Problem 19 - Calculus for Biology and Medicine

Question

In Problems 19-26, solve the given initial-value problem. \(\left[\begin{array}{c}\frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t}\end{array}\right]=\left[\begin{array}{rr}-3 & 0 \\ 4 & 2\end{array}\right]\left[\begin{array}{l}x_{1}(t) \\\ x_{2}(t)\end{array}\right]\) with \(x_{1}(0)=-5\) and \(x_{2}(0)=5\).

Step-by-Step Solution

Answer

The solution is \( x_1(t) = -5e^{-3t} \) and \( x_2(t) = e^{2t} + 4e^{-3t} \).

Step 1: Formulate the System of Equations

The given problem is a system of first-order linear differential equations. We have \( \frac{dx_1}{dt} = -3x_1 \) and \( \frac{dx_2}{dt} = 4x_1 + 2x_2 \). These are derived from the matrix equation by expanding the multiplication.

Step 2: Solve the First Differential Equation

Solve \( \frac{dx_1}{dt} = -3x_1 \). Assuming \( x_1(t) = Ce^{\lambda t} \), we have \( \frac{d}{dt}(Ce^{-3t}) = -3Ce^{-3t} \). Hence, \( \lambda = -3 \) and \( x_1(t) = C_1 e^{-3t} \). Using the initial condition \( x_1(0) = -5 \), we find that \( C_1 = -5 \). Therefore, \( x_1(t) = -5e^{-3t} \).

Step 3: Solve the Second Differential Equation

Substitute \( x_1(t) = -5e^{-3t} \) into \( \frac{dx_2}{dt} = 4x_1 + 2x_2 \) to get \( \frac{dx_2}{dt} = 4(-5e^{-3t}) + 2x_2 = -20e^{-3t} + 2x_2 \). This is a first-order linear inhomogeneous differential equation for \( x_2 \).

Step 4: Solve the Homogeneous Part of the Second Equation

Consider the homogeneous part \( \frac{dx_2}{dt} = 2x_2 \), which gives \( x_2^h(t) = C_2 e^{2t} \).

Step 5: Solve the Particular Solution of the Second Equation

Assume a particular solution of the form \( x_2^p(t) = A e^{-3t} \). Substitute into the differential equation: \( -3A e^{-3t} = -20e^{-3t} + 2A e^{-3t} \). Solving \( -A = -20 + 2A \) gives \( A = 4 \). Thus, \( x_2^p(t) = 4e^{-3t} \).

Step 6: Form the General Solution for \( x_2 \)

Combine the homogeneous and particular solutions: \( x_2(t) = C_2 e^{2t} + 4e^{-3t} \).

Step 7: Apply Initial Conditions for \( x_2 \)

Use the initial condition \( x_2(0) = 5 \), substitute into the general solution: \( 5 = C_2 e^{0} + 4e^{0} \) gives \( 5 = C_2 + 4 \). Solving this, \( C_2 = 1 \). Therefore, \( x_2(t) = e^{2t} + 4e^{-3t} \).

Step 8: Write the Final Solution

The solution to the initial-value problem is \( x_1(t) = -5e^{-3t} \) and \( x_2(t) = e^{2t} + 4e^{-3t} \).

Key Concepts

System of Differential Equations

A system of differential equations involves multiple equations that describe how various quantities change with respect to each other. In such a system, each differential equation involves derivatives of one or more functions. In this particular exercise, the differential equations describe changes in two variables,\(x_1(t)\) and \(x_2(t)\). These types of equations are often presented in matrix form, allowing us to see the interdependencies more clearly.
The system we are dealing with is:

• \(\frac{dx_1}{dt} = -3x_1\)
• \(\frac{dx_2}{dt} = 4x_1 + 2x_2\)

This is a classic example of a coupled system where \(x_1\) influences \(x_2\). Understanding how the variables are interconnected is crucial in predicting the behavior of the system over time.
Solving these systems typically involves finding each function's expression over time, usually using initial conditions given in the problem.

First-Order Linear Differential Equations

A first-order linear differential equation is an equation that involves the first derivative of a function and is linear in the unknown function and its derivatives. Such equations can describe a wide variety of phenomena in science and engineering. For each function like\(x_1(t)\) and \(x_2(t)\), the solution process involves separating these equations individually.
In this problem:

• \(\frac{dx_1}{dt} = -3x_1\) is treated as a straightforward linear differential equation.
• The second equation \(\frac{dx_2}{dt} = 4x_1 + 2x_2\) is considered a linear inhomogeneous equation because it includes a term that is not multiplied by \(x_2(t)\), namely \(4x_1(t)\).

The process involves identifying these equations and applying methods like separation of variables or using integrating factors to solve them. These techniques help us find specific solutions or expressions for \(x_1(t)\) and \(x_2(t)\).

Homogeneous Solution

A homogeneous solution refers to the solution of the corresponding homogeneous differential equation, where all terms independent of the function are set to zero. This represents just one part of the complete solution for the differential equation in question.
For the second differential equation \(\frac{dx_2}{dt} = 4x_1 + 2x_2\), we first consider the homogeneous part:

• \(\frac{dx_2}{dt} = 2x_2\)

This equation is solved similarly to other simple first-order linear differential equations, giving us:

• \(x_2^h(t) = C_2 e^{2t}\)

Here, \(C_2\) is a constant determined later using initial conditions. This homogeneous solution accounts for the system's natural response without external interference or non-homogeneous terms.

Particular Solution

The particular solution focuses on addressing the non-homogeneous part of a differential equation, which involves terms not multiplied by the dependent variable. In our system, the particular solution is required for the equation involving \(x_2(t)\):

• \(\frac{dx_2}{dt} = -20e^{-3t} + 2x_2\)

To solve this, we assume a form for the particular solution, often using a guessed form that matches the inhomogeneous part's structure. In this case, we use \(x_2^p(t) = A e^{-3t}\). Substituting this guess into the equation helps us find the value of \(A\).

• Solving gives \(A = 4\).

The particular solution thus becomes \(x_2^p(t) = 4e^{-3t}\). This particular solution adds to the complete expression, ensuring that the equation's non-homogeneous part is satisfied alongside the contribution from the homogeneous solution.