/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Determine three linearly indepen... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 2

Determine three linearly independent solutions of each of the following equations: (a) \(x^{2} y^{\prime \prime \prime}+2 x y^{\prime \prime}-2 y^{\prime}=0\) (b) \(x^{3} y^{\prime \prime \prime}-3 x y^{\prime}+3 y=0\) (c) \(x^{3} y^{\prime \prime \prime}+2 x^{2} y^{\prime \prime}-3 x y^{\prime}+3 y=0\) (d) \(x^{3} y^{\prime \prime \prime}-6 x^{2} y^{\prime \prime}+7 x y^{\prime}-7 y=0\)

Short Answer

Expert verified
The solutions are: (a) 1, x, 1/x; (b) x^3, 1/x, x; (c) x^3, x, 1/x; (d) x, x^2, 1/x.

Step by step solution

01

Analyze the given differential equations

Every equation mentioned seems to be of the Euler-Cauchy type, usually solved by assuming a solution of the form \( y = x^m \). This assumption allows us to reduce the equation to an algebraic equation by substituting this form into the differential equation.
02

Solve equation (a)

For the equation \(x^{2} y^{\prime \prime \prime}+2 x y^{\prime \prime}-2 y^{\prime}=0\), assume \(y = x^m\), and compute necessary derivatives: - \(y' = mx^{m-1}\)- \(y'' = m(m-1)x^{m-2}\)- \(y''' = m(m-1)(m-2)x^{m-3}\)Substitute these into the differential equation and simplify: \[ x^2[m(m-1)(m-2)x^{m-3}] + 2x[m(m-1)x^{m-2}] - 2[mx^{m-1}] = 0 \] Which simplifies to: \[ m(m-1)(m-2) + 2m(m-1) - 2m = 0 \] Solving this polynomial gives roots \(m = 0, 1, -1\). Thus, three linearly independent solutions are \(y_1 = x^0 = 1\), \(y_2 = x^1 = x\), \(y_3 = x^{-1} = \frac{1}{x}\).
03

Solve equation (b)

For the equation \(x^{3} y^{\prime \prime \prime}-3 x y^{\prime}+3 y=0\), assume \(y = x^m\) similarly:Substitute as done in step 2:\[ x^3[m(m-1)(m-2)x^{m-3}] - 3x[mx^{m-1}] + 3x^m = 0 \]Simplify to:\[ m(m-1)(m-2) - 3m + 3 = 0 \]Solve for \(m\): the characteristic equation yields \(m = 3, -1, 1\). Therefore, the linearly independent solutions are \(y_1 = x^3\), \(y_2 = \frac{1}{x}\), \(y_3 = x\).
04

Solve equation (c)

For the equation \(x^{3} y^{\prime \prime \prime}+2 x^{2} y^{\prime \prime}-3 x y^{\prime}+3 y=0\), again assume \(y = x^m\):Substitute as before:\[ x^3[m(m-1)(m-2)x^{m-3}] + 2x^2[m(m-1)x^{m-2}] - 3x[mx^{m-1}] + 3x^m = 0 \]Simplify to:\[ m(m-1)(m-2) + 2m(m-1) - 3m + 3 = 0 \]This polynomial has roots \(m = 3, 1, -1\), providing solutions \(y_1 = x^3\), \(y_2 = x\), \(y_3 = \frac{1}{x}\).
05

Solve equation (d)

For the equation \(x^{3} y^{\prime \prime \prime}-6 x^{2} y^{\prime \prime}+7 x y^{\prime}-7 y=0\):Assume \(y = x^m\) and substitute similarly:\[ x^3[m(m-1)(m-2)x^{m-3}] - 6x^2[m(m-1)x^{m-2}] + 7x[mx^{m-1}] - 7x^m = 0 \]Simplify to:\[ m(m-1)(m-2) - 6m(m-1) + 7m - 7 = 0 \]The characteristic polynomial gives roots \(m = 1, 2, -1\), leading to solutions \(y_1 = x\), \(y_2 = x^2\), and \(y_3 = \frac{1}{x}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Euler-Cauchy Equation
The Euler-Cauchy equation is a type of linear differential equation that can appear challenging at first.
This equation is typically identified by exponents of the dependent variable being multiplied by corresponding powers of the independent variable, x.
To solve an Euler-Cauchy equation, we often assume a solution of the form: \( y = x^m \).
  • This assumption allows for the transformation of the differential equation into an algebraic equation.
  • It simplifies the overall problem-solving process by reducing the complexity of differential equations.
Once this assumption is substituted back into the original equation, we derive a characteristic polynomial.
Solving the polynomial provides values of \( m \), which give us the linearly independent solutions of the equation.
This method is especially useful because it turns a difficult problem into a fairly straightforward algebraic one.
Linearly Independent Solutions
Linearly independent solutions are essential in forming the general solution to a differential equation.
They ensure that the solution can capture the full behavior described by the differential equation.
For an Euler-Cauchy equation, once the characteristic polynomial is found:
  • Each distinct root of the polynomial corresponds to a different linearly independent solution.
  • The form \( x^m \) ensures that all solutions contribute uniquely to the overall general solution.
The significance of these solutions cannot be overstated since they represent the fundamental modes in which the system can behave.
With these independent solutions, one can construct any solution to the differential equation through a linear combination.
Characteristic Polynomial
The characteristic polynomial arises during the solution of differential equations like the Euler-Cauchy type.
This polynomial is derived once we substitute \( y = x^m \) and its derivatives into the equation.
By simplifying terms, the differential equation becomes an algebraic equation:
  • The characteristic polynomial is then formed from this simplification and typically looks something like: \( m(m-1)(m-2) + \, \text{other terms} = 0 \).
  • Solving this polynomial provides the exponents \( m \) for the potential solutions \( y = x^m \).
The solutions to the characteristic polynomial are the keys to unlocking the different linearly independent solutions of the equation.
Each solution shows how the function y behaves as a power of x, clearly indicating the role of exponents in the solution's variability.
Algebraic Technique
An algebraic technique involves manipulation of algebraic expressions and equations, mainly employed in this context to derive a characteristic equation from a differential equation.
By treating solutions \( y = x^m \), we translate a complicated differential form into an algebraic one.
This simplification is achieved as follows:
  • Pay attention to differentials and derivatives such as \( y' = mx^{m-1} \), \( y'' = m(m-1)x^{m-2} \), and so on.
  • Insert these derivatives into the Euler-Cauchy equation to convert it to polynomial form.
  • Simplify to isolate powers of m.
The elegance of this technique lies in converting the more abstract notion of change (via calculus) into a more intuitive algebraic structure.
Thus, it's a powerful approach for solving complex differential equations efficiently by leveraging familiar algebraic processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine two linearly independent solutions in power series near the origin for each of the following equations: (a) \(x^{2} y^{\prime \prime}+x^{2} y^{\prime}-2 y=0\) (b) \(x y^{\prime \prime}-(1+x) y^{\prime}+2(1-x) y=0\)

Consider the Hermite equation $$ y^{\prime \prime}-2 x y^{\prime}+\gamma y=0 $$ Determine two linearly independent solutions of this equation in power series near the origin and show that one of them terminates if \(\gamma=2 n\), where \(n=0,1\), \(2, \cdots\)

Determine two linearly independent solutions in power series near the origin of the Tschebycheff equation $$ \left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}+\gamma y=0 $$ and show that one of them terminates if \(\gamma=n^{2}, n=0,1,2, \cdots\).

Determine two linearly independent solutions in power series near the origin for each of the following equations: (a) \(4 x y^{\prime \prime}+2 y^{\prime}-y=0\) (b) \(\left(2 x+x^{2}\right) y^{\prime \prime}+y^{\prime}-6 x y=0\) (c) \(9 x(1-x) y^{\prime \prime}-12 y^{\prime}+4 y=0\) (d) \(2 x(1-x) y^{\prime \prime}+(1-x) y^{\prime}+3 y=0\)

Determine two linearly independent solutions of each of the following equations: (a) \(x y^{\prime \prime}+y^{\prime}=0\) (b) \(x^{2} y^{\prime \prime}-y=0\) (c) \(x^{2} y^{\prime \prime}+x y^{\prime}-y=0\) (d) \(x^{2} y^{\prime \prime}+2 x y^{\prime}-4 y=0\) (e) \(x^{2} y^{\prime \prime}-x y^{\prime}+y=0\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.