Question

Solve the initial value problem. \(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}3 & -1 & 0 \\ 4 & -2 & 0 \\ 4 & -4 & 2\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}7 \\ 10 \\ 2\end{array}\right]\)

Short answer

Question: Determine the solution to the initial value problem for the system of linear differential equations with the given matrix and initial condition. Matrix: \[\begin{bmatrix} 3 & -1 & 0\\ 4 & -2 & 0\\ 4 & -4 & 2\end{bmatrix}\] Initial Condition: \[\mathbf{y}(0)=\begin{bmatrix} 7\\ 10\\ 2\end{bmatrix}\] Answer: The solution to the initial value problem is: \[\mathbf{y}(t) = 3e^{t}\begin{bmatrix} 1\\ 2\\ 0\end{bmatrix}+2e^{t}\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}+4e^{2t}\begin{bmatrix} 1\\ 1\\ 0\end{bmatrix}\]

Step-by-step solution

Find the eigenvalues and eigenvectors

To find the eigenvalues of the matrix, we need to solve the characteristic equation: \[|A - \lambda I| = 0\] Where A is the given matrix, \(\lambda\) is the eigenvalue, and I is the identity matrix. Thus, the characteristic equation of the given matrix is, \[\left|\begin{array}{ccc} 3-\lambda & -1 & 0 \\\ 4 & -2-\lambda & 0 \\\ 4 & -4 & 2-\lambda \end{array}\right|=0\] Expanding the determinant, we get \[(3-\lambda)((-2 -\lambda)(2 - \lambda)) - (-1)(4(2 - \lambda)) = 0\] Simplifying the equation and obtaining the eigenvalues, \[\lambda_1 = 1, \lambda_2 = 1, \lambda_3 = 2\] Next, we find the eigenvectors corresponding to these eigenvalues by solving the equation, \[(A - \lambda I)\mathbf{v} = 0\] For \(\lambda_1 = 1\), we have \[\begin{bmatrix} 2 & -1 & 0\\ 4 & -3 & 0\\ 4 & -4 & 1\end{bmatrix}\mathbf{v}_1 = 0\] Solving the above system, we get \[\mathbf{v}_1 = \begin{bmatrix} 1\\ 2\\ 0\end{bmatrix}\] For \(\lambda_2 = 1\), we have \[\begin{bmatrix} 2 & -1 & 0\\ 4 & -3 & 0\\ 4 & -4 & 1\end{bmatrix}\mathbf{v}_2 = 0\] Solving the above system, we get \[\mathbf{v}_2 = \begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}\] For \(\lambda_3 = 2\), we have \[\begin{bmatrix} 1 & -1 & 0\\ 4 & -4 & 0\\ 4 & -4 & 0\end{bmatrix}\mathbf{v}_3 = 0\] Solving the above system, we get \[\mathbf{v}_3 = \begin{bmatrix} 1\\ 1\\ 0\end{bmatrix}\]

Form general solution for the system of differential equations

Using the eigenvalues and eigenvectors, we write the general solution of the system of differential equations as: \[\mathbf{y}(t) = C_1 e^{\lambda_1 t}\mathbf{v}_1 + C_2 e^{\lambda_2 t}\mathbf{v}_2 + C_3 e^{\lambda_3 t}\mathbf{v}_3\] Now, substitute the eigenvalues and eigenvectors into the above equation, \[\mathbf{y}(t) = C_1 e^{t}\begin{bmatrix} 1\\ 2\\ 0\end{bmatrix}+C_2 e^{t}\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}+C_3 e^{2t}\begin{bmatrix} 1\\ 1\\ 0\end{bmatrix}\]

Apply Initial Conditions

We are given the initial condition, \(\mathbf{y}(0)=\begin{bmatrix} 7\\ 10\\ 2\end{bmatrix}\). Substitute \(t=0\) into the general solution and set the result equal to the initial condition vector: \[\mathbf{y}(0) = C_1\begin{bmatrix} 1\\ 2\\ 0\end{bmatrix}+C_2 \begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}+C_3 \begin{bmatrix} 1\\ 1\\ 0\end{bmatrix}\] Setting this vector equal to the initial condition vector, we get the following system of equations: \[C_1+C_3=7\] \[2C_1+C_3=10\] \[C_2=2\] Solving the system above, we find the values of constants: \[C_1=3, \quad C_2=2, \quad C_3=4\]

Write the final solution

Using the values of constants \(C_1, C_2,\) and \(C_3\), the solution to the initial value problem is: \[\mathbf{y}(t) = 3e^{t}\begin{bmatrix} 1\\ 2\\ 0\end{bmatrix}+2e^{t}\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}+4e^{2t}\begin{bmatrix} 1\\ 1\\ 0\end{bmatrix}\]

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra that play a pivotal role in the analysis of linear transformations. They are particularly important when dealing with linear differential equations.

In the context of an initial value problem for a system of differential equations, eigenvalues are special numbers associated with a square matrix, which in this case is the matrix from the differential equation. An eigenvalue, denoted by \( \lambda \), signifies the factor by which an eigenvector is scaled during the transformation described by the matrix.

An eigenvector is a nonzero vector that changes at most by a scalar factor when that linear transformation is applied. In other words, if \( A \) is the matrix and \( \mathbf{v} \) is the eigenvector, after the transformation, \( A\mathbf{v} \) will be equivalent to \( \lambda\mathbf{v} \). The direction of the eigenvector remains unchanged, although its magnitude may be stretched or shrunk.

Finding Eigenvalues and Eigenvectors

To find them, a characteristic equation of the form \( |A - \lambda I| = 0\) is solved, where \( A \) is the matrix in question, and \( I \) is the identity matrix. Solving this equation provides us with eigenvalues, and substituting these into the equation \( (A - \lambda I)\mathbf{v} = 0\) yields the corresponding eigenvectors.

For our initial value problem, once we have both eigenvalues and eigenvectors, we can construct the solution to the system based on these parameters. The eigenvalues dictate the behaviour of the solution, causing exponential growth or decay, while eigenvectors determine the direction in which the growth or decay occurs.

System of Differential Equations

A system of differential equations consists of multiple equations that relate the derivatives of several unknown functions. These systems appear frequently in engineering, physics, and other scientific disciplines where multiple related quantities change over time or space.

In the case of the given initial value problem, we are working with a system that can be expressed in matrix notation as \( \mathbf{y}^\prime = A\mathbf{y} \), where \( A \) is a square matrix containing the coefficients of the system, and \( \mathbf{y} \) is a vector composed of the functions that we are attempting to solve for.

The overall solution of the system is based on the individual solutions corresponding to each eigenvalue and eigenvector pair. For each distinct eigenvalue, there will be a corresponding term of the form \( e^{\lambda t}\mathbf{v} \) in the general solution, where \( e^{\lambda t} \) represents the exponential function indicating the rate and direction of change, while \( \mathbf{v} \) is the eigenvector that preserves the direction.

General Solution

The general solution incorporates these terms along with constants that are later determined by the initial conditions. These initial conditions integrate the specific circumstances of the problem, in this case, the initial state of our system quantified by the vector \( \mathbf{y}(0) \). The goal is to find the particular solution satisfying the differential equations and the initial conditions simultaneously.

Characteristic Equation

The characteristic equation is a critical step in finding the eigenvalues of a matrix, which are integral to solving a system of differential equations. This equation arises from the determinant equation \( |A - \lambda I| = 0\), representing a condition required for the matrix \( A \) to have a non-trivial solution for the eigenvector corresponding to the eigenvalue \( \lambda \).

For our problem, the characteristic polynomial is obtained by calculating the determinant of the matrix \( A - \lambda I \) and setting it to zero, which results in an algebraic equation in terms of \( \lambda \). Solving this polynomial yields the eigenvalues.

Simplifying the Characteristic Equation

While the determinant can be expanded using various methods such as cofactor expansion, for larger matrices, Leverrier-Faddeev's algorithm or other determinant simplifying techniques might be applied. The roots of the characteristic equation (eigenvalues) are then used to find the corresponding eigenvectors. Together, these provide the modal matrix or the ingredients to form the general solution to a system of linear constant coefficient differential equations.