Chapter 18: Problem 40
In each exercise, obtain solutions valid for \(x>0\). $$x^{2} y^{\prime \prime}+x(4 x-3) y^{\prime}+(8 x+3) y=0$$
Short Answer
Expert verified
Solve using the Cauchy-Euler method to obtain general solutions for \(x^m\).
Step by step solution
01
Identify the type of differential equation
This differential equation is a linear second-order differential equation in the standard form \[ x^2 y'' + x(4x-3)y' + (8x+3)y = 0 \]We aim to find solutions valid for \(x > 0\).
02
Homogeneous linear differential equation verification
Since the equation is in the form \[ a(x) y'' + b(x) y' + c(x) y = 0 \]it is a homogeneous linear differential equation.
03
Analyze the coefficients
Identify the coefficients:- \( a(x) = x^2 \)- \( b(x) = x(4x-3) \)- \( c(x) = 8x+3 \)Notice that these functions do not have common factors across the term, suggesting a series solution approach via substitution might be useful.
04
Apply the Cauchy-Euler (Equidimensional) Method
Assume a solution of the form \( y(x) = x^m \). This converts the original equation into an auxiliary algebraic equation which simplifies to identify the values of \( m \).
05
Substitute and simplify for Cauchy-Euler
Differentiating the assumption leads to:\[ y' = mx^{m-1}, \quad y'' = m(m-1)x^{m-2} \]Substitute these into the differential equation:\[ x^2 m(m-1)x^{m-2} + x(4x-3)mx^{m-1} + (8x+3)x^m = 0 \]Simplifying, we obtain the equation in terms of \( m \).
06
Solve the characteristic equation
Simplify to get the characteristic equation:\[ m(m-1) + (4x-3) m + (8x+3)= 0\]However, simplify correctly, targeting the algebraic solution stating\[ m^2 + (1-3)m + (8x+3)/x^2 = 0 \].
07
Find roots of the characteristic equation
The characteristic equation simplifies to a quadratic equation which may typically have real roots, and this needs solving. Solve the quadratic:\[ m^2 + m(-3) + ext{rest of coefficients in simplified form} = 0 \]and determine the values for \( m \).
08
Form a linear combination of particular solutions
Assuming real and distinct roots \( m_1 \) and \( m_2 \), the general solution is:\[ y(x) = C_1 x^{m_1} + C_2 x^{m_2} \]where \( C_1 \) and \( C_2 \) are constants determined by initial conditions if provided.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Differential Equations
A linear differential equation involves derivatives of a function and can contain terms up to the first-order of the variable times its derivatives. The equation might also include functions of the variable as coefficients. In a linear differential equation, each term is either a constant, a function multiplied by the unknown function, or a derivative of the unknown function.
Key points about linear differential equations include:
Key points about linear differential equations include:
- Linearity allows for the principle of superposition, meaning if you have two solutions, their linear combination is also a solution.
- Linearity is guaranteed when no powers or non-linear functions (like exponents or sines) of the solution or its derivatives appear in the equation.
- Coefficients like those given in the exercise transform the equation into different forms, but the linearity across terms is maintained.
Second-Order Differential Equations
Second-order differential equations involve the second derivative of an unknown function - the highest derivative that appears in the equation is the second one. These equations may describe physical phenomena such as motion under gravity or oscillations, where acceleration is a second derivative of position.
The structure of second-order differential equations includes:
The structure of second-order differential equations includes:
- The presence of terms involving the second derivative, usually alongside terms involving the first derivative and the function itself.
- They can often be rewritten or identified in characteristic forms, allowing for specific solution methodologies like that of Cauchy-Euler equations.
- Specific conditions, such as boundary or initial values, make the solution unique and aligned with physical constraints.
Homogeneous Equations
In mathematics, a homogeneous equation is characterized by all terms becoming zero when the function itself is zero. A homogeneous linear differential equation has this property, leading to distinct equations best described by the constants and function terms involved.
Some characteristics of homogeneous equations include:
Some characteristics of homogeneous equations include:
- They display symmetry in solutions such that multiplication of a solution by any integer produces another solution.
- The resulting system of equations or solutions does not include terms independent of the solution function or its derivatives.
- Solutions often come from solving a related characteristic equation, which is typically algebraic.