Problem 19 - Advanced Engineering Mathematics

Question

Verify that the indicated expression is an implicit solution of the given first-order differential equation. Find at least one explicit solution \(y=\phi(x)\) in each case. Use a graphing utility to obtain the graph of an explicit solution. Give an interval \(I\) of definition of each solution \(\phi\). $$ \frac{d X}{d t}=(X-1)(1-2 X) ; \quad \ln \left(\frac{2 X-1}{X-1}\right)=t $$

Step-by-Step Solution

Answer

The explicit solution is \( X = \frac{e^t - 1}{e^t - 2}\) with an interval \((-\infty, \ln 2) \cup (\ln 2, \infty)\).

Step 1: Differentiate the Implicit Solution

Start by differentiating the given implicit solution with respect to the independent variable, which is typically time \(t\). The given implicit equation is \(\ln\left(\frac{2X-1}{X-1}\right) = t\). Differentiate both sides with respect to \(t\) to obtain \(\frac{d}{dt}\left(\ln\left(\frac{2X-1}{X-1}\right)\right) = \frac{d}{dt}(t)\). This yields \(\frac{1}{\frac{2X-1}{X-1}} \cdot \frac{d}{dt}\left(\frac{2X-1}{X-1}\right) = 1\).

Step 2: Apply the Chain Rule

Differentiate the internal function \(\frac{2X-1}{X-1}\) using the quotient rule: \(\frac{d}{dt}\left(\frac{2X-1}{X-1}\right) = \frac{(X-1)(2\frac{dX}{dt}) - (2X-1)\frac{dX}{dt}}{(X-1)^2}\). This simplifies to: \[ \frac{(X-1)(2) - (2X-1)}{(X-1)^2} \cdot \frac{dX}{dt} = 1 \].

Step 3: Simplify the Expression

Simplify the differentiated expression: \(\frac{(2X-2 - 2X+1)}{(X-1)^2} = \frac{-1}{(X-1)^2}\). Hence, the equation becomes \(-\frac{1}{(X-1)^2}\frac{dX}{dt} = 1\), leading to \(\frac{dX}{dt} = -(X-1)^2\).

Step 4: Relate to the Differential Equation

Check if \(\frac{dX}{dt} = (X-1)(1-2X)\) is satisfied by transforming the equation obtained. From Step 3, rewrite \(-\frac{1}{(X-1)^2} = (1-2X)\) to verify the implicit solution by substituting into the original differential equation.

Step 5: Find Explicit Solution

Based on the implicit solution, solve for \(X\) explicitly. The equation \(\ln\left(\frac{2X-1}{X-1}\right) = t\) suggests that \(\frac{2X-1}{X-1} = e^t\). Cross-multiplying and simplifying gives \(2X - 1 = e^t(X - 1)\), leading to \[X = \frac{e^t - 1}{e^t - 2}\] as an explicit solution.

Step 6: Determine Interval of Definition

Solve for the interval \(I\) where the explicit solution \(X = \frac{e^t - 1}{e^t - 2}\) is defined. The denominator cannot be zero, thus solve \(e^t - 2 eq 0\), giving the interval \(t eq \ln 2\). Therefore, the interval of definition is \(I: (-\infty, \ln 2) \cup (\ln 2, \infty)\).

Key Concepts

Implicit Solution

An implicit solution of a differential equation is a relation involving both the dependent and independent variables where the dependent variable is not isolated. In other words, the dependent variable is not expressed explicitly in terms of the independent variable. Instead, it is contained within an equation that also includes the independent variable.

In the original exercise, we have the implicit equation \(\ln\left(\frac{2X-1}{X-1}\right) = t\). Here, the variable \(X\) cannot be easily isolated on one side of the equation. The equation implies a relationship between \(X\) and \(t\), requiring further manipulation or graphing to find a specific solution. Often, implicit solutions are useful because they hold complex relationships that can be solved or simplified to find actual solutions.

When verifying an implicit solution, it involves differentiating the implicit function and ensuring that it satisfies the differential equation provided in the problem.

Explicit Solution

An explicit solution is a form where the dependent variable is expressed solely in terms of the independent variable. This kind of solution is often preferred as it makes evaluating and analyzing the function straightforward.

In this exercise, an explicit solution is derived from the implicit solution \(\ln\left(\frac{2X-1}{X-1}\right) = t\). By solving algebraically, we found that \(X\) could be expressed as \(X = \frac{e^t - 1}{e^t - 2}\). This means that \(X\) is now isolated and fully expressed as a function of \(t\).

If you have an explicit solution, it can be easily plugged into any function to get its value, and graphing utilities can readily plot the function to get a visual representation of how \(X\) changes with \(t\).

First-Order Differential Equation

A first-order differential equation is a type of equation involving the first derivative of the function. Typically, it relates the rate of change of a variable to the variable itself or other variables.

• It is called 'first-order' because it includes only the first derivative (such as \(\frac{dX}{dt}\) in our problem).
• Example: \(\frac{dX}{dt} = (X-1)(1-2X)\) is a first-order differential equation. Here, the rate of change of \(X\) with respect to \(t\) is dependent on the value of \(X\).

First-order differential equations are crucial in modeling real-world phenomena where variables gradually change over time. They can often be solved directly or through substitution, integration, and sometimes graphically, as shown in this problem.

Interval of Definition

In mathematics, the interval of definition for a solution describes the set of values over which the function is valid or meaningful. This is important for ensuring that a solution doesn't approach an undefined state, such as division by zero.For the explicit solution \(X = \frac{e^t - 1}{e^t - 2}\), we calculated the interval of definition by setting the denominator not equal to zero, leading to \(t eq \ln 2\). As a result, \(X\) is undefined when \(e^t = 2\), hence the interval is \((-\infty, \ln 2) \cup (\ln 2, \infty)\).Understanding the interval of definition helps to avoid points or regions where the mathematical model breaks down, ensuring the solution is practical and applicable within correct limits.