Question

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

Short answer

In this exercise, we solved the initial value problem for a system of two linear ordinary differential equations. We found the eigenvalues and eigenvectors of the given matrix (λ1 = 9 and λ2 = -5) and applied the diagonalization method. The solution to the given initial value problem is: $$ \mathbf{y}(t) = \begin{bmatrix} 5e^{9t} + 5e^{-5t} \\ 2e^{9t} + 4e^{-5t} \end{bmatrix} $$

Step-by-step solution

Find Eigenvalues and Eigenvectors

We are given the matrix A: $$ A = \begin{bmatrix} 7 & -15 \\ 3 & -5 \end{bmatrix} $$ To find the eigenvalues, we need to solve the characteristic equation, which is given by the determinant of (A - λI) and is as follows: $$ \det{(A - λI)} = \det{\begin{bmatrix} 7-\lambda & -15 \\ 3 & -5-\lambda \end{bmatrix}} $$ Solving the above determinant: \((7-\lambda)(-5-\lambda)-(-15)(3) = 0\) Expanding the equation, we get: \(\lambda^2 - 2\lambda - 45 = 0\) Factorizing this equation, we find the eigenvalues λ: \((\lambda - 9)(\lambda + 5) = 0\) Thus, we have λ1 = 9 and λ2 = -5 as eigenvalues. Now, let's find the corresponding eigenvectors. For λ1 = 9: \((A - 9I)\mathbf{v}_1 = \mathbf{0}\) $\begin{bmatrix} -2 & -15 \\ 3 & -14 \end{bmatrix} \mathbf{v}_1 = \mathbf{0}$ From the resulting system of linear equations, we can conclude: \(\mathbf{v}_1 = k\begin{bmatrix} 5 \\ 2 \end{bmatrix}\), where k is a constant. For λ2 = -5: \((A - (-5)I)\mathbf{v}_2 = \mathbf{0}\) $\begin{bmatrix} 12 & -15 \\ 3 & 0 \end{bmatrix} \mathbf{v}_2 = \mathbf{0}$ Again, we get a system of linear equations, and the eigenvector is: \(\mathbf{v}_2 = k\begin{bmatrix} 5 \\ 4 \end{bmatrix}\), where k is a constant.

Diagonalization

Now, we will apply the diagonalization method to rewrite matrix A as a product of 3 matrices. Let P be the matrix whose columns are the eigenvectors of A: $$ P = \begin{bmatrix} 5 & 5 \\ 2 & 4 \end{bmatrix} $$ The diagonal matrix D is given by: $$ D = \begin{bmatrix} 9 & 0 \\ 0 & -5 \end{bmatrix} $$ We also need P inverse, which can be found as follows: $$ P^{-1} = \frac{1}{\det{(P)}} \begin{bmatrix} 4 & -5 \\ -2 & 5 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 4 & -5 \\ -2 & 5 \end{bmatrix} $$

Solve the Initial Value Problem

Now, let's rewrite the system of differential equations using the diagonalization: $$ \mathbf{y}^\prime = A \mathbf{y} = (PDP^{-1})\mathbf{y} $$ Now, let \(\mathbf{z} = P^{-1}\mathbf{y}\). Thus, the system of differential equations becomes: $$ P\mathbf{z}^\prime = D\mathbf{z} $$ It is easy to see that the solution for this system is given by: $$ \mathbf{z}(t) = \begin{bmatrix} e^{9t} & 0 \\ 0 & e^{-5t} \end{bmatrix} \mathbf{z}(0) $$ We know that \(\mathbf{y}(0) = \begin{bmatrix} 17 \\ 7 \end{bmatrix}\) and \(\mathbf{z}(0) = P^{-1}\mathbf{y}(0)\). So, $$ \mathbf{z}(0) =\frac{1}{10}\begin{bmatrix}4 & -5 \\ -2 & 5\end{bmatrix}\begin{bmatrix} 17 \\ 7 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$ Thus, the solution for the initial value problem is given by: $$ \mathbf{y}(t) = P\mathbf{z}(t) = \begin{bmatrix} 5 & 5 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} e^{9t} & 0 \\ 0 & e^{-5t} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$ Multiplying the matrices and simplifying, we get: $$ \mathbf{y}(t) = \begin{bmatrix} 5e^{9t} + 5e^{-5t} \\ 2e^{9t} + 4e^{-5t} \end{bmatrix} $$ So, the solution to the given initial value problem is: $$ \mathbf{y}(t) = \begin{bmatrix} 5e^{9t} + 5e^{-5t} \\ 2e^{9t} + 4e^{-5t} \end{bmatrix} $$

Key concepts

Eigenvalues and Eigenvectors

When tackling a system of differential equations, a fundamental step is to find the eigenvalues and eigenvectors of the matrix associated with the system. The eigenvalues are essentially constants that arise from solving the characteristic equation derived from the matrix.

• To find the eigenvalues, we solve the equation \( \det{(A - \lambda I)} = 0 \), where \( A \) is the given matrix and \( I \) is the identity matrix of the same dimension.
• The solutions to this equation, \( \lambda \), are the eigenvalues.
• For each eigenvalue, the eigenvector can be found by solving \((A - \lambda I)\mathbf{v} = \mathbf{0}\).

In simpler terms, think of eigenvalues as the scaling factors for eigenvectors, which are direction vectors that don't change their course under the transformation associated with the matrix. In this example, we calculated eigenvalues \( 9 \) and \( -5 \), with eigenvectors \([5, 2]^T\) and \([5, 4]^T\), respectively.

Diagonalization of Matrices

Diagonalization is a powerful method in linear algebra used to simplify matrix expressions, specifically beneficial for solving differential equations. Through diagonalization, we express a square matrix \( A \) as the product of three matrices: \( A = PDP^{-1} \).

• \( P \) is composed of the eigenvectors.
• \( D \) is a diagonal matrix containing eigenvalues on its diagonal.
• \( P^{-1} \) is the inverse of the matrix \( P \).

Why diagonalize? Diagonal matrices are simpler to handle, particularly when raising them to a power, as it involves simply exponentiation of the elements along the diagonal. In our example, the matrix \( A \) was diagonalized as \( P = \begin{bmatrix}5 & 5 \ 2 & 4\end{bmatrix}, D = \begin{bmatrix}9 & 0 \ 0 & -5\end{bmatrix}, and P^{-1} \), simplifying the task of solving a system of differential equations.

System of Differential Equations

A system of differential equations involves finding an unknown function that satisfies multiple differential equations simultaneously. These systems often model natural or engineered systems where various factors change over time. Solving them can be complex, but using matrix methods like diagonalization helps simplify the process greatly.

• The system \( \mathbf{y}^\prime = A \mathbf{y} \) can be transformed by introducing a variable \( \mathbf{z} = P^{-1}\mathbf{y} \), leading to an equivalent but simplified form.
• Using \( \mathbf{z}(t) = e^{Dt} \mathbf{z}(0) \), where \( e^{Dt} \) is an exponentiated diagonal matrix, allows for straightforward solutions.
• Back-transforming \( \mathbf{z}(t) \) gives solutions in terms of \( \mathbf{y}(t) \), readily interpretable in context.

In practice, this means we express the solution as \( \mathbf{y}(t) = P e^{Dt} \mathbf{z}(0) \), leading directly to the answer. For the problem at hand, this method simplifies and yields the solution \(\mathbf{y}(t) = \begin{bmatrix}5e^{9t} + 5e^{-5t} \ 2e^{9t} + 4e^{-5t}\end{bmatrix} \), which adheres to the initial conditions. It's a systematic way to approach what might otherwise seem a daunting task.