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Chapter 4: Problem 11

Solve, if possible, the given system of differential equations by either systematic elimination or determinants.\(\left(D^{2}-1\right) x-y=0\) \((D-1) x+D y=0\)

Short Answer

Expert verified
Solutions for the system are \(x(t) = C_1 e^{-t} + C_2 e^{\frac{1+\sqrt{5}}{2}t} + C_3 e^{\frac{1-\sqrt{5}}{2}t}\) and corresponding \(y(t)\) expressions.

Step by step solution

01

Understand the System

We have a system of two linear differential equations:1. \((D^{2} - 1)x - y = 0\)2. \((D - 1)x + Dy = 0\)where \(D\) denotes differentiation with respect to \(t\). Our goal is to solve for \(x\) and \(y\).
02

Solve for y in the First Equation

Rearrange the first equation to solve for \(y\):\(y = (D^{2} - 1)x\).
03

Substitute y into the Second Equation

Substitute \(y = (D^{2} - 1)x\) into the second equation:\((D - 1)x + D((D^{2} - 1)x) = 0\).
04

Simplify the Substituted Equation

Simplify the equation from the previous step:\((D - 1)x + (D^{3} - D)x = 0\).Combine terms:\(D^{3}x - 2Dx - x = 0\).This simplifies to:\((D^{3} - 2D - 1)x = 0\).
05

Solve the Simplified Differential Equation

Solve the differential equation \((D^{3} - 2D - 1)x = 0\) by finding the characteristic equation:\(m^{3} - 2m - 1 = 0\).Solve for roots. Factors: \((m + 1)(m^{2} - m - 1) = 0\).Roots: \(m = -1\) and roots of \(m^{2} - m - 1 = 0\) are \(m = \frac{1 \pm \sqrt{5}}{2}\).
06

Write General Solution for x

The general solution for \(x\) involves the roots found:\[ x(t) = C_1 e^{-t} + C_2 e^{\frac{1+\sqrt{5}}{2}t} + C_3 e^{\frac{1-\sqrt{5}}{2}t} \]where \(C_1, C_2, C_3\) are constants.
07

Solve for y Using the General Solution for x

Since \(y = (D^2 - 1)x\), apply this to \(x(t)\):- Find the second derivative of \(x(t)\), substituting back to get \( y(t) \). This yields:\[ y(t) = C_1 (-e^{-t}) + C_2 ((\frac{1+\sqrt{5}}{2})^{2} e^{\frac{1+\sqrt{5}}{2}t} - e^{\frac{1+\sqrt{5}}{2}t}) + C_3 ((\frac{1-\sqrt{5}}{2})^{2} e^{\frac{1-\sqrt{5}}{2}t} - e^{\frac{1-\sqrt{5}}{2}t}) \]

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Key Concepts

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

Systematic Elimination
Systematic elimination is a method used to solve a system of differential equations by eliminating one of the variables to simplify the system into a single differential equation. This approach is particularly handy when dealing with multiple equations that comprise a system. The basic steps involve:
  • Identifying the given equations and the variable to eliminate.
  • Rearranging one of the equations to express one variable in terms of the other.
  • Substituting this expression into the other equation(s) to eliminate the chosen variable.
  • Simplifying the resulting equation into a single differential equation to solve for the remaining variable.
In the problem above, we eliminated the variable \( y \) by first expressing it in terms of \( x \) and its derivatives from the first equation, and then substituting this expression into the second equation. This leads to a more straightforward equation that can be managed more easily through integration techniques.
Determinants
Determinants often make solving systems of differential equations more manageable, especially when using matrices to represent the system. Here's a simplified look at determinants:
  • A determinant is a special number calculated from a square matrix.
  • It is instrumental in determining if a system of equations has a unique solution.
  • If the determinant of the matrix is non-zero, the system is invertible, indicating a unique solution exists.
  • If the determinant is zero, the system could be either consistent with infinitely many solutions or inconsistent with no solution.
In the context of differential equations, determinants help in applying systematic elimination or integration. Although in this particular example, systematic elimination was more directly used, understanding determinants can crucially support solving more complex systems.
Characteristic Equation
The characteristic equation emerges from converting a differential equation into an algebraic form. It is a critical tool when solving linear differential equations with constant coefficients. Here's how it works:
  • Replace derivatives in the differential equation with powers of \( m \), assuming a solution of the form \( e^{mt} \).
  • The differential equation transforms into a polynomial equation in terms of \( m \).
  • The roots of this polynomial are vital as they determine the behavior of the solution.
For instance, in the solution process, the transformed equation \( (D^3 - 2D - 1)x = 0 \) becomes the characteristic equation \( m^3 - 2m - 1 = 0 \). Solving this gave us the roots \( m = -1 \) and \( m = \frac{1\pm \sqrt{5}}{2} \), from which the general solution is constructed, dictating the system's dynamics.
General Solution
The general solution of a differential equation system incorporates all possible solutions of the equation. It accounts for arbitrary constants, reflecting the system's initial conditions. The process to draft a general solution involves:
  • Using the roots obtained from the characteristic equation.
  • Forming a linear combination of exponential solutions corresponding to each root.
  • Including arbitrary constants for every independent solution derived.
In this exercise, solving the differential equation using the roots \( m = -1 \) and \( m = \frac{1 \pm \sqrt{5}}{2} \) leads to the general solution for \( x(t) \): \[ x(t) = C_1 e^{-t} + C_2 e^{\frac{1+\sqrt{5}}{2}t} + C_3 e^{\frac{1-\sqrt{5}}{2}t} \]. From this, \( y(t) \) can then be constructed based on the relationship to \( x(t) \) identified from the original set of equations

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Most popular questions from this chapter

Find an interval around \(x=0\) for which the given initial-value problem has a unique solution. $$ (x-2) y^{\prime \prime}+3 y=x, \quad y(0)=0, y^{\prime}(0)=1 $$

In Problems 1-36 find the general solution of the given differential equation. \(4 y^{\prime \prime}+y^{\prime}=0\)

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In Problems 1-36 find the general solution of the given differential equation. \(3 y^{\prime \prime}+y=0\)

Solve each differential equation by variation of parameters. State an interval on which the general solution is defined. $$y^{\prime \prime}-y=\sinh 2 x$$
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