Problem 7 - Advanced Engineering Mathematics

Question

(a) A model is \\[\begin{array}{ll}\frac{d x_{1}}{d t}=3 \cdot \frac{x_{2}}{100-t}-2 \cdot \frac{x_{1}}{100+t}, & x_{1}(0)=100 \\ \frac{d x_{2}}{d t}=2 \cdot \frac{x_{1}}{100+t}-3 \cdot \frac{x_{2}}{100-t}, & x_{2}(0)=50 \end{array}\\] (b) since the system is closed, no salt enters or leaves the system and \(x_{1}(t)+x_{2}(t)=100+50=150\) for all time. Thus \(x_{1}=150-x_{2}\) and the second equation in part (a) becomes \\[\frac{d x_{2}}{d t}=\frac{2\left(150-x_{2}\right)}{100+t}-\frac{3 x_{2}}{100-t}=\frac{300}{100+t}-\frac{2 x_{2}}{100+t}-\frac{3 x_{2}}{100-t}\\] or \\[\frac{d x_{2}}{d t}+\left(\frac{2}{100+t}+\frac{3}{100-t}\right) x_{2}=\frac{300}{100+t}\\] Using integration by parts, we obtain \\[(100+t)^{2}(100-t)^{-3} x_{2}=300\left[\frac{1}{2}(100+t)(100-t)^{-2}-\frac{1}{2}(100-t)^{-1}+c\right]\\] Thus \\[\begin{aligned}x_{2} &=\frac{300}{(100+t)^{2}}\left[c(100-t)^{3}-\frac{1}{2}(100-t)^{2}+\frac{1}{2}(100+t)(100-t)\right] \\\ &=\frac{300}{(100+t)^{2}}\left[c(100-t)^{3}+t(100-t)\right]\end{aligned}\\] Using \(x_{2}(0)=50\) we find \(c=5 / 3000 .\) At \(t=30, x_{2}=\left(300 / 130^{2}\right)\left(70^{3} c+30 \cdot 70\right) \approx 47.4\) lbs.

Step-by-Step Solution

Answer

At \(t = 30\), \(x_2 \approx 47.4\) lbs.

Step 1: Explain the initial system of equations

We start with two differential equations that describe the dynamics of two variables \(x_1\) and \(x_2\) over time \(t\). Given the initial values \(x_1(0) = 100\) and \(x_2(0) = 50\), these equations model how \(x_1\) and \(x_2\) change due to interactions described by the given first-order linear differential equations.

Step 2: Simplify the system using conservation law

Since the system is closed with no entry or exit, the sum \(x_1(t) + x_2(t) = 150\) remains constant over time. This allows us to write \(x_1 = 150 - x_2\), simplifying the system from two variables to one.

Step 3: Derive a single equation for \(x_2\)

By substituting \(x_1 = 150 - x_2\) from the conservation law into the second differential equation, we get the expression \(\frac{d x_2}{d t} = \frac{300}{100+t} - \frac{2 x_2}{100+t} - \frac{3 x_2}{100-t}\). This further simplifies to \(\frac{d x_2}{d t} + \left(\frac{2}{100+t} + \frac{3}{100-t}\right) x_2 = \frac{300}{100+t} \).

Step 4: Solve the differential equation using integration by parts

Applying integration by parts to solve the differential equation finds an expression for \(x_2\). The solution involves integrating the terms and results in \((100+t)^2 (100-t)^{-3} x_2 = 300\left[\frac{1}{2}(100+t)(100-t)^{-2} - \frac{1}{2}(100-t)^{-1} + c\right]\).

Step 5: Express \(x_2(t)\) fully

Simplifying the result from integration, \(x_2\) is expressed as:\[x_2 = \frac{300}{(100+t)^2}\left[c(100-t)^3 + t(100-t)\right]\]. The constant \(c\) is found using the initial condition.

Step 6: Determine the constant \(c\) using initial conditions

Using the condition \(x_2(0) = 50\), we substitute this value to solve for \(c\). It results in \(c = \frac{5}{3000}\).

Step 7: Calculate \(x_2\) at a specific time

Finally, calculating \(x_2\) at \(t = 30\) using the determined values and expression:\[ x_2(30) = \frac{300}{130^2} \left(70^3 \cdot \frac{5}{3000} + 30 \cdot 70 \right) \approx 47.4\] lbs.

Key Concepts

First-order Linear Differential Equations

Understanding first-order linear differential equations is crucial as they form the backbone of many dynamic systems. They are called linear because the dependent variable (in our case, the variables \(x_1\) and \(x_2\)) and its derivative appear linearly, implying they are not raised to any power other than one nor multiplied together.
A general form of a first-order linear differential equation is \( \frac{dy}{dt} + P(t)y = Q(t) \). Here, \(P(t)\) and \(Q(t)\) are functions of \(t\) (time), while \(y\) is the function we want to solve for.
In the provided system, we observe the expressions for \( \frac{dx_1}{dt} \) and \( \frac{dx_2}{dt} \) neatly depict this structure as they describe how each variable changes over time due to the interaction described by other functions.

Initial Value Problem

An initial value problem (IVP) provides the starting point for solving differential equations. It features an equation coupled with an initial condition, which specifies the value of the function at a certain point, often when time \( t = 0 \).
In our scenario, we observe the initial conditions \( x_1(0) = 100 \) and \( x_2(0) = 50 \). These conditions ensure that any solution to the differential equations reflects not only the behavior of the system but starts from a specific real-world scenario.
Solving an IVP involves not just finding a general solution to the differential equation, but determining a specific solution that satisfies these initial conditions. This makes it easier to relate the mathematical model to the physical system it represents.

Integration by Parts

Integration by parts is a mathematical technique used to simplify the integration process, especially when dealing with products of functions. It is based on the product rule for differentiation and expresses the integral of a product of functions in terms of the integral of their derivative and antiderivative.
The formula is typically written as: \[ \int u \, dv = uv - \int v \, du \]Here, \(u\) and \(dv\) are parts of the integrand that we suitably choose to simplify the problem.
In the exercise, integration by parts is applied to solve the differential equation for \(x_2\). This method is pivotal in breaking down the intricate expressions involving \( (100+t)^2 \) and \( (100-t)^{-3} \), facilitating the step to arrive at the function describing the dynamics of \(x_2(t)\).

Closed System

A closed system in the context of differential equations signifies that there is no exchange of mass across the boundaries of the system. In our exercise, this means that the total quantity, \( x_1(t) + x_2(t) = 150 \), remains constant over time.
This conservation principle is vital because it allows us to reduce the complexity of the system by expressing one variable in terms of the other (\(x_1 = 150 - x_2\)), transforming a two-equation system into a single equation problem.
This simplification not only conserves mathematical resources but enhances comprehension of how the dynamics are governed purely by the internal interactions within the system, free from external influences. It represents the mass balance constraint that one must account for when dealing with closed systems.