Chapter 17: Problem 19
Find the general solution to the given Euler equation. Assume \(x>0\) throughout. $$x^{2} y^{\prime \prime}+x y^{\prime}=0$$
Short Answer
Expert verified
The general solution is \( y(x) = C_1 + C_2 \log(x) \).
Step by step solution
01
Identify the Type of Differential Equation
The given equation is a second-order homogeneous linear differential equation with variable coefficients: \[x^{2} y^{\prime \prime}+x y^{\prime}=0\]This type of equation is known as an Euler or Cauchy-Euler equation with the general form:\[ x^n y^{(n)} + a_{n-1} x^{n-1} y^{(n-1)} + \, \dots \, + a_0 y = 0 \]
02
Apply the Cauchy-Euler Method for Homogeneous Equations
For a second-order Cauchy-Euler equation of the form \[ x^{2} y^{\prime \prime}+b x y^{\prime}+c y=0 \]the approach involves assuming a solution of the form \[ y = x^m \].We substitute this form into the differential equation to find the characteristic equation.
03
Substitute the Assumed Solution
First, calculate the derivatives assuming \( y = x^m \):- \( y^{\prime} = m x^{m-1} \)- \( y^{\prime\prime} = m(m-1) x^{m-2} \)Substitute these into the original equation:\[ x^2 \cdot m(m-1) x^{m-2} + x \cdot m x^{m-1} = 0 \]This simplifies to:\[ m(m-1) x^m + m x^m = 0 \]
04
Formulate the Characteristic Equation
Combine and factor the equation:\[ (m^2 - m + m) x^m = 0 \]This simplifies to:\[ m^2 = 0 \]So, the characteristic equation is \( m^2 = 0 \).
05
Solve the Characteristic Equation
The characteristic equation \( m^2 = 0 \) has a double root, \( m = 0 \). This means the solution involves a repeated root, leading to a specific form for the general solution when we solve such equations.
06
Write the General Solution for Repeated Roots
For a repeated root \( m = 0 \), the general solution for the differential equation is given by:\[ y(x) = C_1 x^m + C_2 x^m \log(x) \]Since \( m = 0 \), this simplifies to:\[ y(x) = C_1 + C_2 \log(x) \]
07
Confirm the Form of the Solution
The final solution, based on substituting back into the original differential equation and confirming no additional terms are generated, is:\[ y(x) = C_1 + C_2 \log(x) \]This confirms the approach and is consistent with the properties of Euler equations with repeated roots.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cauchy-Euler equations
Cauchy-Euler equations, sometimes referred to simply as Euler equations, are a special class of differential equations that display interesting mathematical properties. These equations often arise in problems involving power-law behaviors and scaling symmetry. They are recognized by their specific feature of having variable coefficients that are powers of the independent variable, commonly represented as \( x^n \). The general form of a Cauchy-Euler equation is:
- \( x^n y^{(n)} + a_{n-1} x^{n-1} y^{(n-1)} + \, \dots \, + a_0 y = 0 \)
homogeneous linear differential equations
Homogeneous linear differential equations are a foundational concept in differential equations. They are characterized by having all terms dependent on the function and its derivatives, without any additional non-zero terms (i.e., the equation equals zero). In general, a second-order homogeneous linear differential equation can be expressed as:
- \( a(x) y'' + b(x) y' + c(x) y = 0 \)
characteristic equation
The characteristic equation is the cornerstone of solving differential equations like the Cauchy-Euler equations. It's formed by substituting an assumed solution form, usually \( y = x^m \), into the differential equation and simplifying the resulting expression. For a Cauchy-Euler equation, this substitution simplifies the problem into solving a polynomial equation:
- \( m(m-1) x^m + m x^m = 0 \)
- This results in \( m^2 = 0 \), which is the characteristic equation in this context.
variable coefficients
Variable coefficients are a key feature of Cauchy-Euler equations that differentiate them from constant coefficient differential equations. In these equations, coefficients are not constants but rather functions of the independent variable (for example, \( x^2, x \)). This characteristic adds complexity, as traditional methods used for constant coefficient equations are not directly applicable.
When handling variable coefficients, the Cauchy-Euler method is particularly useful because it transforms the equation into a form that is easier to work with, creating a characteristic equation as seen in the problem solution. This method assumes that the solution can be represented as a power of the independent variable (\( y = x^m \)), which simplifies the variable nature of the coefficients into a manageable algebraic form. Understanding how to navigate variable coefficients is crucial when tackling a range of applied problems in fields like physics and engineering.
When handling variable coefficients, the Cauchy-Euler method is particularly useful because it transforms the equation into a form that is easier to work with, creating a characteristic equation as seen in the problem solution. This method assumes that the solution can be represented as a power of the independent variable (\( y = x^m \)), which simplifies the variable nature of the coefficients into a manageable algebraic form. Understanding how to navigate variable coefficients is crucial when tackling a range of applied problems in fields like physics and engineering.