Chapter 12: Problem 3
\(x^{2} y^{\prime \prime}+2 x y^{\prime}-6 y=0\)
Short Answer
Expert verified
The general solution is \( y = C_1 x^2 + C_2 x^{-3} \).
Step by step solution
01
Identify the type of differential equation
The given differential equation is a second-order linear homogeneous differential equation with variable coefficients: \( x^{2} y^{\rprime \rprime} + 2x y^{\rprime} - 6y = 0 \).
02
Assume a solution of the form
Assume a solution of the form \( y = x^m \). This method works for Cauchy-Euler differential equations. Compute the derivatives: \[ y = x^m \] \[ y^{\rprime} = m x^{m-1} \] \[ y^{\rprime \rprime} = m(m-1) x^{m-2} \]
03
Substitute into the original equation
Substitute \( y, y^{\rprime}, y^{\rprime \rprime} \) into the original differential equation: \[ x^{2} \big[m(m-1) x^{m-2}\big] + 2x \big[m x^{m-1}\big] - 6 x^m = 0 \]
04
Simplify the equation
Simplify the equation: \[ m(m-1) x^m + 2m x^m - 6 x^m = 0 \] Combine the like terms: \[ [m(m-1) + 2m - 6] x^m = 0 \]
05
Solve for the roots of the characteristic equation
Solve the characteristic equation: \[ m(m-1) + 2m - 6 = 0 \] Expand and simplify: \[ m^2 -m + 2m -6 = 0 \] \[ m^2 + m - 6 = 0 \]Factorize the quadratic: \[ (m-2)(m+3) = 0 \] So, the roots are: \[ m_1 = 2 \] \[ m_2 = -3 \]
06
Write the general solution
The general solution to the differential equation is a linear combination of the two solutions found: \[ y = C_1 x^2 + C_2 x^{-3} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-Order Linear Differential Equations
A second-order linear differential equation is a type of differential equation that involves the second derivative of an unknown function with respect to an independent variable. In mathematical form, it looks like this: \[ a(x) y^{\rprime \rprime} + b(x) y^{\rprime} + c(x) y = 0 \]Where \( y \) is the unknown function of \( x \), \( y^{\rprime} \) is its first derivative, and \( y^{\rprime \rprime} \) is its second derivative.
When dealing with second-order linear differential equations, identifying the type is crucial. They differ based on whether they are homogeneous or non-homogeneous and whether the coefficients are constant or variable. In this exercise, the given differential equation is a second-order linear homogeneous differential equation with variable coefficients, which leads us to look for a specific type of solution.
When dealing with second-order linear differential equations, identifying the type is crucial. They differ based on whether they are homogeneous or non-homogeneous and whether the coefficients are constant or variable. In this exercise, the given differential equation is a second-order linear homogeneous differential equation with variable coefficients, which leads us to look for a specific type of solution.
- Order: Refers to the highest derivative in the equation, which is the second derivative in this case.
- Linearity: Means all the terms involving \( y \) and its derivatives are linear.
Homogeneous Equations
Homogeneous equations are a special type of differential equation where all terms are a function of the unknown variable and its derivatives. The standard form is:
\[ a_n(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + ... + a_1(x) y^{\rprime} + a_0(x) y = 0 \]
This means there is no term that is independent of the unknown function \( y \).
In this exercise:
\[ x^2 y^{\rprime \rprime} + 2x y^{\rprime} - 6 y = 0 \]The equation is homogeneous because every term includes a derivative of \( y \) or \( y \) itself. Homogeneous equations are often easier to solve because they suggest certain solutions based on the characteristic equation.
Homogeneous differential equations often allow for simplifications since the same function (or its powers, derivatives, etc.) is repeated. Therefore, the methods for solving them, like assuming solutions of the form \( y = x^m \), can be employed effectively.
\[ a_n(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + ... + a_1(x) y^{\rprime} + a_0(x) y = 0 \]
This means there is no term that is independent of the unknown function \( y \).
In this exercise:
\[ x^2 y^{\rprime \rprime} + 2x y^{\rprime} - 6 y = 0 \]The equation is homogeneous because every term includes a derivative of \( y \) or \( y \) itself. Homogeneous equations are often easier to solve because they suggest certain solutions based on the characteristic equation.
Homogeneous differential equations often allow for simplifications since the same function (or its powers, derivatives, etc.) is repeated. Therefore, the methods for solving them, like assuming solutions of the form \( y = x^m \), can be employed effectively.
Variable Coefficients
Variable coefficients mean that the coefficients (numbers multiplying the derivatives) are not constants but functions of the independent variable, often making these equations more complex to solve than their constant coefficient counterparts.
This characteristic is illustrated in the provided exercise:
\[ x^2 y^{\rprime \rprime} + 2x y^{\rprime} - 6 y = 0 \]Here, the coefficients \( x^2 \) and \( 2x \) are functions of the variable \( x \), making traditional solution methods less straightforward.
However, for Cauchy-Euler equations, which have polynomial-variable coefficients like above, a substitution approach (assuming \( y = x^m \)) simplifies to a related algebraic problem, the characteristic equation.
Understanding how the coefficients vary with the independent variable is crucial for determining which solution methods are applicable. For the given equation type, exploiting the form of these coefficients allows efficient solution-finding strategies.
This characteristic is illustrated in the provided exercise:
\[ x^2 y^{\rprime \rprime} + 2x y^{\rprime} - 6 y = 0 \]Here, the coefficients \( x^2 \) and \( 2x \) are functions of the variable \( x \), making traditional solution methods less straightforward.
However, for Cauchy-Euler equations, which have polynomial-variable coefficients like above, a substitution approach (assuming \( y = x^m \)) simplifies to a related algebraic problem, the characteristic equation.
Understanding how the coefficients vary with the independent variable is crucial for determining which solution methods are applicable. For the given equation type, exploiting the form of these coefficients allows efficient solution-finding strategies.
Characteristic Equation
The characteristic equation is derived from the assumed solution of the differential equation. It transforms the differential problem into an algebraic one.
In this problem, we assumed a solution \( y = x^m \). By substituting \( y = x^m \) and its derivatives back into the original differential equation, we obtain the characteristic equation:
\[ m(m-1) x^m + 2m x^m - 6 x^m = 0 \]
Simplifying, we combine like terms:
\[ (m(m-1) + 2m - 6) x^m = 0 \]
Because \( x^m \) is never zero, we solve the quadratic characteristic equation:
\[ m^2 + m - 6 = 0 \]
This factors further to find the roots:
\[ (m - 2)(m + 3) = 0 \]
Giving the solutions \( m_1 = 2 \) and \( m_2 = -3 \).
The general solution of the differential equation is then a linear combination of these solutions:
\[ y = C_1 x^2 + C_2 x^{-3} \]
The characteristic equation is a vital step in solving second-order linear differential equations, providing a straightforward way to find the general solution.
In this problem, we assumed a solution \( y = x^m \). By substituting \( y = x^m \) and its derivatives back into the original differential equation, we obtain the characteristic equation:
\[ m(m-1) x^m + 2m x^m - 6 x^m = 0 \]
Simplifying, we combine like terms:
\[ (m(m-1) + 2m - 6) x^m = 0 \]
Because \( x^m \) is never zero, we solve the quadratic characteristic equation:
\[ m^2 + m - 6 = 0 \]
This factors further to find the roots:
\[ (m - 2)(m + 3) = 0 \]
Giving the solutions \( m_1 = 2 \) and \( m_2 = -3 \).
The general solution of the differential equation is then a linear combination of these solutions:
\[ y = C_1 x^2 + C_2 x^{-3} \]
The characteristic equation is a vital step in solving second-order linear differential equations, providing a straightforward way to find the general solution.
