Problem 1 - Advanced Engineering Mathematics

Question

(a) Solving \(N(1-0.0005 N)=0\) for \(N\) we find the equilibrium solutions \(N=0\) and \(N=2000\) When 0 From the phase portrait we see that \(\lim _{t \rightarrow \infty} N(t)=2000.\) (b) Separating variables and integrating we have \\[ \frac{d N}{N(1-0.0005 N)}=\left(\frac{1}{N}-\frac{1}{N-2000}\right) d N=d t \\] and \\[ \ln N-\ln (N-2000)=t+c \\] Solving for \(N\) we get \(N(t)=2000 e^{c+t} /\left(1+e^{c+t}\right)=2000 e^{c} e^{t} /\left(1+e^{c} e^{t}\right) .\) Using \(N(0)=1\) and solving for \(e^{c}\) we find \(e^{c}=1 / 1999\) and so \(N(t)=2000 e^{t} /\left(1999+e^{t}\right) .\) Then \(N(10)=1833.59,\) so 1834 companies are expected to adopt the new technology when \(t=10.\)

Step-by-Step Solution

Answer

Approximately 1834 companies will adopt the technology at \( t=10 \).

Step 1: Identify Equilibrium Solutions

To find equilibrium solutions, we set the growth equation \( N(1-0.0005 N)=0 \). This equation is equal to zero when either \( N=0 \) or \( 1-0.0005 N=0 \). Solving \( 1-0.0005 N=0 \) gives \( N=2000 \), so the equilibrium solutions are \( N=0 \) and \( N=2000 \).

Step 2: Analyze Phase Portrait

The phase portrait helps us determine the behavior of \( N(t) \) as \( t \to \infty \). Given the differential equation and the equilibria, the phase portrait indicates that \( \lim_{t \to \infty} N(t) = 2000 \). This means that over time, the number of companies adopting the technology approaches 2000.

Step 3: Separate Variables

To solve the differential equation, we first separate variables: \( \frac{dN}{N(1-0.0005 N)} = dt \). This separation allows us to integrate each side independently.

Step 4: Perform Partial Fraction Decomposition

We decompose \( \frac{1}{N(1-0.0005 N)} \) into partial fractions: \( \frac{1}{N(1-0.0005 N)} = \frac{1}{N} - \frac{1}{N-2000} \). This form makes the integral easier to solve.

Step 5: Integrate Both Sides

Integrate both sides: \( \int \left( \frac{1}{N} - \frac{1}{N-2000} \right) dN = \int dt \). Solving the integrals gives \( \ln N - \ln (N-2000) = t + c \), where \( c \) is the integration constant.

Step 6: Solve for \(N\)

Rearrange the equation \( \ln \frac{N}{N-2000} = t + c \) to solve for \( N \), resulting in \( N(t) = \frac{2000 e^{c+t}}{1 + e^{c+t}} \) or rearranged to \( N(t) = \frac{2000 e^{c} e^{t}}{1 + e^{c} e^{t}} \).

Step 7: Determine Constant \( e^{c} \)

To find \( e^{c} \), use the initial condition \( N(0) = 1 \), so \( \frac{2000 e^{c}}{1999} = 1 \). Solving for \( e^{c} \) gives \( e^{c} = \frac{1}{1999} \).

Step 8: Define Final Solution for \(N(t)\)

Substitute \( e^{c} = \frac{1}{1999} \) into the equation for \( N(t) \), resulting in the specific solution \( N(t) = \frac{2000 e^{t}}{1999 + e^{t}} \).

Step 9: Calculate \(N(10)\)

To find \( N(10) \), substitute \( t=10 \) into the equation: \( N(10) = \frac{2000 e^{10}}{1999 + e^{10}} \). Calculating gives \( N(10) = 1833.59 \), or approximately 1834 companies.

Key Concepts

Phase Portrait

Understanding phase portraits is essential when dealing with differential equations, as they provide a graphical representation of the system dynamics. A phase portrait displays the behavior of solutions to a differential equation for different initial conditions.

In the context of our exercise, the phase portrait reveals the equilibrium solutions where the system becomes stable. It indicates how the solution behaves as time progresses. For an equation like our growth differential equation, we can draw trajectories that show how the population of companies changes over time.

The phase portrait helps visualize where solutions converge. In this case, it shows the solutions moving towards 2000, meaning over time, the number of companies adopting the technology approaches this number. This visual tool is crucial for understanding long-term behavior predictions in differential equations.

Equilibrium Solutions

Equilibrium solutions are points where the system represented by a differential equation remains constant over time. To find these, we set the equation equal to zero. For our exercise, the equation is \( N(1-0.0005N)=0 \).

Solving this involves finding \( N \) values that make the equation zero:

• \( N = 0 \) - This is an equilibrium where no companies adopt the technology.
• \( N = 2000 \) - A stable equilibrium where approximately 2000 companies adopt the technology.

Equilibrium solutions are crucial as they provide insights into steady states of the system. They help identify if these points are stable or unstable. In this exercise, the equilibrium at \( N = 2000 \) is stable, meaning the system will tend to return to this point if slightly perturbed. Equilibria form important foundations for analyzing the behavior of differential equations mathematically.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify complex fractions in order to make integration manageable. In our exercise, we use it for the fraction \( \frac{1}{N(1-0.0005N)} \). This complex fraction is decomposed into simpler fractions:

\[ \frac{1}{N(1-0.0005 N)} = \frac{1}{N} - \frac{1}{N-2000} \]

This decomposition allows easier integration of each part separately, transforming the original integral into a sum of simpler integrals.

Partial fraction decomposition breaks down rational expressions into components that are more straightforward to integrate or differentiate, a vital tool in calculus used to simplify many complex algebraic expressions.

Separation of Variables

Separation of variables is a powerful method for solving differential equations. It involves rearranging the equation so that each variable and its derivative are on opposite sides. In our exercise, the differential equation is separated as follows:

• \( \frac{dN}{N(1-0.0005N)} = dt \)

This separation allows each side to be integrated independently. The left-hand side includes the dependent variable \( N \), and the right-hand side solely includes \( t \).

By integrating both sides, we transform the differential equation into an expression relating \( N \) and \( t \). This step is crucial because it provides the means to solve for \( N \) as a function of \( t \), allowing us to predict behavior over time.

In this exercise, separation of variables leads us to integrate and solve equations, ultimately finding an analytical solution for the model's behavior. This method underlines many processes in mathematical modeling and solving differential systems.