Problem 54 - Advanced Engineering Mathematics

Question

Find a linear second-orderdifferential equation \(F\left(x, y, y^{\prime}, y^{\prime \prime}\right)=0\) for which \(y=c_{1} x+c_{2} x^{2}\) is a two- parameter family of solutions. Make sure that your equation is free of the arbitrary parameters \(c_{1}\) and \(c_{2}\).

Step-by-Step Solution

Answer

The differential equation is \( xy' - \frac{1}{2} x^2 y'' - y = 0 \).

Step 1: Differentiate the given function

We are given the function \( y = c_1 x + c_2 x^2 \). To find the differential equation, we first compute the first and second derivatives with respect to \( x \). The first derivative is \( y' = \frac{d}{dx}(c_1 x + c_2 x^2) = c_1 + 2c_2 x \).The second derivative is \( y'' = \frac{d^2}{dx^2}(c_1 x + c_2 x^2) = 2c_2 \).

Step 2: Eliminate constants by forming equation

To eliminate \( c_1 \) and \( c_2 \), we express these constants in terms of \( y, y' \), and \( y'' \). From the second derivative, we have \( c_2 = \frac{1}{2} y'' \). We substitute \( c_2 \) back into the expression for \( y' \) to eliminate \( c_2 \):\[ y' = c_1 + 2c_2 x = c_1 + xy'' \].Solve for \( c_1 \) by rearranging:\[ c_1 = y' - xy'' \].

Step 3: Substitute and organize the equation

Now that we have expressions for \( c_1 \) and \( c_2 \), substitute these back into the original function \( y = c_1 x + c_2 x^2 \):\[ y = (y' - xy'')x + \left(\frac{1}{2}y''\right)x^2 \].Simplify the above expression:\[ y = xy' - x^2 y'' + \frac{1}{2}x^2 y'' \].Combine like terms:\[ y = xy' - \frac{1}{2} x^2 y'' \].

Step 4: Form the differential equation

From Step 3, we equate the expression to zero to obtain the differential equation:\[ xy' - \frac{1}{2} x^2 y'' - y = 0 \].Now, this is the required linear second-order differential equation free of the arbitrary parameters \( c_1 \) and \( c_2 \).

Key Concepts

Linear Differential Equations

A linear differential equation is one in which the dependent variable and all its derivatives appear linearly in the equation; that is, no powers, products, or other nonlinear functions of the variable or its derivatives are present. In the context of our second-order equation, we are looking at a differential equation involving the function, its first derivative, and second derivative, all in a linear format.

When dealing with a second-order differential equation, it typically has the form:

• \[ a(x)y'' + b(x)y' + c(x)y = f(x) \]

Here, \( a(x), b(x), \) and \( c(x) \) are functions of \( x \) and \( f(x) \) is a given function. In our exercise, \( f(x) = 0 \), making it a homogeneous linear differential equation.

The objective is to find an equation that holds for all solutions of the given form, \( y = c_1 x + c_2 x^2 \). This means the task involved writing the equation without reference to any particular values of \( c_1 \) or \( c_2 \), by utilizing linear combinations of the function and its derivatives.

Elimination of Constants

The elimination of constants is a fundamental process when deriving a differential equation from a given family of solutions. Constants such as \( c_1 \) and \( c_2 \) in our solution function normally represent arbitrary parameters. These constants need to be eliminated to form a differential equation that is universally valid for all possible values of \( c_1 \) and \( c_2 \).

To do this, we perform differentiation:

• First, by differentiating \( y = c_1 x + c_2 x^2 \) to obtain its first derivative: \( y' = c_1 + 2c_2 x \).
• Then differentiate again to find the second derivative: \( y'' = 2c_2 \).

From these, we solve for \( c_2 = \frac{1}{2}y'' \) and substitute it back into \( y' \) to express in terms of the derivatives only, removing \( c_2 \).

Next, solve the adjusted \( y' \) to find \( c_1 = y' - xy'' \). This successfully expresses both constants in terms of \( y \) and its derivatives, leading to their elimination.

Two-Parameter Family of Solutions

The concept of a two-parameter family of solutions is important in understanding differential equations. When we refer to such solutions, it means that the differential equation initially has a general solution that involves two arbitrary constants. These constants, \( c_1 \) and \( c_2 \), provide flexibility, representing a wide family of curves described by the equation.

The form \( y = c_1 x + c_2 x^2 \) indicates a solution generated by any value of these constants. Different pairs \( (c_1, c_2) \) will result in different specific solutions, each still valid within the broader solution family. This is an inherent property of solutions to second-order differential equations.

By manipulating the differential equation through differentiation and substitution, the initial general expression involving \( c_1 \) and \( c_2 \) becomes an equation void of them, indicating a specific relationship among the derivatives of \( y \) that must hold for any allowed solution within this family.