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Chapter 4: Problem 3

$$z^{\prime \prime \prime}-6 z^{\prime}+10 z=0$$

Short Answer

Expert verified
After solving the characteristic equation, place the roots into the general solution \(z = Ae^{m_1x} + Be^{m_2x} + Ce^{m_3x}\). This is the solution for the differential equation.

Step by step solution

01

Formulate the Characteristic Polynomial

To formulate the characteristic (auxiliary) polynomial, replace every \(z^{\prime \prime \prime}\) by \(m^3\), every \(z^{\prime}\) by \(m\) and every \(z\) by one constant in the proposed equation. This results in the quadratic equation: \(m^3 - 6m + 10 = 0\).
02

Solve the Characteristic Polynomial

To solve this cubic equation, use either the standard methods or a solver tool. The roots of this equation will provide the solutions for the differential equation. It might be complex as well, given the cubic nature of the polynomial.
03

Write down the General Solution

The general solution of a homogeneous differential equation with constant coefficients is a linear combination of terms of the form \(y = e^{mx}\), where m is a root of the characteristic polynomial. The solution will be in the form of \(z = Ae^{m_1x} + Be^{m_2x} + Ce^{m_3x}\), where \(A\), \(B\) and \(C\) are arbitrary constants and \(m_1\), \(m_2\), \(m_3\) are the roots of the characteristic equation.

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Key Concepts

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

Characteristic Polynomial
In the context of differential equations, the characteristic polynomial is an essential tool. It helps transform a differential equation into a form that is easier to solve. Here, the given differential equation is \(z''' - 6z' + 10z = 0\). We form the characteristic polynomial by substituting derivatives with terms involving \(m\).
  • \(z^{\prime \prime \prime}\) becomes \(m^3\)
  • \(z^{\prime}\) becomes \(m\)
  • \(z\) becomes a constant (just 1)

Thus, the characteristic polynomial is \(m^3 - 6m + 10 = 0\). This polynomial is integral in determining the behavior and solutions for the differential equation. It's important because the roots of this polynomial give us insights into the system's response.
Cubic Equation
A cubic equation is a polynomial equation of degree three, such as \(m^3 - 6m + 10 = 0\) seen in our characteristic polynomial.

Solving cubic equations can be more complex than quadratic ones because they might have one or more real roots, and potentially complex roots. There are several methods to tackle these equations:
  • Factorization: If possible, factor the cubic equation. This method works nicely if the equation has obvious factors.
  • Cardano's Formula: This specific mathematical formula helps find roots of cubic equations.
  • Numerical Methods: Solutions like the Newton-Raphson method approximate roots efficiently.

Finding the roots of the cubic characteristic polynomial is crucial because these roots form the foundation for the differential equation鈥檚 general solution.
General Solution
Once we have the roots of the characteristic polynomial, we can build the general solution to the original differential equation. In the case of a third-order linear homogeneous differential equation with constant coefficients, the general solution is expressed as a linear combination of exponential terms.

The form of the solution is \(z = Ae^{m_1x} + Be^{m_2x} + Ce^{m_3x}\), where:
  • \(A\), \(B\), and \(C\) are constants determined by initial conditions.
  • \(m_1\), \(m_2\), and \(m_3\) are the roots of the characteristic polynomial, which can be real or complex.

This solution structure captures the essence of the response of the system represented by our differential equation. Each term in the solution corresponds to a mode of behavior dictated by each root, encompassing everything from exponential growth/decay to oscillatory motion.

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Most popular questions from this chapter

$$2 \omega^{\prime \prime}(x)-3 \omega(x)=4 x \sin ^{2} x+4 x \cos ^{2} x$$

$$z^{\prime \prime}-2 z^{\prime}-2 z=0 ; \quad z(0)=0, \quad z^{\prime}(0)=3$$

$$\theta^{\prime \prime}(t)-\theta(t)=t \sin t$$

$$3 y^{\prime}-7 y=0$$

Use Abel's formula (Problem 32) to determine (up to a constant multiple) the Wronskian of two solutions on \((0, \infty)\) to \(\quad t y^{\prime \prime}+(t-1) y^{\prime}+3 y=0\)
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