Problem 29 - Biocalculus Calculus for the Life Sciences

Question

Second-order linear differential equations take the form $$y^{\prime \prime}(t)+p(t) y^{\prime}(t)+q(t) y(t)=g(t)$$ where \(p, q,\) and \(g\) are continuous functions of \(t .\) Suppose we have initial conditions \(y(0)=a\) and \(y^{\prime}(0)=b .\) Show that this equation can be rewritten as a system of two first-order linear differential equations having the form \(\frac{d \mathbf{x}}{d t}=\left[ \begin{array}{cc}{0} & {1} \\ {-q(t)} & {-p(t)}\end{array}\right] \mathbf{x}+\left[ \begin{array}{c}{0} \\\ {g(t)}\end{array}\right]\) \(\begin{aligned} \text { with } & \mathbf{x}(0)=\left[ \begin{array}{l}{a} \\\ {b}\end{array}\right] \\ \text { where } x_{1}(t)=y(t) \text { and } x_{2}(t) &=y^{\prime}(t) \end{aligned}\)

Step-by-Step Solution

Answer

The equation is rewritten as a system: \( \frac{d\mathbf{x}}{dt} = \begin{bmatrix} 0 & 1 \\ -q(t) & -p(t) \end{bmatrix} \mathbf{x} + \begin{bmatrix} 0 \\ g(t) \end{bmatrix} \).

Step 1: Identify Variables

To rewrite the second-order differential equation as a system of first-order equations, we first define two new variables. Let \( x_1(t) = y(t) \) and \( x_2(t) = y'(t) \). These variables will help translate the second-order equation into a system of first-order equations.

Step 2: Express the Original Equation in Terms of New Variables

Using the new variables, express the derivatives: the derivative of \( x_1(t) \) is simply \( x_2(t) \) because \( x_1(t) = y(t) \). Therefore, \( \frac{dx_1}{dt} = x_2(t) \). Similarly, since \( x_2(t) = y'(t) \), the second derivative is \( \frac{dx_2}{dt} = y''(t) \). Substitute this into the original differential equation \( y''(t) + p(t)y'(t) + q(t)y(t) = g(t) \).

Step 3: Substitute New Variables into Original Equation

Substitute \( y(t) = x_1(t) \) and \( y'(t) = x_2(t) \) into the original second-order differential equation: \( \frac{dx_2}{dt} + p(t)x_2(t) + q(t)x_1(t) = g(t) \). This becomes the second equation of our system.

Step 4: Write the System of First-Order Equations

From previous steps, we now have the system of equations: \( \frac{dx_1}{dt} = x_2(t) \) and \( \frac{dx_2}{dt} = -q(t)x_1(t) - p(t)x_2(t) + g(t) \). These can be combined in matrix form as \( \frac{d \mathbf{x}}{dt} = \begin{bmatrix} 0 & 1 \ -q(t) & -p(t) \end{bmatrix} \mathbf{x} + \begin{bmatrix} 0 \ g(t) \end{bmatrix} \).

Step 5: Verify Initial Conditions

For the initial conditions, substitute \( x_1(0) = y(0) = a \) and \( x_2(0) = y'(0) = b \). Therefore, the initial condition matrix is \( \mathbf{x}(0) = \begin{bmatrix} a \ b \end{bmatrix} \).

Key Concepts

Second-order differential equations

Second-order differential equations are equations involving the second derivative of a function. They specify how the function evolves in terms of its rate of change and acceleration. In a typical form, the equation is written as follows: \[ y''(t) + p(t)y'(t) + q(t)y(t) = g(t) \]Here, \( y''(t) \) is the second derivative, \( y'(t) \) is the first derivative, and \( y(t) \) is the function itself. Functions \( p(t), q(t), \) and \( g(t) \) are continuous in terms of \( t \).

This type of differential equation is prevalent in physics and engineering. They often describe systems like harmonic oscillators or circuits with inductance and resistance. Solving these requires finding the function \( y(t) \) that satisfies both the equation and any accompanying initial conditions at a specified point.

Initial conditions

Initial conditions are critical when solving differential equations as they allow us to find a unique solution. For second-order differential equations, we need two initial conditions:

• The value of the function at a starting point, say \( y(0) = a \).
• The value of the first derivative at that point, \( y'(0) = b \).

These conditions provide the specific starting values that the function and its rate of change must satisfy.

Without these conditions, a differential equation can have many potential solutions, since the general solution will include arbitrary constants that are determined by the initial conditions.

First-order systems

First-order systems simplify the problem by breaking down complex equations into simpler ones. A second-order differential equation can often be transformed into a system of first-order equations. This provides ease in both theoretical analysis and numerical computations.

To adjust a second-order equation into a first-order system:

• Introduce new variables to represent the function and its first derivative. For example, let \( x_1(t) = y(t) \) and \( x_2(t) = y'(t) \).
• Express the original differential equations using these new variables, thus transforming it into a system with two first-order equations.

The system can then be solved using matrix operations, benefiting from more straightforward solution techniques.

Matrix representation

Matrix representation is a powerful tool used to express and solve systems of linear differential equations. By arranging equations in matrix form, you leverage the capabilities of linear algebra to find solutions. For a system of equations derived from a second-order differential equation, begin by framing them in matrix notation like:\[\frac{d \mathbf{x}}{d t} = \begin{bmatrix} 0 & 1 \ -q(t) & -p(t) \end{bmatrix} \mathbf{x} + \begin{bmatrix}0 \ g(t)\end{bmatrix} \]This matrix form simplifies expressing complex equations and utilizes vector operations for solution finding.

Using matrices not only provides clarity but also efficiency in calculations. Initial conditions are seamlessly integrated using column vectors, shown as:\[\mathbf{x}(0) = \begin{bmatrix} a \ b \end{bmatrix} \]In this representation, the solution process involves finding functions \(x_1(t)\) and \(x_2(t)\) that solve the matrix equation, which corresponds back to the original problem in terms of the function \(y(t)\).