Chapter 5: Problem 8
$$x y^{\prime \prime}-y=0$$
Short Answer
Expert verified
The general solution is \( y(x) = C_1 x^{\frac{1 + \sqrt{5}}{2}} + C_2 x^{\frac{1 - \sqrt{5}}{2}} \).
Step by step solution
01
Recognize the Type of Equation
The given differential equation \( x y'' - y = 0 \) is a second-order linear homogeneous differential equation with variable coefficients.
02
Simplify and Analyze
Rewrite the equation in standard form by dividing the entire equation by \( x \), resulting in \( y'' = \frac{y}{x} \). This is the same as saying \( y'' - \frac{1}{x} y = 0 \).
03
Identify if Special Techniques Apply
Recognize that this differential equation matches the Cauchy-Euler form, \( a_n x^n y'' + a_{n-1} x^{n-1} y' + a_{n-2} x^{n-2} y = 0 \), with \( n = 1 \). We can apply the substitution \( y = x^m \) to find a solution.
04
Substitute and Solve for m
Substitute \( y = x^m \) into the equation. Then, \( y' = mx^{m-1} \) and \( y'' = m(m-1)x^{m-2} \). Substituting into the equation gives \( x(m(m-1)x^{m-2}) - x^m = 0 \), simplifying to \( m(m-1) - 1 = 0 \).
05
Solve the Characteristic Equation
Solve the characteristic equation \( m(m-1) = 1 \) which simplifies to \( m^2 - m - 1 = 0 \). Using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = -1 \), find \( m = \frac{1 \pm \sqrt{5}}{2} \).
06
Write the General Solution
The general solution, based on the roots \( m_1 = \frac{1 + \sqrt{5}}{2} \) and \( m_2 = \frac{1 - \sqrt{5}}{2} \), is \( y(x) = C_1 x^{m_1} + C_2 x^{m_2} \), where \( C_1 \) and \( C_2 \) are arbitrary constants.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-Order Differential Equations
Second-order differential equations involve derivatives up to the second degree. These equations can describe a wide range of phenomena, from the motion of a pendulum to electrical circuits. A general form of a second-order differential equation can be written as \(a(x)y'' + b(x)y' + c(x)y = g(x)\), where:
- \(y''\) is the second derivative of the unknown function \(y\).
- \(y'\) is the first derivative of \(y\).
- \(a(x), b(x), c(x)\) are coefficients that can be constants or functions of \(x\).
- \(g(x)\) is the non-homogeneous term which is generally dependent on \(x\).
Linear Homogeneous Differential Equations
Linear homogeneous differential equations are a special type of differential equation where the non-homogeneous part \(g(x)\) is zero. This means that the equation relies entirely on the terms involving \(y\), \(y'\), and \(y''\). The general form is \(a(x)y'' + b(x)y' + c(x)y = 0\). Here, the task is to find a solution for \(y\) that satisfies the equation:
- "Linear" indicates that each term involving \(y\) and its derivatives is to the power of one, ensuring no products or nonlinear operations involve \(y\) and its derivatives together.
- "Homogeneous" signifies that the solution has no outside forces or terms affecting the equation.
Characteristic Equations
Characteristic equations are fundamental in solving linear homogeneous differential equations. They arise from substituting a trial solution \(y = x^m\) into the differential equation, leading to a polynomial equation. This equation is pivotal in the process because finding its roots gives us crucial information about the solution to the differential equation.For the Cauchy-Euler equation, after substitution, the differential equation simplifies into a characteristic equation of form \(a_n m(m-1) + a_{n-1} m + a_{n-2} = 0\). Solving this polynomial for \(m\) provides the exponents used in the solutions. The roots can be:
- Real and distinct: Resulting in solutions that are power functions of the form \(x^m\).
- Real and repeated: Leading to terms like \(x^m, \log(x)x^m\).
- Complex: Resulting in functions involving sinusoids, such as \(x^a \cos(b\log x), x^a \sin(b\log x)\).