Question

Each initial value problem was obtained from an initial value problem for a higher order scalar differential equation. What is the corresponding scalar initial value problem? $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 0 & 1 \\ -3 & 2 \end{array}\right] \mathbf{y}+\left[\begin{array}{c} 0 \\ 2 \cos 2 t \end{array}\right], \quad \mathbf{y}(-1)=\left[\begin{array}{l} 1 \\ 4 \end{array}\right] $$

Short answer

Based on the provided step by step solution, the corresponding scalar initial value problem for the given matrix equation is: $$ y_1'(t) = y_2(t), \ y_1(-1) = 1 $$ $$ y_2'(t) = -3y_1(t) + 2y_2(t) + 2\cos 2t, \ y_2(-1) = 4 $$

Step-by-step solution

Identify the matrices and vectors involved in the problem

First, let's identify the given matrices and vectors: Matrix A: $$ \begin{bmatrix} 0 & 1 \\ -3 & 2 \end{bmatrix} $$ Vector y(t): $$ \begin{bmatrix} y_1(t) \\ y_2(t) \end{bmatrix} $$ Vector of input functions: $$ \begin{bmatrix} 0 \\ 2 \cos 2t \end{bmatrix} $$ Vector of initial conditions: $$ \begin{bmatrix} 1 \\ 4 \end{bmatrix} $$

Perform matrix-vector multiplication

Next, perform the matrix-vector multiplication: $$ \mathbf{y}^{\prime} = \left[\begin{array}{rr} 0 & 1 \\\ -3 & 2 \end{array}\right] \left[\begin{array}{c} y_1(t) \\\ y_2(t) \end{array}\right] + \left[\begin{array}{c} 0 \\\ 2 \cos 2t \end{array}\right] $$ This results in the following system of equations: $$ \begin{bmatrix} y_1'(t) \\ y_2'(t) \end{bmatrix} = \begin{bmatrix} y_2(t) \\ -3y_1(t) + 2y_2(t) +2 \cos 2t \end{bmatrix} $$

Write down the corresponding scalar initial value problem

Now we can write down the separate scalar initial value problems: $$ y_1'(t) = y_2(t), \ y_1(-1) = 1 $$ $$ y_2'(t) = -3y_1(t) + 2y_2(t) + 2\cos 2t, \ y_2(-1) = 4 $$ These two scalar equations and their respective initial conditions represent the original scalar initial value problem for the given matrix equation.

Key concepts

Scalar Differential Equation

A scalar differential equation is a type of differential equation where the unknown function depends on a single variable, typically time. These equations involve relationships between the function and its derivatives.
They are called 'scalar' since they consist of one single variable entity, as opposed to vector or matrix variables. Scalar differential equations are frequently used in physics, engineering, and mathematics to model various phenomena.
In the context of our initial value problem, the scalar differential equations are given by the separate equations describing the change in each variable with respect to time. These are extracted from the system:

• \( y_1'(t) = y_2(t) \) with the initial condition \( y_1(-1) = 1 \).
• \( y_2'(t) = -3y_1(t) + 2y_2(t) + 2\cos 2t \) with the initial condition \( y_2(-1) = 4 \).

Together, they represent the related scalar initial value problems derived from the matrix differential equation.

Matrix-Vector Multiplication

Matrix-vector multiplication is a mathematical operation involving a matrix and a vector to produce another vector. In differential equations, it is used to express complex systems in a compact form.
To perform matrix-vector multiplication, each element in the resulting vector is calculated as the dot product of the rows of the matrix with the columns of the vector.
For our initial value problem, the matrix \( \begin{bmatrix} 0 & 1 \ -3 & 2 \end{bmatrix} \) is multiplied by the vector \( \begin{bmatrix} y_1(t) \ y_2(t) \end{bmatrix} \), which produces:

• \( 0 \times y_1(t) + 1 \times y_2(t) = y_2(t) \)
• \(-3 \times y_1(t) + 2 \times y_2(t) \)

This matrix-vector multiplication results in the system of equations which describes the dynamics of the system. Matrix representations are powerful because they simplify complex systems and are convenient for computation.

System of Equations

A system of equations is a set of equations with multiple variables, which we solve simultaneously to find values that satisfy all equations in the system.
In differential equations, these systems often result from modeling processes that involve more than one interacting quantity.
The given matrix-vector form leads to a system of first-order linear differential equations:

• \( y_1'(t) = y_2(t) \)
• \( y_2'(t) = -3y_1(t) + 2y_2(t) + 2\cos 2t \)

These equations describe how one component depends on another. Solving this system lets us understand how the entire system evolves over time.To solve them, techniques like substitution, elimination, or matrix methods can be used.

Higher Order Differential Equations

Higher order differential equations involve derivatives of an unknown function that are greater than the first derivative. They arise in systems where the relationship involves more than a simple rate of change.
Often, initial value problems for higher order equations are transformed into equivalent systems of first-order equations, like in our example.
Each higher order differential equation can be rewritten as a system of first-order equations by defining additional variables to represent the higher derivatives.
In our case, the system derived from the matrix equation is equivalent to a second-order scalar differential equation in terms of one variable, stemming from writing one differential equation for each element in the vector.