Chapter 5: Problem 5
Solve the following differential equations: \((3-x) \frac{d^{2} y}{d x^{2}}-(9-4 x) \frac{d y}{d x}+(6-3 x) y=0\)
Short Answer
Expert verified
The general solution is \(y(x) = C_1 x^{m_1} + C_2 x^{m_2}\).
Step by step solution
01
Recognize the Structure
The given differential equation is a linear differential equation with constant coefficients, suggestive of Cauchy-Euler or Equidimensional equations, since it involves terms where the power of each derivative is matched by a power of the multiplicative variable.
02
Transform to Homogeneous Linear Form
Identify the form of the equation as \[(3-x)\frac{d^{2} y}{d x^{2}} - (9-4x)\frac{d y}{d x} + (6-3x)y = 0\]and start by assuming a solution of the form \(y = x^m\). Then compute the first and second derivatives, \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\). Plug these into the original equation.
03
Substitute and Simplify Equation
Replace \(y\), \(y'\), and \(y''\) in the differential equation with their expressions in terms of \(x^m\). Simplify to find the characteristic equation by matching coefficients of like powers (assuming \(x eq 0\)). This leads to a polynomial in terms of \(m\).
04
Solve the Characteristic Equation
The characteristic equation obtained is a polynomial in \(m\). Solve this polynomial to find the values of \(m\), which gives the exponents in the general solution. For instance, solving a second-degree polynomial \(a m^2 + b m + c = 0\) yields roots \(m_1\) and \(m_2\).
05
Construct the General Solution
Using the roots \(m_1\) and \(m_2\) obtained from the characteristic equation, the general solution of the differential equation is constructed as \(y(x) = C_1 x^{m_1} + C_2 x^{m_2}\), where \(C_1\) and \(C_2\) are arbitrary constants determined by initial conditions or boundaries.
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Linear Equations
A homogeneous linear equation is a type of differential equation where all terms are accounted for in such a way that they add up to zero. In other words, there are no constant or non-zero terms on the right side of the equation. When you see a homogeneous linear equation, it often appears with variable coefficients that relate to the derivatives of a function. In the case of the given problem, we have:
- Each term of the equation \[(3-x)\frac{d^{2} y}{d x^{2}} - (9-4x)\frac{d y}{d x} + (6-3x)y = 0\]ends with "= 0", showcasing that it is homogeneous.
- No term in the equation is independent, meaning every part of the equation involves the function or its derivatives.
Characteristic Equation
The characteristic equation plays a pivotal role when solving differential equations like the Cauchy-Euler type. It arises when substituting a trial solution into a differential equation and simplifying the terms. In our exercise, after inserting the proposed solution \(y = x^m\) along with its derivatives into the equation, we focus on matching powers of \(x\) to build a characteristic equation in terms of \(m\).
- This typically results in a polynomial equation: \[a m^2 + b m + c = 0\],which is derived by equating the coefficients of the powers of \(x\) to zero.
- Solving this polynomial gives us specific values for \(m\), representing the exponents of the solution.
General Solution
The general solution of a differential equation brings together the insights obtained from the characteristic equation. Once the characteristic equation is solved, and roots \(m_1\) and \(m_2\) are determined, these roots are used to express the solution. For the Cauchy-Euler equation, this solution takes the form:
- \(y(x) = C_1 x^{m_1} + C_2 x^{m_2}\),where the constants \(C_1\) and \(C_2\) can be arbitrary.
- If the roots are complex, the solution may also include trigonometric functions, and if the roots are repeated, a logarithmic term might appear.
- These constants are often determined by additional conditions, such as initial or boundary conditions provided in the problem.