Problem 23 - Advanced Engineering Mathematics

Question

The indicated functions are known linearly independent solutions of the associated homogeneous differential equation on the interval \((0, \infty)\). Find the general solution of the given nonhomogeneous equation. $$ \begin{aligned} &x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{1}{4}\right) y=x^{3 / 2} ; \quad y_{1}=x^{-1 / 2} \cos x, \\ &y_{2}=x^{-1 / 2} \sin x \end{aligned} $$

Step-by-Step Solution

Answer

The general solution is: \(y = C_1 x^{-1/2} \cos x + C_2 x^{-1/2} \sin x + x^{-1/2} (x^2 \sin x + x^2 \cos x)\).

Step 1: Identify the Differential Equation and Given Functions

The given nonhomogeneous differential equation is \(x^2 y'' + x y' + \left(x^2 - \frac{1}{4}\right) y = x^{3/2}\). The linearly independent solutions for the associated homogeneous equation are \(y_1 = x^{-1/2} \cos x\) and \(y_2 = x^{-1/2} \sin x\).

Step 2: Write the General Solution for the Homogeneous Equation

The general solution of the associated homogeneous differential equation is a linear combination of the two linearly independent solutions: \(y_h = C_1 y_1 + C_2 y_2 = C_1 x^{-1/2} \cos x + C_2 x^{-1/2} \sin x\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step 3: Set Up Variation of Parameters Method

To find the particular solution \(y_p\) for the nonhomogeneous equation, we use the variation of parameters method. Assume \(y_p = u_1 y_1 + u_2 y_2\), where \(u_1\) and \(u_2\) are functions to be determined.

Step 4: Compute Wronskian of \(y_1\) and \(y_2\)

Find the Wronskian W of \(y_1\) and \(y_2\): \[W = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} x^{-1/2} \cos x & x^{-1/2} \sin x \ -x^{-1/2} \sin x - \frac{1}{2}x^{-3/2} \cos x & x^{-1/2} \cos x - \frac{1}{2}x^{-3/2} \sin x \end{vmatrix}\]This simplifies to \(W = \frac{1}{x}\).

Step 5: Determine \(u_1\) and \(u_2\)

Using the formulas from variation of parameters: \(u_1 = \int \frac{-y_2(x^{3/2})}{W} dx\), and \(u_2 = \int \frac{y_1(x^{3/2})}{W} dx\), calculate:\(u_1 = \int -x^{-1/2} \sin x \, x^{3/2} dx = \int -x \sin x \, dx\)\(u_2 = \int x^{-1/2} \cos x \, x^{3/2} dx = \int x \cos x \, dx\)

Step 6: Evaluate Integrals for \(u_1\) and \(u_2\)

Evaluate the integrals:For \(u_1 = \int -x \sin x \, dx\), use integration by parts:\(u_1 = x \cos x - \int \cos x \, dx = x \cos x - \sin x\).For \(u_2 = \int x \cos x \, dx\), again use integration by parts:\(u_2 = x \sin x + \int \sin x \, dx = x \sin x - \cos x\).

Step 7: Form Particular Solution \(y_p\)

Substitute \(u_1\) and \(u_2\) back into \(y_p = u_1 y_1 + u_2 y_2\):\(y_p = (x \cos x - \sin x) x^{-1/2} \cos x + (x \sin x - \cos x) x^{-1/2} \sin x\).

Step 8: Combine Homogeneous and Particular Solutions

The general solution of the nonhomogeneous differential equation is the sum of the homogeneous solution \(y_h\) and the particular solution \(y_p\):\(y = y_h + y_p = C_1 x^{-1/2} \cos x + C_2 x^{-1/2} \sin x + y_p\).

Step 9: Simplify and Write Final General Solution

Combine and simplify terms from \(y_h\) and \(y_p\) to write the final general solution:\(y = C_1 x^{-1/2} \cos x + C_2 x^{-1/2} \sin x + x^{-1/2} (x^2 \sin x + x^2 \cos x)\).

Key Concepts

Linear Differential Equations

Linear differential equations form the foundation of many mathematical and physical models, making them crucial in understanding complex systems. A linear differential equation is one in which all occurrences of the unknown function and its derivatives appear to the first power, without any products of these terms. These equations can often be expressed in the form:

• \( a_n(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + imes imes imes + a_1(x) y' + a_0(x) y = f(x) \)

Here, the coefficients \(a_i(x)\) are functions of \(x\), and \(f(x)\) is a given function which defines the nonhomogeneous part of the equation. Understanding whether a linear differential equation is homogeneous or nonhomogeneous is key in solving them efficiently.

Homogeneous vs. Nonhomogeneous

A homogeneous linear differential equation has \(f(x) = 0\) while a nonhomogeneous one has \(f(x) eq 0\). Solving a nonhomogeneous linear differential equation typically involves finding a particular solution for the nonhomogeneous term \(f(x)\), in addition to the general solution of the associated homogeneous equation. This is crucial because the difficulty of solving these equations generally lies in addressing the nonhomogeneous terms.

In our exercise above, the given equation is a nonhomogeneous second-order linear differential equation. This implies the need for integrating the homogeneous solution with a particular solution obtained through methods like variation of parameters.

Wronskian

The Wronskian is an invaluable determinant in determining whether two functions are linearly independent, specifically in the context of differential equations. It provides insight into the relationship and independence of solutions to a differential equation.
Imagine you have two functions \(y_1\) and \(y_2\). You construct the Wronskian, \(W\), as follows:

• \[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_2 y_1' \]

If \(W(y_1, y_2) eq 0\) within an interval, the functions are linearly independent on that interval. This independence is crucial for solving differential equations through methods such as the method of variation of parameters.

Wronskian in Practice

For the given equation in our problem, the linearly independent solutions \(y_1 = x^{-1/2} \cos x \) and \(y_2 = x^{-1/2} \sin x \) allow us to find the Wronskian explicitly, which was calculated as \( W = \frac{1}{x} \). This non-zero Wronskian confirms the independence of \(y_1\) and \(y_2\), ensuring our method of variation of parameters will work correctly to find the particular solution. Detecting such independence is vital to applying this and similar techniques effectively.

Particular Solution

Finding a particular solution is often the main challenge when dealing with nonhomogeneous linear differential equations. The particular solution addresses the specific nonhomogeneous term \(f(x)\) in the differential equation.
A standard technique for finding the particular solution is the method of variation of parameters. This method leverages the knowledge of linearly independent solutions of the associated homogeneous equation.

Method of Variation of Parameters

For our given differential equation, we seek a particular solution, \(y_p = u_1 y_1 + u_2 y_2\), where \(u_1\) and \(u_2\) are functions to be determined. The functions are derived using:

• \( u_1(x) = \int \frac{-y_2(x)f(x)}{W(x)} \, dx \)
• \( u_2(x) = \int \frac{y_1(x)f(x)}{W(x)} \, dx \)

Thus, based on our specific equation and previously found linearly independent solutions \(y_1\) and \(y_2\), we calculate these integrals to determine the particular solution. This step involves integration by parts and algebraic manipulation. Once found, the particular solution is combined with the homogeneous solution to build the general solution of the equation, successfully addressing both homogeneous and nonhomogeneous components.