Problem 19 - Advanced Engineering Mathematics

Question

In Problems 19 and 20, 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 the 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 \( X = \frac{e^t - 1}{2 - e^t} \), valid for \( t \in (-\infty, \ln(2)) \cup (\ln(2), \infty) \).

Step 1: Verify Implicit Solution

To verify that the given expression is an implicit solution, let's differentiate both sides of \[ \ln \left(\frac{2X-1}{X-1}\right) = t \]with respect to \( t \). First, differentiate the left side using the chain rule:\[ \frac{d}{dt} \left(\ln \left(\frac{2X-1}{X-1}\right)\right) = \frac{1}{\frac{2X-1}{X-1}} \cdot \frac{d}{dt}\left(\frac{2X-1}{X-1}\right) = \frac{1}{\frac{2X-1}{X-1}} \cdot \frac{(2)(X-1)-(2X-1)(1)}{(X-1)^2} \cdot \frac{dX}{dt} \]Simplifying the numerators and factoring out \( \frac{dX}{dt} \):\[ = \frac{1}{\frac{2X-1}{X-1}} \cdot \frac{-X+1}{(X-1)^2} \cdot \frac{dX}{dt} = \frac{(X-1)^2}{2X-1} \cdot \frac{-X+1}{(X-1)^2} \cdot \frac{dX}{dt} \]\[ = \frac{1-2X}{2X-1} \cdot \frac{dX}{dt} \]Since the derivative of \( t \) with respect to \( t \) is 1, we equate and solve for \( \frac{dX}{dt} \):\[ \frac{1-2X}{2X-1} \cdot \frac{dX}{dt} = 1 \quad \Rightarrow \quad \frac{dX}{dt} = (X-1)(1-2X) \]Thus, verifying that the expression given satisfies the differential equation \( \frac{d X}{d t}=(X-1)(1-2 X) \).

Step 2: Find Explicit Solution

The provided implicit equation is \[ \ln \left(\frac{2X-1}{X-1}\right) = t \].Solving for \( X \) in terms of \( t \), we exponentiate both sides:\[ \frac{2X-1}{X-1} = e^t \]Cross-multiply to obtain:\[ 2X - 1 = e^t (X - 1) \]This simplifies to:\[ 2X - 1 = e^t X - e^t \]\[ 2X - e^t X = e^t - 1 \]Factor \( X \) out:\[ X (2 - e^t) = e^t - 1 \]Solve for \( X \):\[ X = \frac{e^t - 1}{2 - e^t} \]

Step 3: Determine Definition Interval

To find the interval \( I \) where the explicit solution \( X = \phi(t) = \frac{e^t - 1}{2 - e^t} \) is valid, consider where the denominator is not zero:\[ 2 - e^t eq 0 \quad \Rightarrow \quad e^t eq 2 \]Thus the interval for \( t \) is \[ t \in \mathbb{R} \setminus \{\ln(2)\} = (-\infty, \ln(2)) \cup (\ln(2), \infty) \]

Step 4: Graph the Explicit Solution

Using a graphing utility like Desmos or a graphing calculator, plot the explicit solution function \( X = \frac{e^t - 1}{2 - e^t} \). Observe the behavior near the undefined point at \( t = \ln(2) \). The graph will have a vertical asymptote at this point, showing that \( X \) approaches infinity or negative infinity depending on the side of the asymptote.

Key Concepts

Implicit Solutions

Implicit solutions in differential equations are solutions that are not expressed as a direct formula or function for the dependent variable. Instead, they represent a relationship between the variables that satisfies the equation. In our problem, the implicit solution given is:

• \[ \ln \left(\frac{2X-1}{X-1}\right) = t \]

This equation is not solved for \( X \) explicitly. It demonstrates a connection between \( X \) and \( t \) through the equation containing a logarithm.
Implicit solutions are useful because they can be simpler or offer insights into the overall behavior of the solution even when a specific formula isn't immediately apparent. Sometimes solving for an explicit solution is complex or impossible, and the implicit form is the most practical representation.

Explicit Solutions

An explicit solution is a solution that specifically delineates the dependent variable as a function of the independent variable. In simpler terms, it shows the dependent variable directly expressed in terms of the independent variable. From our problem, we derived the explicit solution for \( X \) in terms of \( t \):

• \[ X = \frac{e^t - 1}{2 - e^t} \]

This expression lets us compute \( X \) directly once we know \( t \).
Explicit solutions are often preferred because they are straightforward and allow for easier computation and analysis. However, not every differential equation can be solved explicitly, and sometimes they only lend themselves to implicit forms.

Definition Intervals

The definition interval for a solution indicates the range of values over which the solution is valid. For explicit solutions, this means ensuring the function doesn’t result in any undefined expressions such as division by zero. In this exercise, the explicit solution:

• \[ X = \frac{e^t - 1}{2 - e^t} \]

requires that \( 2 - e^t eq 0 \).
This condition ensures \( X \) remains valid and avoids division by zero, leading to an interval of validity:

• \( t \in \mathbb{R} \setminus \{ \ln(2) \} = (-\infty, \ln(2)) \cup (\ln(2), \infty) \)

Knowing the definition interval is key when solving real-world problems, as it informs the domains over which the solutions are applicable.

Graphical Solutions

Graphical solutions provide a visual representation of the behavior of solutions over their domain. In the context of the differential equation problem we're discussing, plotting the explicit solution:

• \[ X = \frac{e^t - 1}{2 - e^t} \]

helps to visualize how the function behaves as \( t \) changes.
Using a graphing calculator or software, you can observe the graph. Notably, you'll see a vertical asymptote at \( t = \ln(2) \), illustrating where the function becomes undefined.
Graphical solutions can be invaluable for understanding the characteristics of the solution, such as growth, decay, periodicity, or singularities, which might not be immediately apparent through algebraic expressions alone.