Chapter 1: Problem 54
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 \(c_{2}\).
Short Answer
Expert verified
The differential equation is \( xy' - \frac{1}{2} x^2 y'' - y = 0 \).
Step by step solution
01
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 \( y'' = \frac{d^2}{dx^2}(c_1 x + c_2 x^2) = 2c_2 \).
02
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:\[ c_1 = y' - xy'' \].
03
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:\[ y = xy' - \frac{1}{2} x^2 y'' \].
04
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 \( c_2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
When dealing with a second-order differential equation, it typically has the form:
- \[ a(x)y'' + b(x)y' + c(x)y = f(x) \]
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:
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.
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: \( y'' = 2c_2 \).
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.
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.