Problem 40 - Advanced Engineering Mathematics

Question

(a) Solving \(k_{1}(M-A)-k_{2} A=0\) for \(A\) we find the equilibrium solution \(A=k_{1} M /\left(k_{1}+k_{2}\right) .\) From the phase portrait we see that \(\lim _{t \rightarrow \infty} A(t)=k_{1} M /\left(k_{1}+k_{2}\right)\) since \(k_{2}>0,\) the material will never be completely memorized and the larger \(k_{2}\) is, the less the amount of material will be memorized over time. (b) Write the differential equation in the form \(d A / d t+\left(k_{1}+k_{2}\right) A=k_{1} M\) Then an integrating factor is \(e^{\left(k_{1}+k_{2}\right) t},\) and $$\begin{aligned} \frac{d}{d t}\left[e^{\left(k_{1}+k_{2}\right) t} A\right] &=k_{1} M e^{\left(k_{1}+k_{2}\right) t} \\ e^{\left(k_{1}+k_{2}\right) t} A &=\frac{k_{1} M}{k_{1}+k_{2}} e^{\left(k_{1}+k_{2}\right) t}+c \\ A &=\frac{k_{1} M}{k_{1}+k_{2}}+c e^{-\left(k_{1}+k_{2}\right) t} \end{aligned}$$ \(\operatorname{Using} A(0)=0\) we find \(c=-\frac{k_{1} M}{k_{1}+k_{2}}\) and \(A=\frac{k_{1} M}{k_{1}+k_{2}}\left(1-e^{-\left(k_{1}+k_{2}\right) t}\right) .\) As \(t \rightarrow \infty\) \(A \rightarrow \frac{k_{1} M}{k_{1}+k_{2}}\).

Step-by-Step Solution

Answer

The equilibrium solution is \( A = \frac{k_1 M}{k_1 + k_2} \), and as \( t \rightarrow \infty \), \( A(t) \rightarrow \frac{k_1 M}{k_1 + k_2} \).

Step 1: Rearrange the given equation

Start with the equation from part (a): \( k_1(M - A) - k_2 A = 0 \). Rearrange the terms to isolate \( A \) on one side. This gives: \( k_1 M - k_1 A = k_2 A \).

Step 2: Collect all terms involving A

Combine all terms that have \( A \) on the same side of the equation: \( k_1 M = k_1 A + k_2 A \). Factor \( A \) out: \( k_1 M = (k_1 + k_2)A \).

Step 3: Solve for A

Divide both sides of the equation by \( k_1 + k_2 \) to solve for \( A \). This results in: \( A = \frac{k_1 M}{k_1 + k_2} \). This is the equilibrium solution for \( A \).

Step 4: Setup the differential equation in required form

Rewrite the equation in the form: \( \frac{dA}{dt} + (k_1 + k_2)A = k_1M \). Now recognize it as a first-order linear differential equation.

Step 5: Determine the integrating factor

Identify the integrating factor for the linear differential equation as \( e^{(k_1 + k_2)t} \). This factor will be used to make the left side of the equation an exact derivative.

Step 6: Solve using the integrating factor

Multiply through by the integrating factor to obtain: \( \frac{d}{dt}(e^{(k_1 + k_2)t}A) = k_1M e^{(k_1 + k_2)t} \). Integrate both sides with respect to \( t \).

Step 7: Integrate and solve for A

Integrate the left side directly and the right side with the respect to \( t \): \( e^{(k_1+k_2)t}A = \frac{k_1M}{k_1+k_2}e^{(k_1+k_2)t} + C \). Solve for \( A \): \( A = \frac{k_1M}{k_1 + k_2} + Ce^{-(k_1 + k_2)t} \).

Step 8: Apply initial condition

Use the initial condition \( A(0) = 0 \) to find \( C \). Substitute into the equation: \( 0 = \frac{k_1M}{k_1 + k_2} + C \). Solve for \( C \): \( C = -\frac{k_1M}{k_1 + k_2} \).

Step 9: Write the final solution

Plug \( C \) back into the equation to find the particular solution: \( A = \frac{k_1M}{k_1+k_2}(1 - e^{-(k_1 + k_2)t}) \). As \( t \rightarrow \infty \), \( A \rightarrow \frac{k_1 M}{k_1+k_2} \).

Key Concepts

Linear Differential Equations

Linear differential equations form the foundation of many problems in calculus and are pivotal to understanding changes over time. They are defined as differential equations where the unknown function and its derivatives appear linearly. For instance, the equation \( \frac{dA}{dt} + (k_1 + k_2)A = k_1M \) is a classic example of a linear differential equation. This structure allows us to use specific techniques, such as the integrating factor, to find solutions more easily. Linear equations can be identified by:

• The presence of terms that add up linearly, meaning they can be represented as a sum or difference of the function and its derivative.
• No powers or higher-degree polynomial terms in the function or its derivatives.

Recognizing the form of a differential equation as linear lets us apply systematic methods to solve them, giving insights into systems that change according to predictable patterns.

Integrating Factor

The integrating factor is a powerful technique used in solving linear first-order differential equations. It simplifies the process by transforming the differential equation into an exact derivative, making it easier to integrate. The integrating factor is found using the exponential of the integral of the coefficient of \( A \) in the equation. For the given equation, the integrating factor is \( e^{(k_1 + k_2)t} \).

Here's a step-by-step approach to using the integrating factor:

• Identify the linear differential equation in the standard form: \( \frac{dA}{dt} + P(t)A = Q(t) \).
• Calculate the integrating factor: \( e^{\int P(t) \, dt} \). In our scenario, \( P(t) = (k_1 + k_2) \) making the factor \( e^{(k_1 + k_2)t} \).
• Multiply the entire differential equation by this factor, making the left-hand side an exact derivative.
• Integrate both sides with respect to \( t \) to find \( A(t) \).

This method converts a complex differential equation into something much more manageable, offering clear paths to understanding the dynamics of the system.

Equilibrium Solution

An equilibrium solution in the context of differential equations refers to a steady-state condition where the system remains constant over time. For the given problem, the equilibrium solution was found to be \( A = \frac{k_1 M}{k_1 + k_2} \). This solution tells us what value the system approaches as time goes to infinity.

Key aspects of equilibrium solutions include:

• They represent a state with no change, meaning \( \frac{dA}{dt} = 0 \).
• They often emerge from setting the derivative equal to zero and solving for the unknown variable.
• In our problem, the equilibrium solution also reveals information about the balance between parameters \( k_1 \) and \( k_2 \) in context to \( M \).

Understanding equilibrium solutions helps in predicting long-term behavior and stability of systems, and shows how different parameters influence the steady state.