Question

In Exercises \(17-24\) solve the initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 7 & 15 \\ -3 & 1 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l} 5 \\ 1 \end{array}\right] $$

Short answer

Question: Solve the initial value problem of the first-order linear differential equation for the vector function \(\mathbf{y}(t)\) with the matrix A = \(\left[\begin{array}{rr} 7 & 15 \\ -3 & 1 \end{array}\right]\) and the initial condition \(\mathbf{y}(0) = \left[\begin{array}{c} 5 \\ 1 \end{array}\right]\). Answer: The particular solution of the initial value problem is \(\mathbf{y}(t) = \frac{2}{5} e^{(4+2i)t}\left[\begin{array}{c}3-2i\\1\end{array}\right] + \frac{3}{5} e^{(4-2i)t}\left[\begin{array}{c}3+2i\\1\end{array}\right]\).

Step-by-step solution

Find the eigenvalues of the matrix A

We are given the matrix A: \(\left[\begin{array}{rr} 7 & 15 \\ -3 & 1 \end{array}\right]\). To find the eigenvalues, we must compute the determinant of \(|A - \lambda I|\) and set it equal to 0, where \(\lambda\) is the eigenvalue and \(I\) is the identity matrix. $$ |A - \lambda I| = \begin{vmatrix} 7 - \lambda & 15 \\ -3 & 1 - \lambda \end{vmatrix} = (7 - \lambda)(1 - \lambda) + 45 $$ Now set it equal to 0 and find the eigenvalues: $$ (7 - \lambda)(1 - \lambda) + 45 = 0 $$

Solve for the eigenvalues

Find the roots of the above equation to get the eigenvalues: $$ \lambda^2 - 8\lambda + 52 = 0 $$ The eigenvalues are \(\lambda_1 = 4 + 2i\) and \(\lambda_2 = 4 - 2i\).

Find the eigenvectors corresponding to each eigenvalue

For each eigenvalue, we will find an eigenvector by solving the equation \((A-\lambda I) \mathbf{v} = 0\). For \(\lambda_1 = 4 + 2i\): $$ (A- (4 + 2i)I) \mathbf{v}=\left[\begin{array}{cc} 3-2i& 15\\-3&-3-2i\end{array}\right] \mathbf{v} =0 $$ Choose one equation (for example, the first one), since they will be linearly dependent: $$ (3-2i)v_1+15v_2=0 $$ Choose an arbitrary value for \(v_2\) (for example, \(v_2=1\)), and then calculate \(v_1\). The eigenvector corresponding to \(\lambda_1\) is \(\mathbf{v}_{1} = \left[\begin{array}{c} 3-2i \\ 1 \end{array}\right]\). For \(\lambda_2 = 4 - 2i\): $$ (A- (4 - 2i)I) \mathbf{v}=\left[\begin{array}{cc} 3+2i& 15\\-3&-3+2i\end{array}\right] \mathbf{v} =0 $$ Choose one equation (for example, the first one), since they will be linearly dependent: $$ (3+2i)v_1+15v_2=0 $$ Choose an arbitrary value for \(v_2\) (for example, \(v_2=1\)), and then calculate \(v_1\). The eigenvector corresponding to \(\lambda_2\) is \(\mathbf{v}_{2} = \left[\begin{array}{c} 3+2i \\ 1 \end{array}\right]\).

Find the general solution

The general solution to a linear system of differential equations is given by: $$ \mathbf{y}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_{1} + c_2 e^{\lambda_2 t}\mathbf{v}_{2} $$ where \(c_1\) and \(c_2\) are constants. In this case: $$ \mathbf{y}(t) = c_1 e^{(4+2i)t}\left[\begin{array}{c}3-2i\\1\end{array}\right] + c_2 e^{(4-2i)t}\left[\begin{array}{c}3+2i\\1\end{array}\right] $$

Apply the initial condition to find the particular solution

The initial condition is \(\mathbf{y}(0) = \left[\begin{array}{c} 5 \\ 1 \end{array}\right]\). Substitute this into the general solution and solve for the constants \(c_1\) and \(c_2\): $$ \left[\begin{array}{c} 5 \\ 1 \end{array}\right]= c_1 \left[\begin{array}{c}3-2i\\1\end{array}\right] + c_2 \left[\begin{array}{c}3+2i\\1\end{array}\right] $$ We get \(c_1 = \frac{2}{5}\) and \(c_2 = \frac{3}{5}\). Finally, the particular solution of the initial value problem is: $$ \mathbf{y}(t) = \frac{2}{5} e^{(4+2i)t}\left[\begin{array}{c}3-2i\\1\end{array}\right] + \frac{3}{5} e^{(4-2i)t}\left[\begin{array}{c}3+2i\\1\end{array}\right] $$

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra with vast applications, notably in solving systems of linear differential equations. Every square matrix has a set of scalar values known as eigenvalues. An eigenvalue, denoted typically by \(\lambda\), is a value for which the matrix equation \(A\mathbf{v} = \lambda\mathbf{v}\) holds true, where \(A\) is a matrix and \(\mathbf{v}\) is a non-zero vector, referred to as the eigenvector.

To find the eigenvalues of a given matrix, a characteristic equation is formed by setting the determinant of \(A - \lambda I\) to zero, where \(I\) is the identity matrix. Solving this equation yields the eigenvalues, and for each eigenvalue, its corresponding eigenvector can be found. These eigenvectors are special because they only scale under the transformation by the matrix and do not change direction. This property makes them particularly useful in simplifying linear transformations.

System of Differential Equations

A system of differential equations is a set of two or more equations involving derivatives of dependent variables with respect to one or more independent variables. These systems can often be expressed in matrix form, especially when dealing with linear differential equations. The solutions to such a system reveal how a set of related variables evolve over time or space when they depend on one another.

In the context of linear algebra, the solutions to a linear system of differential equations can be expressed using eigenvalues and eigenvectors of the associated matrix, especially when dealing with homogeneous systems. The general solution is a linear combination of terms formed by multiplying eigenvectors by exponential functions of eigenvalues and independent variables (e.g., time). The aforementioned process is incredibly efficient when dealing with systems that otherwise would be cumbersome to solve by traditional methods.

Matrix Determinant

The determinant is a scalar value that can be calculated from the elements of a square matrix. The determinant provides important information about the matrix, such as whether it is invertible or not, and plays a critical role in solving systems of linear equations, evaluating eigenvalues, and determining volume changes under the linear transformation described by the matrix.

In the context of differential equations, the determinant is used to construct the characteristic polynomial to find eigenvalues. The determinant of the matrix \(A - \lambda I\) is set to zero, which leads to an equation whose solutions are the eigenvalues. The determinant can be visualized as a scaling factor for volume when the matrix is considered as a transformation of space. A determinant of zero implies that the matrix will flatten space into a lower dimension, eliminating the possibility of inverses and unique solutions.

Linear Algebra in Differential Equations

Linear algebra provides a robust framework for addressing and solving differential equations. When these differential equations are linear and their coefficients are constant, they can be represented and analyzed as matrix equations. The intertwining of linear algebra with differential equations simplifies problems by transforming a system of differential equations into an algebraic one using matrices.

The solution to such problems involves finding eigenvalues and eigenvectors, constructing the general solution of the system, and then applying any initial conditions to find a particular solution. The use of matrices allows for the concise representation of the system and frequently enables the use of computers for solving complicated differential equations. This interplay is a quintessential example of how abstract algebraic concepts have direct application in understanding dynamic systems.