Problem 21 - Calculus for Biology and Medicine

Question

Solve the given initial-value problem. \(\left[\begin{array}{c}\frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t}\end{array}\right]=\left[\begin{array}{rr}3 & -2 \\ 0 & 1\end{array}\right]\left[\begin{array}{l}x_{1}(t) \\\ x_{2}(t)\end{array}\right]\) with \(x_{1}(0)=1\) and \(x_{2}(0)=1\).

Step-by-Step Solution

Answer

\( x_1(t) = e^t \), \( x_2(t) = e^t \). Solution: \( \mathbf{x}(t) = \begin{bmatrix} e^t \\ e^t \end{bmatrix} \).

Step 1: Write the System of Differential Equations

The given vector differential equation can be broken down into two separate differential equations based on the matrix multiplication: \( \frac{dx_1}{dt} = 3x_1 - 2x_2 \) and \( \frac{dx_2}{dt} = x_2 \).

Step 2: Solve the Equation for x_2

The differential equation \( \frac{dx_2}{dt} = x_2 \) is a first-order linear differential equation. It can be solved using separation of variables: \( \frac{dx_2}{x_2} = dt \). Integrate both sides to obtain \( \ln|x_2| = t + C_2 \). Exponentiating both sides gives \( x_2(t) = C_3e^t \).

Step 3: Apply Initial Condition for x_2

Using the initial condition \( x_2(0) = 1 \), we substitute into \( x_2(t) = C_3e^t \) to find \( 1 = C_3e^0 \Rightarrow C_3 = 1 \). Hence, \( x_2(t) = e^t \).

Step 4: Solve the Equation for x_1

The differential equation \( \frac{dx_1}{dt} = 3x_1 - 2x_2 \) is a first-order linear differential equation. Using the solution for \( x_2(t) \), substitute \( x_2(t) = e^t \) into the equation to get \( \frac{dx_1}{dt} = 3x_1 - 2e^t \).

Step 5: Use an Integrating Factor for x_1

Rearrange the equation to the standard linear form: \( \frac{dx_1}{dt} - 3x_1 = -2e^t \). The integrating factor is \( \mu(t) = e^{\int -3 dt} = e^{-3t} \). Multiply the entire differential equation by this integrating factor: \( e^{-3t}\frac{dx_1}{dt} - 3e^{-3t}x_1 = -2e^{-2t} \).

Step 6: Solve for x_1

Notice the left-hand side is the derivative of \( e^{-3t}x_1 \). Integrate both sides: \( \int d(e^{-3t}x_1) = \int -2e^{-2t} dt \). The right-hand side integrates to \( -2 \cdot e^{-2t} / (-2) = e^{-2t} + C_1 \). Thus, \( e^{-3t}x_1 = e^{-2t} + C_1 \).

Step 7: Solve for x_1(t)

Solving for \( x_1(t) \) gives \( x_1(t) = e^{3t}(e^{-2t} + C_1) = e^t + C_1 e^{3t} \).

Step 8: Apply Initial Condition for x_1

Using the initial condition \( x_1(0) = 1 \), substitute to find \( 1 = e^0 + C_1 e^0 \), which simplifies to \( 1 = 1 + C_1 \). Hence, \( C_1 = 0 \), and \( x_1(t) = e^t \).

Step 9: Find the Solution to the Initial Value Problem

The solutions are \( x_1(t) = e^t \) and \( x_2(t) = e^t \). Therefore, the solution to the initial-value problem is \( \mathbf{x}(t) = \begin{bmatrix} e^t \ e^t \end{bmatrix} \).

Key Concepts

Initial Value Problem

An initial value problem in differential equations involves finding a function that satisfies a differential equation and also meets specified values at a particular point. These problems are fundamental in fields like physics, engineering, and mathematics because they can model dynamic systems.

In these problems, you're given a differential equation and an initial condition. The initial condition specifies the value of the solution at a specific point. This makes it possible to find a unique solution. For instance, in the problem we're tackling, both components of the vector function are initially set to the value of 1 at time zero, specifically,

• \( x_1(0) = 1 \)
• \( x_2(0) = 1 \)

The initial conditions provide the starting point for solving the system of differential equations which define how the variables change over time. By keeping track of initial values, you ensure that the solution you find is relevant to the specific scenario being modelled.

Linear Differential Equation

A linear differential equation is an equation that involves a function and its derivatives. It can have constant or variable coefficients, and it is linear with respect to the unknown function and its derivatives. It has a variety of forms, but one classic form resembles:

• \( rac{dy}{dt} + p(t)y = g(t) \)

where \( p(t) \) and \( g(t) \) are given functions.

In this exercise, the given second equation, \( \frac{dx_2}{dt} = x_2 \), is a simple linear first-order differential equation. Such equations are typically solved using methods like separation of variables, as we separated \( \frac{dx_2}{x_2} = dt \) and integrated to find \( x_2(t) \). Similarly, the equation \( \frac{dx_1}{dt} = 3x_1 - 2x_2 \) is modified using the integrating factor method to make it more solvable.

The integrating factor is a specific function that simplifies the equation into a form where the left side becomes the derivative of the product of an integrating factor and the unknown function. It turns complex-looking equations into more manageable ones, as seen when solving for \( x_1(t) \). By using integration, the solution becomes relatively straightforward to find.

Matrix Differential Equation

Matrix differential equations involve matrices. These equations are common in systems of linear differential equations, where multiple functions are related to each other. Matrices provide a compact way to express systems of equations and help in modeling situations with several interdependent variables.

The given exercise represents a simple matrix differential equation:

• \( \left[\begin{array}{c}\frac{d x_{1}}{d t} \ \frac{d x_{2}}{d t}\end{array}\right]=\left[\begin{array}{rr}3 & -2 \ 0 & 1\end{array}\right]\left[\begin{array}{l}x_{1}(t) \ x_{2}(t)\end{array}\right] \)

In this equation, the rate of change of the vector \( \mathbf{x}(t) \) is given by a product of a matrix \( A \) and the vector \( \mathbf{x} \) itself. Each element in the resulting vector equation corresponds to one of the system of differential equations outlined earlier. In essence, this representation allows us to work more efficiently with systems of equations, highlighting the connections and dependencies between variables. Such representations are particularly useful in higher dimensions or complex systems and are widely used in areas like control theory and physics.