Chapter 4: Problem 7
Solve the given differential equation. $$x^{2} y^{\prime \prime}-3 x y^{\prime}-2 y=0$$
Short Answer
Expert verified
The solution is \( y = c_1 x^{2 + \sqrt{6}} + c_2 x^{2 - \sqrt{6}} \).
Step by step solution
01
Recognize the Type of Differential Equation
The given differential equation is \( x^2 y'' - 3x y' - 2y = 0 \). This is a second-order linear differential equation with variable coefficients. Specifically, it's an Euler-Cauchy or equidimensional equation, as the powers of x match the order of the derivatives.
02
Assume a Power Series Solution
For Euler-Cauchy equations, assume a solution of the form \( y = x^m \). This transforms the differential equation into an algebraic equation when derivative expressions are substituted back into the equation.
03
Substitute and Simplify
Find the first derivative \( y' = mx^{m-1} \) and the second derivative \( y'' = m(m-1)x^{m-2} \). Substitute into the differential equation: \[ x^2(m(m-1)x^{m-2}) - 3x(mx^{m-1}) - 2x^m = 0 \]Simplify: \[ m(m-1)x^m - 3mx^m - 2x^m = 0 \] \[ (m(m-1) - 3m - 2)x^m = 0 \] For this equation to be true, the coefficient of \( x^m \) must be zero.
04
Solve the Characteristic Equation
The characteristic equation is:\[ m(m-1) - 3m - 2 = 0 \] Simplifies to:\[ m^2 - 4m - 2 = 0 \]Solve this quadratic equation using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), \( c = -2 \).
05
Use the Quadratic Formula
Plug in the values into the quadratic formula:\[ m = \frac{4 \pm \sqrt{16 + 8}}{2} \]\[ m = \frac{4 \pm \sqrt{24}}{2} \]\( \sqrt{24} = 2\sqrt{6} \), so:\[ m = \frac{4 \pm 2\sqrt{6}}{2} \] This simplifies to:\[ m = 2 \pm \sqrt{6} \]
06
Write the General Solution
The general solution for the Euler-Cauchy equation, given the roots \( m_1 = 2 + \sqrt{6} \) and \( m_2 = 2 - \sqrt{6} \), is:\[ y = c_1 x^{2 + \sqrt{6}} + c_2 x^{2 - \sqrt{6}} \]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.
Euler-Cauchy equation
An Euler-Cauchy equation, often referred to as an equidimensional equation, is a special type of differential equation. It has the form \[ a_n x^n y^{(n)} + a_{n-1} x^{n-1} y^{(n-1)} + \cdots + a_0 y = 0 \]where the powers of \( x \) correspond to the order of the derivative. This characteristic makes the equation especially suitable for solutions involving power series or assuming solutions of a form like \( y = x^m \). By doing so, the differential equation can be transformed into an algebraic equation in terms of \( m \), which is much easier to solve. Recognizing an Euler-Cauchy equation is critical because it suggests a method of solution that leverages this symmetry. In practical terms, if you see an equation with variable coefficients where the power of \( x \) matches the derivative's order, you are dealing with an Euler-Cauchy equation.
second-order linear differential equation
A second-order linear differential equation takes the general form:\[ a(x) y'' + b(x) y' + c(x) y = f(x) \]For our exercise, the equation simplifies since it equals zero, making it homogeneous. These types of equations describe a wide array of physical phenomena, including oscillations and waves.
- Order: The highest derivative is considered the order of the differential equation; here, it's the second derivative \( y'' \).
- Linearity: The way the equation is structured tells us it's linear, meaning any linear combination of solutions is also a solution.
power series solution
Power series solutions are particularly effective for equations like the Euler-Cauchy equation. They transform a differential equation into an algebraic format by assuming a series expansion or a simple form such as \( y = x^m \). This technique can handle variable coefficient equations and is vital when typical solution methods don't apply. - **First Derivative** \( y' = m x^{m-1} \)- **Second Derivative** \( y'' = m(m-1)x^{m-2} \)Substituting these derivatives back into the equation results in an expression that depends only on \( m \), allowing us to solve algebraically. This approach is essential when faced with differential equations that don't fit neatly into other standard solving techniques.
quadratic formula
The quadratic formula is a universal solver for quadratic equations of the form \( ax^2 + bx + c = 0 \). For our problem, the characteristic equation emerges as a quadratic, resulting from transforming the Euler-Cauchy differential equation: \[ m^2 - 4m - 2 = 0 \]The formula\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]provides the solution by substituting \( a = 1 \), \( b = -4 \), and \( c = -2 \). In such applications:- **Root Calculation**: Calculate the discriminant \( b^2 - 4ac \) first to determine the roots' nature.- **Solution Simplicity**: Ensures that all solutions, real or complex, emerge accurately. The quadratic formula reveals that roots \( m_1 = 2 + \sqrt{6} \) and \( m_2 = 2 - \sqrt{6} \) exist, and these directly inform the structure of the solution set for the differential equation.