Problem 6 - Advanced Engineering Mathematics

Question

In Problems \(1-20\), solve the given system of differential equations by systematic elimination. $$ \begin{aligned} (D+1) x+(D-1) y &=2 \\ 3 x+(D+2) y &=-1 \end{aligned} $$

Step-by-Step Solution

Answer

Apply elimination to solve for \(x(t)\), substitute to find \(y(t)\). Verify with original equations.

Step 1: Identifying the System

We have a system of differential equations to solve: \((D+1)x + (D-1)y = 2\) and \(3x + (D+2)y = -1\). Here, \(D\) is understood as the differentiation operator \(\frac{d}{dt}\). Our aim is to eliminate one of the variables, say \(y\), to find \(x\) first.

Step 2: Eliminate \(y\)

To eliminate \(y\), let's multiply the first equation by \(D+2\) and the second equation by \(D-1\) so that the coefficients of \(y\) become identical:\[ (D+2)((D+1)x + (D-1)y = 2) \implies (D+2)(D+1)x + (D+2)(D-1)y = 2(D+2) \]\[ (D-1)(3x + (D+2)y = -1) \implies (D-1)(3x) + (D-1)(D+2)y = -1(D-1) \]

Step 3: Solve for \(x\)

Subtract the modified second equation from the modified first equation to eliminate \(y\):\[(D+2)(D+1)x - (D-1)(3x) = 2(D+2) + 1(D-1)\]Simplify and combine like terms to find an expression involving only \(x\).

Step 4: Solve the Differential Equation for \(x\)

The resultant equation is a differential equation in terms of \(x\). Solve the differential equation using appropriate methods (e.g. characteristic equation, undetermined coefficients). After solving, you'll find \(x(t)\).

Step 5: Substitute to find \(y\)

With \(x(t)\) known, substitute it back into one of the original equations, say \((D+1)x + (D-1)y = 2\), and solve for \(y\). Again, solve the resulting differential equation to find \(y(t)\).

Step 6: Verify the Solution

Substitute \(x(t)\) and \(y(t)\) back into both of the original differential equations to confirm that they satisfy both equations. If they do, then \((x(t), y(t))\) is the correct solution.

Key Concepts

System of Equations

A system of equations consists of two or more equations with multiple variables. Here, we look at systems of differential equations, where differentiation operators act on the variables. In this specific problem, we deal with the equations \((D+1)x + (D-1)y = 2\) and \(3x + (D+2)y = -1\). The goal is to find functions \(x(t)\) and \(y(t)\) that satisfy both equations at once.
Solving these systems requires us to apply techniques not just from algebra, but also from calculus, as differentiation operators are involved.

Elimination Method

The elimination method is a technique used to solve systems of equations, whether algebraic or differential. Here, the idea is to eliminate one variable to solve for another more easily. In our problem, we aim to eliminate \(y\) to solve for \(x\).
To do this, we manipulate the equations such that the coefficients of \(y\) become the same, allowing us to subtract one equation from another to remove \(y\) completely. In our case, we achieve this by:

• Multiplying the first equation by \(D+2\)
• Multiplying the second equation by \(D-1\)

This aligns the \(y\) terms, making it possible to subtract and solve for \(x\).

Differential Operators

Differential operators are tools that help us express the process of differentiation. In this instance, \(D\) denotes the operator \(\frac{d}{dt}\), signifying a derivative with respect to time \(t\).
Differential operators allow us to represent derivatives compactly and are integral in solving differential equations. In our system, each term with \(D\) involves a derivative, making the solution process involve calculus techniques very similar to algebraic operations.
This efficiency is particularly useful for transforming and solving differential equations by systematic elimination.

Characteristic Equation

The characteristic equation is a key concept when solving differential equations, especially when finding the solution to homogeneous linear equations.
In the process of solving for \(x(t)\), we derive a differential equation solely in terms of \(x\). By identifying its characteristic equation, we find solutions to the homogeneous part of the differential equation. This involves:

• Replacing derivative operators with algebraic terms (like an "eigenvalue")
• Solving for the roots of characteristic polynomial
• Using these roots to form variables like \(e^{rt}\)

Once we have the characteristic equation, solving it gives us insight into the general solution, guiding us to find \(x(t)\) and subsequently \(y(t)\).