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Chapter 4: Problem 20

\(x^{3} y^{(4)}+6 x^{2} y^{\prime \prime \prime}+7 x y^{\prime \prime}+y^{\prime}=0\)

Short Answer

Expert verified
The general solution is \( y(x) = C_1 x^{m_1} + C_2 x^{m_2} + C_3 x^{m_3} + C_4 x^{m_4} \).

Step by step solution

01

Identify the Given Differential Equation

The given differential equation is \( x^{3} y^{(4)} + 6 x^{2} y''' + 7 x y'' + y' = 0 \).
02

Identify the Order of the Differential Equation

Observe the highest derivative in the equation. The highest derivative is \( y^{(4)} \), indicating that this is a fourth-order differential equation.
03

Recognize the Type of Differential Equation

Notice that all the terms include either the dependent variable \( y \) or its derivatives, each multiplied by a power of the independent variable \( x \). This indicates the equation is a linear homogeneous ordinary differential equation.
04

Check for Euler-Cauchy Form

Identify if the equation fits the Euler-Cauchy form, given by the general form: \[ x^n y^{(n)} + a_{n-1} x^{n-1} y^{(n-1)} + \text{...} + a_0 y = 0 \]. The given equation fits this form with coefficients expressed as powers of \( x \).
05

Solve Using the Substitution Method

For Euler-Cauchy equations, use the substitution \( y = x^m \). Subsequently, \( y' = mx^{m-1} \), \( y'' = m(m-1)x^{m-2} \), \( y''' = m(m-1)(m-2)x^{m-3} \), \( y^{(4)} = m(m-1)(m-2)(m-3)x^{m-4} \). Substitute these into the equation.
06

Derive the Characteristic Equation

Substitute \( y = x^m \) and its derivatives into the given equation: \[ x^3 (m(m-1)(m-2)(m-3)x^{m-4}) + 6x^2 (m(m-1)(m-2)x^{m-3}) + 7x (m(m-1)x^{m-2}) + m x^{m-1} = 0 \]. This simplifies to: \[ m(m-1)(m-2)(m-3) + 6m(m-1)(m-2) + 7m(m-1) + m = 0 \].
07

Simplify and Solve the Polynomial

Factor out common terms: \[ m(m-1)(m-2)(m-3) + 6m(m-1)(m-2) + 7m(m-1) + m = 0 \] becomes: \[ m [ (m-1)(m-2)(m-3) + 6(m-1)(m-2) + 7(m-1) + 1 ] = 0 \]. Solve the polynomial for roots (values of \( m \)).
08

Find the General Solution

Based on the roots \( m_1, m_2, m_3, m_4 \), the general solution of the differential equation is: \[ y(x) = C_1 x^{m_1} + C_2 x^{m_2} + C_3 x^{m_3} + C_4 x^{m_4} \], where \( C_1, C_2, C_3, C_4 \) are arbitrary constants.

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Key Concepts

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

Euler-Cauchy Equation
Euler-Cauchy equations are a special class of linear differential equations. They have the general form: Euler-Cauchy equations are typically solved using a substitution method. This equation is particularly interesting because it connects to polynomial roots, making it a gateway to deeper mathematical concepts.
Homogeneous Differential Equation
A homogeneous differential equation is one where every term is a function of the dependent variable and its derivatives. In simpler terms, if all terms can be grouped to zero without including any external functions or constants, it is homogeneous. For the given differential equation: The Euler-Cauchy equation discussed above is homogeneous because all of its terms involve either the function y or its derivatives, multiplied by powers of the independent variable x. This type of equation is crucial in mathematical modeling, as it often appears in physical systems without external influences.
Characteristic Equation
The characteristic equation helps us to find particular solutions to differential equations. For Euler-Cauchy equations, we use the substitution method, where we assume a solution of the form y = x^m and then derive the characteristic (or auxiliary) equation. For the given fourth-order differential equation: After simplifying, this transforms into a polynomial equation in m, known as the characteristic equation: Solving this polynomial will give us the roots, which are crucial for finding the general solution. These roots can be real, complex, or repeated, each affecting the form of the solution differently.
General Solution
The general solution to a differential equation includes all possible specific solutions. Once we have found the roots of the characteristic equation (m_1, m_2, m_3, m_4), we can express the general solution as: This means we sum up all the particular solutions, each multiplied by an arbitrary constant. For our fourth-order Euler-Cauchy equation, it looks like this: The constants (C_1, C_2, C_3, C_4) are determined based on initial conditions or boundary values provided in a specific problem scenario. By understanding and applying these concepts, you're well-equipped to tackle not only fourth-order differential equations but a wide range of mathematical problems!

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Most popular questions from this chapter

\(x^{2} y^{\prime \prime}+6 y=0\)

(Abel's Formula for Higher Order Equations) Consider the \(n\)th order linear homogeneous equation \(y^{(n)}+p_{n-1}(t) y^{(n-1)}+\cdots+\) \(p_{0}(t) y=0\) with \(n\) linearly independent solutions \(y_{1}, y_{2}, \ldots, y_{n}\) on an interval \(I\). We can show that the Wronskian of these \(n\) solutions satisfies the same identity as that presented in Exercises 4.1. We do this for the third order differential equation, \(y^{\prime \prime \prime}+\) \(p_{2}(t) y^{\prime \prime}+p_{1}(t) y^{\prime}+p_{0}(t) y=0\), with linearly independent solutions \(y_{1}, y_{2}\), and \(y_{3}\) on an interval \(I\). (a) Show that $$ \frac{d}{d t}\left(W\left(\left\\{y_{1}, y_{2}, y_{3}\right\\}\right)\right)=\left|\begin{array}{ccc} y_{1} & y_{2} & y_{3} \\ y_{1}^{\prime} & y_{2}^{\prime} & y_{3}^{\prime} \\ y_{1}^{\prime \prime \prime} & y_{2}^{\prime \prime \prime} & y_{3}^{\prime \prime \prime} \end{array}\right| . $$ (b) Use the differential equation to solve for \(y_{1}^{\prime \prime \prime}, y_{2}^{\prime \prime \prime}\), and \(y_{3}^{\prime \prime \prime}\). Substitute these values to obtain \(d W / d t+p_{2}(t) W=0\). (c) Solve this differential equation to find that \(W\left(\left\\{y_{1}, y_{2}, y_{3}\right\\}\right)=C e^{-\int p_{2}(t) d t}\). (d) Indicate how to generalize this result to the \(n\)th order linear homogeneous equation.

Show that \(y=J_{0}(k x)\) where \(k\) is a constant is a solution of the parametric Bessel equation of order zero, \(x y^{\prime \prime}+y^{\prime}+k^{2} x y=0\).

Use the change of variables \(y=v(x) x^{-1 / 2}\) to transform Bessel's equation \(x^{2} y^{\prime \prime}+x y^{\prime \prime}+\) \(\left(x^{2}-k^{2}\right) y=0\) into the equation \(v^{\prime \prime}+\) \(\left[1+\left(\frac{1}{4}-k^{2}\right) x^{-2}\right] v=0\). By substituting \(k=1 / 2\) into the transformed equation, derive the solution to Bessel's equation with \(k=1 / 2\).

Chebyshev's equation is given by $$ \left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}+k^{2} y=0, \quad k \geq 0 . $$ Using a power series expansion about the ordinary point \(x=0\), obtain a general solution of this equation for (a) \(k=1\) and (b) \(k=\) 3. Show that if \(k\) is a nonnegative integer, then one of the solutions is a polynomial of degree \(k\).
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