Problem 13 - Advanced Engineering Mathematics

Question

In Problems \(1-20\), solve the given system of differential equations by systematic elimination. $$ \begin{aligned} &2 \frac{d x}{d t}-5 x+\frac{d y}{d t}=e^{t} \\ &\frac{d x}{d t}-x+\frac{d y}{d t}=5 e^{t} \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( x(t) = \frac{4}{3} e^t + C e^{4t} \) and \( y(t) = 5 e^t - \frac{3C}{4} e^{4t} + D \).

Step 1: Write down the system of equations

The given system of differential equations is:1. \( 2 \frac{dx}{dt} - 5x + \frac{dy}{dt} = e^t \)2. \( \frac{dx}{dt} - x + \frac{dy}{dt} = 5 e^t \).

Step 2: Eliminate \( \frac{dy}{dt} \) from the equations

Subtract the second equation from the first equation to eliminate \( \frac{dy}{dt} \):\[ (2 \frac{dx}{dt} - \frac{dx}{dt}) - (5x - x) = e^t - 5e^t \]This simplifies to:\[ \frac{dx}{dt} - 4x = -4e^t \].

Step 3: Solve for \( x(t) \)

The differential equation \( \frac{dx}{dt} - 4x = -4e^t \) can be solved using an integrating factor. The integrating factor is \( e^{-4t} \).Multiply through by this integrating factor:\[ e^{-4t} \frac{dx}{dt} - 4 e^{-4t} x = -4 e^{-3t} \]. The left-hand side becomes the derivative of \( e^{-4t}x \):\[ \frac{d}{dt}(e^{-4t} x) = -4 e^{-3t} \].Integrate both sides:\[ e^{-4t} x = \int -4 e^{-3t} dt \]\[ e^{-4t} x = \frac{-4}{-3} e^{-3t} + C \]\[ e^{-4t} x = \frac{4}{3} e^{-3t} + C \]. Therefore,\( x(t) = \frac{4}{3} e^t + C e^{4t} \).

Step 4: Substitute \( x(t) \) into one of the original equations

Substitute \( x(t) = \frac{4}{3} e^t + C e^{4t} \) into the second equation of the original system:\( \frac{dx}{dt} - x + \frac{dy}{dt} = 5 e^t \).First, calculate \( \frac{dx}{dt} \):\( \frac{dx}{dt} = \frac{4}{3} e^t + 4C e^{4t} \).Substitute these expressions into the equation:\[ \left( \frac{4}{3} e^t + 4C e^{4t} \right) - \left( \frac{4}{3} e^t + C e^{4t} \right) + \frac{dy}{dt} = 5 e^t \].This simplifies to:\[ 3C e^{4t} + \frac{dy}{dt} = 5 e^t \].

Step 5: Solve for \( y(t) \)

Rearrange the equation from the previous step to solve for \( \frac{dy}{dt} \):\( \frac{dy}{dt} = 5e^t - 3Ce^{4t} \).Integrate both sides to find \( y(t) \):\[ y(t) = \int (5e^t - 3Ce^{4t}) dt \]\[ y(t) = 5e^t - \frac{3C}{4} e^{4t} + D \].

Step 6: Write the final solution

The solution to the system of differential equations is:\[ x(t) = \frac{4}{3} e^t + C e^{4t} \]\[ y(t) = 5 e^t - \frac{3C}{4} e^{4t} + D \], where \( C \) and \( D \) are constants determined by initial conditions.

Key Concepts

Integrating Factor Method

The integrating factor method is a nifty trick for solving linear first-order differential equations. It relates to the equation form \( \frac{dy}{dt} + p(t)y = g(t) \). To make it easier to solve, we use an integrating factor. This is usually given by \( e^{\int p(t)} \), which transforms the left-hand side of the differential equation into the derivative of a product.

The goal is to make the equation exact, which means we can integrate both sides easily. Multiplying through by the integrating factor enables us to reorganize the equation where:

• the left-hand side becomes a single derivative, allowing us to perform straightforward integrations.
• ensures both sides of the differential equation are easier to manage.

Integrating both sides directly gives the solution up to a constant. In the problem we're tackling, the integrating factor used is \( e^{-4t} \). This method simplifies the otherwise complex calculations.

Systematic Elimination

Systematic elimination is a key tool for solving systems of equations. The main idea is to reduce the system to a single equation in one variable. This often involves adding, subtracting, or manipulating the original equations.

In our original problem, systematic elimination was used by subtracting the two given equations. This approach is effective because:

• It removes one of the variables, allowing focus on solving for the remaining one.
• The new equation is often simpler to solve.

This method is quite similar to elimination methods used in algebra, ensuring that one solution paves the way for finding the other.

First-Order Differential Equations

First-order differential equations involve derivatives of a function. These are among the simplest forms of differential equations because they only involve the first derivative and no higher ones.

In the given exercise, both equations are linear first-order because:

• Each incorporates up to the first derivative \( \frac{dx}{dt} \) or \( \frac{dy}{dt} \).
• They are linear combinations of \( x \) and \( y \), their derivatives, and functions of \( t \).

These equations form the foundation for more complex differential systems and are commonly used to describe basic rates of change in various fields.

Constant of Integration

The constant of integration completes the solution to a differential equation and arises when integrating. It's important because it accounts for all possible initial conditions, reflecting the fact that solving a differential gives a family of solutions.

When you integrate, you must add a constant:

• This constant allows the solution to be adjusted for different specific conditions.
• It ensures the equation's generality, making it applicable to various scenarios.

In our problem, the constants \( C \) and \( D \) appear in solutions for \( x(t) \) and \( y(t) \). Once initial conditions are known, these constants are crucial in determining the precise, unique solution for the problem.