Problem 3 - Biocalculus Calculus for the Life Sciences

Question

Show that \(x_{1}(t)\) and \(x_{2}(t)\) are solutions to the system of differential equations \(d \mathbf{x} / d t=A \mathbf{x}\) . \(\begin{array}{l}{A=\left[ \begin{array}{rr}{-1} & {-4} \\ {1} & {-1}\end{array}\right]} \\ {x_{1}(t)=e^{-t} \cos 2 t,} & {x_{2}(t)=\frac{1}{2} e^{-t} \sin 2 t}\end{array}\)

Step-by-Step Solution

Answer

\(x_1(t)\) and \(x_2(t)\) are solutions as their differentiated forms match \(A\mathbf{x}(t)\).

Step 1: Differentiate \( x_1(t) \)

First, differentiate \( x_1(t) = e^{-t} \cos(2t) \) with respect to \( t \). Using the product rule for differentiation, \( d(uv)/dt = u'v + uv' \), and chain rule where needed:\[\frac{d}{dt} \left( e^{-t} \cos(2t) \right) = -e^{-t} \cos(2t) + e^{-t}(-2\sin(2t)) = e^{-t} (-\cos(2t) - 2\sin(2t)).\]

Step 2: Differentiate \( x_2(t) \)

Now, differentiate \( x_2(t) = \frac{1}{2} e^{-t} \sin(2t) \) with respect to \( t \). Similarly using the product rule and chain rule:\[\frac{d}{dt} \left( \frac{1}{2} e^{-t} \sin(2t) \right) = -\frac{1}{2} e^{-t} \sin(2t) + \frac{1}{2} e^{-t}(2\cos(2t)) = \frac{1}{2} e^{-t} (-\sin(2t) + 2\cos(2t)).\]

Step 3: Formulate the Vector Equation

Express the solution vector \( \mathbf{x}(t) = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \) and its derivative \( \frac{d\mathbf{x}}{dt} = \begin{bmatrix} \frac{d}{dt}x_1(t) \ \frac{d}{dt}x_2(t) \end{bmatrix} \). With previous derivatives, this becomes:\[\frac{d\mathbf{x}}{dt} = \begin{bmatrix} e^{-t} (-\cos(2t) - 2\sin(2t)) \ \frac{1}{2} e^{-t} (-\sin(2t) + 2\cos(2t)) \end{bmatrix}.\]

Step 4: Multiply Matrix \( A \) with \( \mathbf{x}(t) \)

Compute \( A \mathbf{x}(t) \) where \( A = \begin{bmatrix} -1 & -4 \ 1 & -1 \end{bmatrix} \) and \( \mathbf{x}(t) = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \):\[A \mathbf{x}(t) = \begin{bmatrix} -1 & -4 \ 1 & -1 \end{bmatrix} \begin{bmatrix} e^{-t} \cos(2t) \ \frac{1}{2} e^{-t} \sin(2t) \end{bmatrix} = \begin{bmatrix} -e^{-t} \cos(2t) - 2e^{-t} \sin(2t) \ e^{-t} \cos(2t) - \frac{1}{2} e^{-t} \sin(2t) \end{bmatrix}.\]

Step 5: Verify Solutions Match

Compare the calculated \( \frac{d\mathbf{x}}{dt} \) from Step 3 and \( A \mathbf{x}(t) \) from Step 4. They should match if \( x_1(t) \) and \( x_2(t) \) are solutions:Both yield:\[\begin{bmatrix} e^{-t} (-\cos(2t) - 2\sin(2t)) \ \frac{1}{2} e^{-t} (-\sin(2t) + 2\cos(2t)) \end{bmatrix},\]confirming that \( x_1(t) \) and \( x_2(t) \) satisfy the differential equation.

Key Concepts

systems of differential equations

In mathematics, a system of differential equations is essentially a set of multiple differential equations that relate to various functions and their derivatives. These systems often emerge in scenarios where multiple interacting quantities evolve over time, such as in physics, engineering, and biology.
For the exercise at hand, we're looking at a system described by \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \) where \( A \) is a matrix and \( \mathbf{x}(t) \) is a vector representing functions \( x_1(t) \) and \( x_2(t) \). Solving such systems involves determining these functions that satisfy the differential equations over some interval.

• These equations provide insights into complex systems where the behavior of variables is interconnected.
• The solution method generally involves linear algebra tools such as matrices and vectors, married with calculus techniques.
• Checking that \( x_1(t) \) and \( x_2(t) \) are solutions to the system requires both differentiation and matrix operations.

Understanding these systems can provide better approaches to predicting the behavior of complex dynamic processes.

matrix multiplication

Matrix multiplication forms a crucial part of resolving systems of differential equations. In our example, we compute \( A \mathbf{x}(t) \), where \( A \) is a matrix and \( \mathbf{x}(t) \) is a vector.
The operation entails specific steps:

• Each element of the resulting vector results from the sum of the products of corresponding elements in the rows of \( A \) and the columns of \( \mathbf{x}(t) \).
• For the entry \( i, j \) in the matrix \( A \), multiply the \( i \)-th row of \( A \) by the \( j \)-th element of \( \mathbf{x}(t) \), and sum these products.

For our exercise, multiplying the matrix \( \left[ \begin{array}{rr}{-1} & {-4} \ {1} & {-1}\end{array}\right] \) with the vector \( \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \) means executing these calculations.
This technique is essential for combining systems of linear equations into a compact matrix form, facilitating easier manipulation and solution.

product rule

The product rule is a vital tool needed when differentiating functions that are products of two or more functions. In our exercise, we used the product rule for both \( x_1(t) \) and \( x_2(t) \) because they are both the product of exponential and trigonometric functions.
The rule states that: \[ \frac{d}{dt}(u(t) v(t)) = u'(t)v(t) + u(t)v'(t) \] where \( u(t) \) and \( v(t) \) are functions of \( t \).
For \( x_1(t) = e^{-t}\cos(2t) \):

• \( u(t) = e^{-t} \) with derivative \( u'(t) = -e^{-t} \).
• \( v(t) = \cos(2t) \) with derivative \( v'(t) = -2\sin(2t) \).

Applying the product rule helps us differentiate the function effectively, ensuring each component's behavior in respect to \( t \) is considered correctly.

chain rule

The chain rule is another fundamental concept when dealing with composite functions, aiding in the differentiation of complex expressions. It is particularly useful in our example, where functions involve combinations such as exponentials and trigonometric terms.
The rule is applied as: \[ \frac{d}{dt}(f(g(t))) = f'(g(t)) \cdot g'(t) \] where \( f \) and \( g \) are functions of \( t \).
In our exercise, we have components like \( \cos(2t) \) and \( \sin(2t) \), where,

• \( g(t) = 2t \) with derivative \( g'(t) = 2 \).
• \( f(g(t)) = \cos(2t) \) or \( \sin(2t) \) requiring differentiation with respect to \( g(t) \).

The chain rule lets us capture the contribution of inner functions' rates of change, ensuring precision in evaluating derivatives of more involved functions.