Problem 22 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The solutions are \(x_1(t) = -e^{-t}\) and \(x_2(t) = \frac{1}{3} e^{-t} - \frac{7}{3} e^{2t}\).

Step 1: Understand the Problem

We're dealing with a system of two first-order linear differential equations. The matrix notation given is a common way to represent such systems, where \(\frac{dx_1}{dt}\) and \(\frac{dx_2}{dt}\) represent the rates of change of \(x_1\) and \(x_2\) respectively. The matrix multiplication gives us a system of equations that must be solved simultaneously.

Step 2: Write the System of Equations

From the given matrix expression, we can write the system of differential equations:1. \(\frac{dx_1}{dt} = -x_1\) 2. \(\frac{dx_2}{dt} = x_1 - 2x_2\)These are our equations to solve for \(x_1(t)\) and \(x_2(t)\).

Step 3: Solve the First Equation

The equation \(\frac{dx_1}{dt} = -x_1\) is a separable differential equation. We separate variables and integrate:\[\int \frac{1}{x_1} \, dx_1 = \int -1 \, dt\]This yields\[\ln |x_1| = -t + C_1\]Exponentiating both sides gives:\[x_1(t) = C_1' e^{-t}\] Using the initial condition \(x_1(0) = -1\), we find \(C_1' = -1\). Hence, \(x_1(t) = -e^{-t}\).

Step 4: Substitute Back into the Second Equation

Substitute \(x_1(t) = -e^{-t}\) into \(\frac{dx_2}{dt} = x_1 - 2x_2\) to get:\[\frac{dx_2}{dt} = -e^{-t} - 2x_2\]

Step 5: Solve for \(x_2(t)\)

The equation \(\frac{dx_2}{dt} = -e^{-t} - 2x_2\) is a linear ordinary differential equation. Use the integrating factor method. The integrating factor here is \(e^{\int -2 \, dt} = e^{-2t}\). Multiply through by the integrating factor: \[e^{-2t} \frac{dx_2}{dt} + 2 e^{-2t} x_2 = -e^{-3t}\]This simplifies to \[\frac{d}{dt}(e^{-2t} x_2) = -e^{-3t}\]Integrate both sides:\[e^{-2t} x_2 = \int -e^{-3t} \, dt = \frac{1}{3} e^{-3t} + C_2\]Multiply by \(e^{2t}\) to get: \[x_2(t) = \frac{1}{3} e^{-t} + C_2 e^{2t}\] Using \(x_2(0) = -2\), we solve for \(C_2\):\[-2 = \frac{1}{3} + C_2\]Thus, \(C_2 = -\frac{7}{3}\). Therefore, \(x_2(t) = \frac{1}{3} e^{-t} - \frac{7}{3} e^{2t}\).

Step 6: Combine Solutions

Now we have both solutions:\(x_1(t) = -e^{-t}\) and \(x_2(t) = \frac{1}{3} e^{-t} - \frac{7}{3} e^{2t}\). These solutions satisfy the differential equations and the initial conditions. Consequently, they provide the complete solution to the system.

Key Concepts

Understanding Initial-Value Problems

An initial-value problem involves finding a function that satisfies a differential equation and an initial condition. In our exercise, the goal is to determine functions for \(x_1(t)\) and \(x_2(t)\) that solve the differential equations given in the matrix form, simultaneously adhering to the initial conditions \(x_1(0) = -1\) and \(x_2(0) = -2\).
These initial conditions are crucial because they allow us to determine the specific solution among the infinite possibilities that satisfy the differential equation. Without them, we could only solve a general solution containing arbitrary constants. Having initial conditions helps us pinpoint exact constants, making the solution unique to the problem given.

Breaking Down Systems of Equations

A system of equations is a set of equations with multiple variables where the solution needs to satisfy all equations simultaneously. In the context of differential equations, particularly linear ones, these involve derivatives of multiple interrelated variables, thus forming a system.
Our exercise specifies the rate of change for two variables, \(x_1\) and \(x_2\). Here's a simplified way of seeing it:

• The first equation, \(\frac{dx_1}{dt} = -x_1\), tells us how \(x_1\) changes over time.
• The second, \(\frac{dx_2}{dt} = x_1 - 2x_2\), describes \(x_2\)'s rate of change depending on both itself and \(x_1\).

This coupled nature means they share a relationship which we must respect while solving them, typically by solving one equation and using that solution to find solutions for the other.

Linear Differential Equations

Linear differential equations are characterized by the linearity of derivatives; none of the functions multiply or divide each other.
Our system includes two first-order linear differential equations. The order refers to the highest derivative (first in this case) present. These equations are linear because each term involving the function \(x_1(t)\) or \(x_2(t)\) and their derivatives occurs in a non-exponential fashion, combined only through addition and multiplication by constants.
Such equations are significant because they have known solution methods, which often involve techniques like separating variables or using integrating factors, as used in our exercise. These techniques simplify the problem by treating each variable independently or integrating directly to find solutions.

Matrix Representation of Systems

Matrix representation transforms multiple linear equations into a concise, visually manageable format.
In our exercise, the matrix form allows us to encapsulate the relationship between \(x_1\) and \(x_2\) in a compact structure. The left-hand side of the equation indicates the rates of change, while the matrix accounts for how each variable influences itself and the other. This is shown as:

• The entry in the first row and column, \(-1\), influences \(x_1\) itself.
• The entry in the second row and column, \(-2\), affects \(x_2\) in a similar self-influencing manner.
• The off-diagonal term, \(1\), signifies \(x_1\)'s influence on \(x_2\).

Matrix forms are very efficient in representing and solving systems of linear equations, especially when extending to multiple variables or when computational methods are leveraged.