Question

Determine whether the given functions form a fundamental set of solutions for the linear system. $$ =\left[\begin{array}{rr} 2 & -2 \\ 0 & 1 \end{array}\right] \mathbf{y}, \quad \mathbf{y}_{1}(t)=\left[\begin{array}{c} 3 e^{2 t} \\ 0 \end{array}\right], \quad \mathbf{y}_{2}(t)=\left[\begin{array}{c} 2 e^{t} \\ e^{t} \end{array}\right] $$

Short answer

Answer: Yes, the given functions form a fundamental set of solutions for the linear system. This is because both functions are solutions of the given linear system and their Wronskian is non-zero for all values of t.

Step-by-step solution

Check if the given functions are solutions of the linear system

Plug in the given functions into the linear system to check if they are solutions: For \(\mathbf{y}_{1}(t)\): $$A\cdot\mathbf{y}_{1}(t)=\left[\begin{array}{rr} 2 & -2 \\ 0 & 1 \end{array}\right]\cdot\left[\begin{array}{c} 3 e^{2 t} \\ 0 \end{array}\right]=\left[\begin{array}{c} 6 e^{2 t} - 0 \\ 0 \end{array}\right]=\left[\begin{array}{c} 6 e^{2 t} \\ 0 \end{array}\right]=\frac{d\mathbf{y}_{1}(t)}{dt}$$ For \(\mathbf{y}_{2}(t)\): $$A\cdot\mathbf{y}_{2}(t)=\left[\begin{array}{rr} 2 & -2 \\ 0 & 1 \end{array}\right]\cdot\left[\begin{array}{c} 2 e^{t} \\ e^{t} \end{array}\right]=\left[\begin{array}{c} 4 e^{t} - 2e^{t} \\ e^{t} \end{array}\right]=\left[\begin{array}{c} 2 e^{t} \\ e^{t} \end{array}\right]=\frac{d\mathbf{y}_{2}(t)}{dt}$$ So both functions are solutions of the given linear system.

Calculate the Wronskian of the given functions

The Wronskian of the two functions is given by: $$W(\mathbf{y}_{1}, \mathbf{y}_{2})(t) = \begin{vmatrix} 3 e^{2 t} & 2 e^{t} \\ 0 & e^{t} \end{vmatrix} =(3 e^{2 t})(e^{t}) - (0)(2 e^{t}) = 3 e^{3 t}$$

Determine if the given functions form a fundamental set of solutions

Since the Wronskian \(W(\mathbf{y}_{1}, \mathbf{y}_{2})(t) = 3 e^{3 t}\) is non-zero for all \(t\), the given functions form a fundamental set of solutions for the linear system.

Key concepts

Linear Systems

Linear systems involve a set of equations where the unknowns appear only to the first power. They can be understood as finding values that satisfy multiple linear equations simultaneously.
In our given exercise, the system is expressed using matrices and vectors:

• The matrix represents constant coefficients of the system.
• The vector \( \mathbf{y} \) represents the functions involved.

We examine whether a set of functions \( \mathbf{y}_1(t) \) and \( \mathbf{y}_2(t) \) are solutions by substituting them back into the system of equations. If they satisfy the equations at all times \( t \), they are solutions. In this case, both functions are confirmed as solutions.

Wronskian

The Wronskian is a determinant used to test whether a set of functions is linearly independent, which is crucial in establishing a fundamental set of solutions. For two functions \( \mathbf{y}_1(t) \) and \( \mathbf{y}_2(t) \), the Wronskian \( W(t) \) is calculated as follows:\[ W(\mathbf{y}_1, \mathbf{y}_2)(t) = \begin{vmatrix} 3e^{2t} & 2e^t \ 0 & e^t \end{vmatrix} = 3e^{3t} \]A non-zero Wronskian implies that the functions are linearly independent. Since we computed that \( W(t) = 3e^{3t} \) is non-zero for all \( t \), the functions form a linearly independent set. This independence confirms that they form a fundamental set.

Differential Equations

Differential equations involve functions and their derivatives, describing how a system changes over time. In our exercise, we deal with a linear system of differential equations. The solutions involve:

• Finding functions whose derivatives satisfy the equation.
• Verifying solutions by differentiating the functions and matching with the system's equations.

Both functions \( \mathbf{y}_1(t) \) and \( \mathbf{y}_2(t) \) are differentiated and found to match their respective equations within the linear system, confirming their role as solutions to the differential equations.

Linear Algebra

Linear algebra is the mathematical study of vectors and linear transformations, often using matrices. It provides tools for solving systems of equations that are crucial in this exercise.

• Matrices represent linear systems compactly.
• Determinants, like the Wronskian, assess linear independence of solutions.

By using linear algebra techniques, we transformed the problem into a matrix format, checked solution validity through multiplications, and confirmed linear independence using determinants. This approach underpins the ability to solve and interpret systems of differential equations effectively.