Chapter 18: Problem 3
Find two linearly independent solutions, valid for \(x>0,\) unless otherwise instructed. $$ 2 x y^{\prime \prime}+6 y^{\prime}+y=0 $$
Short Answer
Expert verified
The solutions are \(y_1(x) = x^{-1 + \frac{\sqrt{2}}{2}}\) and \(y_2(x) = x^{-1 - \frac{\sqrt{2}}{2}}\).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \(2x y'' + 6y' + y = 0\). We notice that it is a second-order linear homogeneous differential equation with variable coefficients.
02
Analyze the Equation for Euler-Cauchy Form
The equation can be categorized as an Euler-Cauchy (or equidimensional) differential equation because it fits the form \(a x^2 y'' + b x y' + c y = 0\). For this equation, it takes the form \(2x^2 y'' + 6x y' + y = 0\).
03
Assume a Power Series Solution
For Euler-Cauchy equations, assume a solution of the form \(y = x^m\). Substituting this into the equation, we proceed to find \(y''\) and \(y'\): \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\).
04
Substitute and Simplify
Substitute \(y = x^m\), \(y' = mx^{m-1}\), and \(y'' = m(m-1)x^{m-2}\) into the equation: \[2x(m(m-1)x^{m-2}) + 6(mx^{m-1}) + x^m = 0\]Simplify: \[2m(m-1)x^m + 6mx^{m-1} + x^m = 0\]\[(2m(m-1) + 6m + 1)x^m = 0\].
05
Solve the Resulting Polynomial Equation
Factor out \(x^m\), resulting in: \[2m^2 + 4m + 1 = 0\].Solve for \(m\) using the quadratic formula: \[m = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2}\]\[m = \frac{-4 \pm \sqrt{8}}{4}\]\[m = \frac{-4 \pm 2\sqrt{2}}{4}\]\[m = -1 \pm \frac{\sqrt{2}}{2}\].Thus, the roots are \(m_1 = -1 + \frac{\sqrt{2}}{2}\) and \(m_2 = -1 - \frac{\sqrt{2}}{2}\).
06
Form the General Solution
The general solution to the Euler-Cauchy equation is a combination of both solutions: \[y(x) = C_1 x^{-1 + \frac{\sqrt{2}}{2}} + C_2 x^{-1 - \frac{\sqrt{2}}{2}}\].This forms two linearly independent solutions for the given differential equation, valid for \(x > 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-Order Linear Differential Equation
A second-order linear differential equation is one where the highest derivative is the second derivative. In general form, it can be represented as \( a(x) y'' + b(x) y' + c(x) y = f(x) \). In this type of equation:
This type of equation is "linear" because each term depends linearly upon the unknown function \( y \) and its derivatives. The equation \( 2x y'' + 6y' + y = 0 \) given in the example is considered a second-order linear differential equation where each term's dependency involves a linear relationship with \( y \), \( y' \), and \( y'' \). The focus here is on finding solutions that satisfy the equation within its domain, for instance, when \( x > 0 \).
- \( y'' \) is the second derivative of \( y \).
- The functions \( a(x) \), \( b(x) \), and \( c(x) \) are known as the coefficients, and they can be constants or functions of \( x \).
- \( f(x) \) is the non-homogeneous part, but in a homogeneous equation, \( f(x) = 0 \).
This type of equation is "linear" because each term depends linearly upon the unknown function \( y \) and its derivatives. The equation \( 2x y'' + 6y' + y = 0 \) given in the example is considered a second-order linear differential equation where each term's dependency involves a linear relationship with \( y \), \( y' \), and \( y'' \). The focus here is on finding solutions that satisfy the equation within its domain, for instance, when \( x > 0 \).
Linearly Independent Solutions
When solving differential equations, especially homogeneous ones, it's crucial to find a set of solutions that are linearly independent. Two solutions \( y_1(x) \) and \( y_2(x) \) are said to be linearly independent if there are no constants \( C_1 \) and \( C_2 \), not both zero, such that: \[ C_1 y_1(x) + C_2 y_2(x) = 0 \]for all values of \( x \) in the interval of interest.
Finding linearly independent solutions means that any solution to the differential equation can be written as a linear combination of these solutions.
Finding linearly independent solutions means that any solution to the differential equation can be written as a linear combination of these solutions.
- In our example, the obtained linearly independent solutions are based on the roots derived from substituting a power series solution into the Euler-Cauchy form.
- The general solution is composed of these independent solutions, ensuring it can solve a wide range of initial conditions for the given differential equation.
Homogeneous Differential Equation
A homogeneous differential equation is one in which every term is a function of \( y \) and its derivatives. There is no term that is a standalone function of \( x \), i.e., \( f(x) = 0 \) in the equation \( a(x) y'' + b(x) y' + c(x) y = 0 \).
For example, the equation \( 2x y'' + 6y' + y = 0 \) is homogeneous because there is no term independent of \( y \) or its derivatives on the right side of the equation.
For example, the equation \( 2x y'' + 6y' + y = 0 \) is homogeneous because there is no term independent of \( y \) or its derivatives on the right side of the equation.
- In a homogeneous equation, the superposition principle applies, which allows for any linear combination of solutions to the equation to also be a solution.
- This property is used to find the general solution by combining linearly independent solutions.
Power Series Solution
For certain differential equations, a power series method is useful, particularly when dealing with variable coefficients, as is the case with an Euler-Cauchy equation. The idea is to assume a solution in the form of a power series: \[ y = x^m \]where \( m \) is determined through substitution into the equation. This approach:
These roots lead to distinct solutions that become the basis for forming the general solution which can solve the Euler-Cauchy differential equation for any initial conditions when \( x > 0 \).
- Identifies potential solutions based on the equation's algebraic structure.
- Helps solve complicated coefficients that vary with \( x \).
These roots lead to distinct solutions that become the basis for forming the general solution which can solve the Euler-Cauchy differential equation for any initial conditions when \( x > 0 \).