solve for Coefficients

Substitute the derivatives and the particular solution back into the original differential equation.The equation becomes:\[ y_p'' - 4y_p = (x^2 - 3) \sin 2x \].Equate the coefficients on both sides for \(\sin 2x\) and \(\cos 2x\) terms to determine the values of A, B, C, D, E, and F. Solve the resulting system of equations.

Step 5: Write the General Solution

Once the coefficients A, B, C, D, E, and F are determined, plug them back into the particular solution \(y_p\). Add this particular solution to the general solution of the homogeneous equation:\[ y = y_h + y_p = C_1e^{2x} + C_2e^{-2x} + (Ax^2 + Bx + C) \cos 2x + (Dx^2 + Ex + F) \sin 2x \].

Key Concepts

Homogeneous Equation

A homogeneous equation is part of the process of solving differential equations where you focus only on the portion without inhomogeneous terms. In this initial step, consider the equation as if the right-hand side doesn't exist. For the given problem, this simplifies the differential equation to:

• \(y'' - 4y = 0\).

This step primarily involves finding solutions that satisfy this new equation, setting the stage for tackling the entire equation.
The solution to a homogeneous equation is often termed the "general solution," which is determined by solving its characteristic equation.

Particular Solution

In contrast to the homogeneous solution, a particular solution solves the entire inhomogeneous differential equation. This involves considering the specific non-zero right-hand side,

• which is \((x^2 - 3) \sin 2x\) in this exercise.

To determine the particular solution, a method called undetermined coefficients is employed. Here, a guess is made based on the form of the inhomogeneous term. A sensible guess for this situation is usually a combination of sinusoidal terms and polynomials,

• such as \((Ax^2 + Bx + C)\cos 2x + (Dx^2 + Ex + F)\sin 2x\).

This guessed solution is then adjusted by solving for coefficients to satisfy the differential equation.

Characteristic Equation

The characteristic equation is a crucial part of determining the general solution for a homogeneous differential equation. It arises by assuming a solution of the form \(e^{rx}\), which leads the original differential equation to a polynomial form. For our scenario,

• the characteristic equation is derived from \(r^2 - 4 = 0\), which simplifies to \((r - 2)(r + 2) = 0\).

By solving this polynomial equation, we get the roots \(r = 2\) and \(r = -2\). These roots provide the exponents for the exponential portions of the general solution,

• resulting in \(C_1e^{2x} + C_2e^{-2x}\).

Sinusoidal Functions

In this exercise, sinusoidal functions such as \(\sin 2x\) play a significant role in forming the particular solution. Because the inhomogeneous term of the differential equation contains sinusoidal parts, our assumed particular solution must include sine and cosine components.
Sinusoidal functions have specific derivative properties that make them reappear in equations of motion and wave forms in mathematics. It is essential to ensure your guessed solution includes terms that match the form of the driving force on the equation,

• here represented by \((Ax^2 + Bx + C)\cos 2x + (Dx^2 + Ex + F)\sin 2x\).

This structured guess helps in equating and solving for unknown coefficients through simplification and derivative computations.