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Chapter 18: Problem 5

Obtain two linearly independent solutions valid for \(x>0\) unless otherwise instructed. $$ x(1+x) y^{\prime \prime}+(1+5 x) y^{\prime}+3 y=0 $$.

Short Answer

Expert verified
The linearly independent solutions are \( y_1(x) = x^{-1} \) and \( y_2(x) = x^{-3} \).

Step by step solution

01

Identify the type of differential equation

The given differential equation is \(x(1+x) y'' + (1+5x) y' + 3y = 0\). It鈥檚 a second-order linear ordinary differential equation. Given its form, it appears to be a Cauchy-Euler equation, which is a specific type of differential equation that can often be solved using substitution methods.
02

Apply substitution for Cauchy-Euler equation

Substitute \( y = x^m \) into the differential equation. Compute \( y' = mx^{m-1} \) and \( y'' = m(m-1)x^{m-2} \). Substitute these expressions into the differential equation to try and find a characteristic equation.
03

Derive and solve the characteristic equation

Substituting \( y = x^m \), \( y' = mx^{m-1} \), and \( y'' = m(m-1)x^{m-2} \) into the differential equation gives us: \( x(1+x)m(m-1)x^{m-2} + (1+5x)mx^{m-1} + 3x^m = 0 \). Simplifying and equating the coefficients, form a quadratic equation in terms of \( m \): \( m^2 + 4m + 3 = 0 \). Solving this gives two roots, \( m_1 = -1 \) and \( m_2 = -3 \).
04

Write the general solution

Using the roots of the characteristic equation, the general solution of the differential equation is a linear combination of the two linearly independent solutions: \( y(x) = C_1x^{-1} + C_2x^{-3} \) where \( C_1 \) and \( C_2 \) are arbitrary constants. This general solution is valid for \( x > 0 \).

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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
These types of equations involve derivatives of a function up to the second derivative. The general form is expressed as:
  • \( a(x) y'' + b(x) y' + c(x) y = 0 \)
Linear means each term in the equation has the function \( y \) or its derivatives raised to the first power, and not multiplied by each other. They're key in describing various natural phenomena, like vibrations and heat conduction.
For our equation, \( x(1+x) y'' + (1+5x) y' + 3y = 0 \), the coefficients depend on \( x \), making it a variable coefficient differential equation, specifically of the second order.
Ordinary Differential Equations
An ordinary differential equation (ODE) involves functions of a single variable and their derivatives. In contrast to partial differential equations, which involve multiple variables, ODEs are simpler in terms of their scope. They're abundant in modeling real-world systems like population growth or mechanical movements.
Our specific ODE is derived from the mathematics of the Cauchy-Euler equation, characterized by its ability to simplify solutions by a particular method of substitution. This makes solving Cauchy-Euler type ODEs a common exercise in advanced calculus.
Characteristic Equation
The characteristic equation is a tool to solve differential equations, especially those resembling the form of Cauchy-Euler equations. By assuming a solution of the form \( y = x^m \), and substituting it back into the equation, we derive an algebraic equation involving \( m \), called the characteristic equation.
For our problem:
  • The given substitution \( y = x^m \), leads to: \( m^2 + 4m + 3 = 0 \)
This quadratic equation results after simplifying the original differential equation. Solving it, we find roots \( m_1 = -1 \) and \( m_2 = -3 \), which are crucial for finding the general solution.
General Solution
Once the characteristic equation is solved, the general solution can be formed. For second-order linear ODEs like the Cauchy-Euler type, the roots of the characteristic equation guide us. If the roots \( m_1 \) and \( m_2 \) are real and distinct, as they are here, the general solution is:
  • \( y(x) = C_1x^{m_1} + C_2x^{m_2} \)
Given our roots \( m_1 = -1 \) and \( m_2 = -3 \), the general solution becomes:
\( y(x) = C_1x^{-1} + C_2x^{-3} \).
This solution holds for \( x > 0 \) and each \( C_1 \) and \( C_2 \) represents arbitrary constants determined by initial conditions, showcasing the flexibility of solutions in differential equations.

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Most popular questions from this chapter

Obtain two linearly independent solutions valid near the origin for \(x>0\). Always state the region of validity of each solution that you obtain. $$ 3 x^{2} y^{\prime \prime}+x y^{\prime}-(1+x) y=0 $$.

The singular points in the finite plane have already been located and classified. For each equation, determine whether the point at infinity is an ordinary point (O.P.), a regular singular point point (R.S.P.), or an irregular singular point (I.S.P.). Do not solve the problems. \(x^{4} y^{\prime \prime}+2 x^{3} y^{\prime}+4 y=0 .\) (Exercise 18, Section 18.1.)

The singular points in the finite plane have already been located and classified. For each equation, determine whether the point at infinity is an ordinary point (O.P.), a regular singular point point (R.S.P.), or an irregular singular point (I.S.P.). Do not solve the problems. \(y^{\prime \prime}+x y=0\). (Exercise 3, Section 18.1.)

In each exercise, obtain solutions valid for \(x>0\). $$x^{2} y^{\prime \prime}+x(2 x-3) y^{\prime}+(4 x+3) y=0$$

Find two linearly independent solutions, valid for \(x>0,\) unless otherwise instructed. $$ x y^{\prime \prime}+(3+2 x) y^{\prime}+8 y=0 $$
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