Question

Solve the initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{cc} -7 & 24 \\ -6 & 17 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l} 3 \\ 1 \end{array}\right] $$

Short answer

Question: Solve the initial value problem for the given system: $$ \frac{d\mathbf{y}}{dt} = \left[\begin{array}{cc} -7 & 24 \\ -6 & 17 \end{array}\right] \mathbf{y}, \hspace{1cm} \mathbf{y}(0) = \left[\begin{array}{c} 3 \\ 1\end{array}\right] $$ Answer: The solution to the given initial value problem is: $$ \mathbf{y}(t) = -2 e^{6 t} \left[\begin{array}{c} -2 \\ 1\end{array}\right] + 3 e^{4t} \left[\begin{array}{c} 3 \\ 1\end{array}\right] $$

Step-by-step solution

Find the eigenvalues and eigenvectors

First, we need to find the eigenvalues and eigenvectors of the given matrix A: $$ A = \left[\begin{array}{cc} -7 & 24 \\ -6 & 17 \end{array}\right] $$ To find the eigenvalues, compute the characteristic equation given by the determinant of \((A - \lambda I)\): $$ \begin{vmatrix} -7-\lambda & 24 \\ -6 & 17-\lambda \end{vmatrix} = (-7 - \lambda)(17 - \lambda) - (24)(-6) = \lambda^2 -10\lambda -144 $$ Solve for the eigenvalues, \(\lambda_1\) and \(\lambda_2\): $$ \lambda^2 -10\lambda -144 = 0 $$ We find that the eigenvalues are \(\lambda_1 = 6\) and \(\lambda_2 = 4\). Next, we must find the eigenvectors associated with these eigenvalues. For each eigenvalue, solve the equation \((A - \lambda_i I)\mathbf{x}_i = 0\). For \(\lambda_1 = 6\): $$ \left[\begin{array}{cc} -13 & 24 \\ -6 & 11 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] = 0 $$ We find the eigenvector \(\mathbf{x}_1 = \left[\begin{array}{c} -2 \\ 1\end{array}\right]\). For \(\lambda_2 = 4\): $$ \left[\begin{array}{cc} -11 & 24 \\ -6 & 13 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] = 0 $$ We find the eigenvector \(\mathbf{x}_2 = \left[\begin{array}{c} 3 \\ 1\end{array}\right]\).

Determine the general expression for the solution

Now that we have the eigenvalues and eigenvectors, we can write a general expression for the solution: $$ \mathbf{y}(t) = C_1 e^{\lambda_1 t} \mathbf{x}_1 + C_2 e^{\lambda_2 t} \mathbf{x}_2 $$ Substitute the eigenvalues and eigenvectors found in step 1: $$ \mathbf{y}(t) = C_1 e^{6 t} \left[\begin{array}{c} -2 \\ 1\end{array}\right] + C_2 e^{4 t} \left[\begin{array}{c} 3 \\ 1\end{array}\right] $$

Use the initial values to find the constants

Given the initial condition, \(\mathbf{y}(0) = \left[\begin{array}{c} 3 \\ 1\end{array}\right]\), we can find the constants \(C_1\) and \(C_2\): $$ \left[ \begin{array}{cc} -2 & 3 \\ 1 & 1 \end{array}\right] \left[\begin{array}{c} C_1 \\ C_2 \end{array}\right] = \left[\begin{array}{c} 3 \\ 1 \end{array}\right] $$ Solving for \(C_1\) and \(C_2\), we get \(C_1 = -2\) and \(C_2 = 3\) Finally, substitute \(C_1\) and \(C_2\) back into the solution: $$ \mathbf{y}(t) = -2 e^{6 t} \left[\begin{array}{c} -2 \\ 1\end{array}\right] + 3 e^{4t} \left[\begin{array}{c} 3 \\ 1\end{array}\right] $$

Key concepts

Eigenvalues

In linear algebra, eigenvalues are fundamental in understanding the properties of matrices. When we have a matrix, it can be transformed into a simpler form using its eigenvalues. To find the eigenvalues, calculate the determinant of matrix \( A \) when subtracted by \( \lambda I \) (where \( I \) is the identity matrix). For our problem, the given matrix is:

• \( A = \left[ \begin{array}{cc} -7 & 24 \ -6 & 17 \end{array} \right] \)

The characteristic equation is obtained by setting the determinant of \( A - \lambda I \) to zero:

• \( \det(A - \lambda I) = 0 \)

Thus, \[(-7 - \lambda)(17 - \lambda) - (24)(-6) = \lambda^2 - 10\lambda - 144 = 0\]Solving this quadratic equation gives us the eigenvalues, \( \lambda_1 = 6 \) and \( \lambda_2 = 4 \). These values are critical as they describe the scale factors along the directions determined by the eigenvectors.

Eigenvectors

Eigenvectors correspond to the eigenvalues and give us insight into the directions where transformations occur unchanged in magnitude. To find an eigenvector, solve \( (A - \lambda I)\mathbf{x} = 0 \). This involves substituting each eigenvalue back into the matrix \( A \) and solving for \( \mathbf{x} \).

For \( \lambda_1 = 6 \):

• Matrix: \( \left[ \begin{array}{cc} -13 & 24 \ -6 & 11 \end{array} \right] \)
• Solution: \( \mathbf{x}_1 = \left[ \begin{array}{c} -2 \ 1 \end{array} \right] \)

For \( \lambda_2 = 4 \):

• Matrix: \( \left[ \begin{array}{cc} -11 & 24 \ -6 & 13 \end{array} \right] \)
• Solution: \( \mathbf{x}_2 = \left[ \begin{array}{c} 3 \ 1 \end{array} \right] \)

These eigenvectors, corresponding to different eigenvalues, point in directions where the system experiences no change in direction but only in magnitude, scaled by the eigenvalues.

Initial Value Problems

Initial value problems are explored in differential equations where a solution is sought that satisfies both the differential equation and a specified initial condition. In our exercise, the equation is

• \( \mathbf{y} ' = A \mathbf{y}, \quad \mathbf{y}(0) = \left[ \begin{array}{c} 2 \ -4 \end{array} \right] \)

This means we need a function \( \mathbf{y}(t) \) that obeys the differential equation at every point \( t \) and matches the initial condition when \( t=0 \). By expressing the solution as a linear combination of the eigenvectors scaled by exponentials of eigenvalues, we form:

• \( \mathbf{y}(t) = C_1 e^{\lambda_1 t} \mathbf{x}_1 + C_2 e^{\lambda_2 t} \mathbf{x}_2 \)
• Substitute known \( \lambda \) and \( \mathbf{x} \) to obtain \( \mathbf{y}(0) = \left[ \begin{array}{c} 2 \ -4 \end{array} \right] \).

Solve for constants \( C_1 \) and \( C_2 \) to satisfy the initial condition. For this problem, \( C_1 = -2 \) and \( C_2 = 3 \) satisfy the initial value.

Matrix Exponentiation

Matrix exponentiation plays a critical role in solving systems of linear differential equations related to initial value problems. It's about raising the matrix \( A \) to the power of \( t \) in terms of exponentiation. The matrix exponential allows for solutions of the form \( e^{At}\), which handle systems like \( \mathbf{y}' = A\mathbf{y} \) efficiently.

For matrices, exponentiation uses the Taylor series expansion conceptually similar to scalar exponentiation. Still, it often leverages the decomposed form of matrix \( A \) using eigenvalues and eigenvectors:

• If \( A \) is diagonalizable, it's expressed as \( A = PDP^{-1} \) where \( D \) is a diagonal matrix.
• Then \( e^{At} = Pe^{Dt}P^{-1} \).

In this problem, without explicit calculation of \( e^{At} \), matrix exponentiation forms the backbone of expressing the solution derived as linear combinations of parts involving \( e^{\lambda_1 t} \) and \( e^{\lambda_2 t} \), multiplied by their corresponding eigenvectors.