Question

\(\mathbf{x}^{\prime}=\left(\begin{array}{cc}{2} & {-1} \\ {3} & {-2}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{1} \\\ {-1}\end{array}\right) e^{t}, \quad \mathbf{x}=\left(\begin{array}{l}{1} \\\ {0}\end{array}\right) e^{t}+2\left(\begin{array}{l}{1} \\\ {1}\end{array}\right) t e^{t}\)

Short answer

Answer: Yes, the given function \(\mathbf{x}\) satisfies the given differential equation.

Step-by-step solution

Calculate the derivative of \(\mathbf{x}\) with respect to t

We have the given function: \(\mathbf{x} = \left(\begin{array}{l}{1}\\\ {0}\end{array}\right) e^{t} + 2\left(\begin{array}{l}{1}\\\ {1}\end{array}\right) t e^{t}\). To calculate \(\mathbf{x'}\), we need to differentiate each component with respect to \(t\): \(\mathbf{x'} = \frac{d}{dt}\left(\begin{array}{l}{1\cdot e^{t}+2\cdot1\cdot t\cdot e^{t}}\\\ {0\cdot e^{t}+2\cdot1\cdot t\cdot e^{t}}\end{array}\right) = \left(\begin{array}{l}{e^{t}+2te^{t}+2e^{t}}\\\ {2te^{t}+2e^{t}}\end{array}\right)\)

Multiply the matrix and function \(\mathbf{x}\)

Now we need to multiply the given matrix with the given function \(\mathbf{x}\): \(\left(\begin{array}{cc}{2} & {-1}\\\ {3}&{-2}\end{array}\right)\cdot \left(\begin{array}{l}{1}\\\ {0}\end{array}\right) e^{t} + \left(\begin{array}{cc}{2} & {-1}\\\ {3}&{-2}\end{array}\right)\cdot 2\left(\begin{array}{l}{1}\\\ {1}\end{array}\right) t e^{t} = e^t\cdot \left(\begin{array}{l}{2}\\\ {3}\end{array}\right) + 2te^t\cdot \left(\begin{array}{l}{1}\\\ {1}\end{array}\right)\)

Add the given vector exponential function

Now add the given vector exponential function to the result from step 2: \(\left(\begin{array}{l}{2}\\\ {3}\end{array}\right) e^{t} + 2\left(\begin{array}{l}{1}\\\ {1}\end{array}\right) t e^{t} + \left(\begin{array}{l}{1}\\\ {-1}\end{array}\right) e^{t} = \left(\begin{array}{l}{(2+1)e^{t}+2te^{t}}\\\ {(3-1)e^{t}+2te^{t}}\end{array}\right) = \left(\begin{array}{l}{e^{t}+2te^{t}+2e^{t}}\\\ {2te^{t}+2e^{t}}\end{array}\right)\)

Compare the results

By comparing the result obtained from step 1 with the result from step 3, we can see that they are equal: \(\mathbf{x'} = \left(\begin{array}{l}{e^{t}+2te^{t}+2e^{t}}\\\ {2te^{t}+2e^{t}}\end{array}\right) = \left(\begin{array}{l}{e^{t}+2te^{t}+2e^{t}}\\\ {2te^{t}+2e^{t}}\end{array}\right)\) Since they are equal, we can conclude that the given function \(\mathbf{x}\) satisfies the given differential equation.

Key concepts

Matrix Multiplication

Understanding matrix multiplication is crucial when dealing with systems of differential equations expressed in matrix form. Suppose we have a matrix \( A \) of size 2x2 and a vector \( \mathbf{v} \) with two components. Matrix multiplication involves multiplying each row of \( A \) by the corresponding component of \( \mathbf{v} \).
For example, with matrix \( A \) given as \( \left(\begin{array}{cc}{2} & {-1} \ {3} &{-2}\end{array}\right) \) and vector \( \mathbf{v} = \left(\begin{array}{c}{1} \ {0}\end{array}\right) \), the resulting product is another vector:

• The first component: \( 2 \times 1 + (-1) \times 0 = 2 \)
• The second component: \( 3 \times 1 + (-2) \times 0 = 3 \)

Answer: \( \left(\begin{array}{c}{2} \ {3}\end{array}\right) \)
Matrix multiplication helps us transform vector functions, maintaining the linear structure crucial for solving differential equations.

Vector Function

Vector functions are essential in representing systems with multiple variables. In this context, a vector function is a function \( \mathbf{x} \) where each component is a function of a common variable, often time \( t \).
Given \( \mathbf{x} = \left(\begin{array}{c}{1} \ {0}\end{array}\right) e^{t} + 2\left(\begin{array}{c}{1} \ {1}\end{array}\right) t e^{t} \), this expression describes how the components of \( \mathbf{x} \) vary with respect to \( t \):

• The first component: \( e^{t} + 2t e^{t} \)
• The second component: \( 0 + 2t e^{t} \)

Each component function depends on \( t \), conveying movement or change over time.
This form allows the system to be analyzed or solved in tandem through differentiation, matrix multiplication, and other operations. Vector functions embody a directed motion through their varying components.

Exponential Function

The exponential function, noted as \( e^{t} \), is omnipresent in differential equations due to its integral growth properties. It captures both the rate and the process of continuous growth or decay.
In the given exercise, the exponential function \( e^{t} \) appears as a factor in both vector components:

• In \( \left(\begin{array}{c}{1}\{0}\end{array}\right) e^{t} \), it describes a natural growth.
• In \( 2\left(\begin{array}{c}{1}\{1}\end{array}\right) t e^{t} \), it portrays growth modified by time \( t \).

Including \( e^{t} \) allows the vector function to model exponentially changing systems, such as populations or financial investments.
Its unique property, where differentiation does not alter its form, simplifies solving systems of differential equations. This characteristic ensures the exponential growth aspect remains consistent after mathematical operations like differentiation or integration.

Differentiation

Differentiation is a mathematical operation that highlights how quantities change. It is vital for analyzing dynamic systems, such as those described by differential equations.
To differentiate a vector function like \( \mathbf{x} = \left(\begin{array}{c}{1} \ {0}\end{array}\right) e^{t} + 2\left(\begin{array}{c}{1} \ {1}\end{array}\right) t e^{t} \), you apply the derivative to each vector component:

• The derivative of \( e^{t} + 2t e^{t} \) is \( e^{t} + 2e^{t} + 2t e^{t} \), using the product rule where necessary.
• The derivative of \( 2t e^{t} \) is \( 2e^{t} + 2t e^{t} \).

By differentiating, we obtain the rate and direction of change for each component of our system, crucial for understanding system dynamics under different conditions.
This approach helps verify whether a given function satisfies given differential equations, enabling effective modeling and prediction of behavior in real-world scenarios.