Question

In Exercises \(17-24\) solve the initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} 1 & 15 & -15 \\ -6 & 18 & -22 \\ -3 & 11 & -15 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{c} 15 \\ 17 \\ 10 \end{array}\right] $$

Short answer

Based on the step-by-step solution, answer the following question: Question: Find the solution of the initial value problem of the given first-order linear system of ordinary differential equations with the initial condition $\mathbf{y}(0)=\left[\begin{array}{c} 15 \\ 17 \\ 10 \end{array}\right]$. Answer: The solution of the initial value problem is: $$\mathbf{y}(t)=\left[\begin{array}{c} (1+5e^{t}-15e^{2t}) \\ (1+19e^{t}-45e^{2t}) \\ 11e^{t}-30e^{2t} \end{array}\right]$$

Step-by-step solution

Find the matrix exponential of the given matrix

To solve the system of ODEs, we first need to find the matrix exponential \(e^{At}\) of the coefficient matrix A, where $$A=\left[\begin{array}{ccc} 1 & 15 & -15 \\ -6 & 18 & -22 \\ -3 & 11 & -15 \end{array}\right]$$ To find the matrix exponential of A, we follow these steps: 1. Find the eigenvalues and eigenvectors of A. 2. Construct the matrix \(D\) with eigenvalues on its diagonal and the matrix \(P\) whose columns are the eigenvectors. 3. Calculate \(P^{-1}AP=D\). 4. Compute \(e^{Dt}\) and then \(e^{At}=Pe^{Dt}P^{-1}\). Let's find the eigenvalues and eigenvectors of A:

Find eigenvalues and eigenvectors of A

To find the eigenvalues of A, we need to solve the characteristic equation $$\det(A-\lambda I_3)=0$$ where \(I_3\) is the 3x3 identity matrix and \(\lambda\) represents the eigenvalues. The characteristic equation is: $$\left| \begin{array}{ccc} 1-\lambda & 15 & -15 \\ -6 & 18-\lambda & -22 \\ -3 & 11 & -15-\lambda \end{array} \right| =0$$ Solving this equation, we find that the eigenvalues are \(\lambda_1=1\), \(\lambda_2=2\), and \(\lambda_3=2\). Now let's find the eigenvectors corresponding to these eigenvalues. For \(\lambda_1=1\), the eigenvector \(v_1=[\frac{1}{3}, 1, 1]\) (found by solving the linear system \((A-\lambda_1 I_3)v_1=0\)). We can normalize this eigenvector, we find the normalized vector \(u_1=[\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{3}{\sqrt{11}}]\). Note that eigenvalue 2 appears twice; this may result in a single eigenvector or a generalized eigenvector. In this case, we find that there are indeed two distinct eigenvectors: for \(\lambda_2=\lambda_3=2\), we have eigenvectors \(v_2=[1, -1, 0]\) and \(v_3=[-3, -3, 2]\). Normalize them to obtain \(u_2=[\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0]\) and \(u_3=[-\frac{3}{\sqrt{18}}, -\frac{3}{\sqrt{18}}, \frac{2}{\sqrt{18}}]\).

Construct matrices D and P

Now that we have the eigenvalues and eigenvectors, we can construct the matrices D and P. The matrix D contains the eigenvalues in the diagonal, and P has the normalized eigenvectors as its columns. $$D=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right], \quad P=\left[\begin{array}{ccc} \frac{1}{\sqrt{11}} & \frac{1}{\sqrt{2}} & -\frac{3}{\sqrt{18}} \\ \frac{3}{\sqrt{11}} & -\frac{1}{\sqrt{2}} & -\frac{3}{\sqrt{18}} \\ \frac{3}{\sqrt{11}} & 0 & \frac{2}{\sqrt{18}} \end{array}\right]$$

Calculate matrix exponential of A

Calculate the inverse of P (\(P^{-1}\)), the matrix exponential of D (\(e^{Dt}\)), and finally the matrix exponential of A (\(e^{At}\)) using the relation \(e^{At}=Pe^{Dt}P^{-1}\): $$P^{-1} = \left[\begin{array}{ccc} \frac{1}{3} & 1 & -1 \\ -1 & -1 & 0 \\ 1 & 1 & 2 \end{array}\right]$$ $$e^{Dt} = \left[\begin{array}{ccc} e^t & 0 & 0 \\ 0 & e^{2t} & 0 \\ 0 & 0 & e^{2t} \end{array}\right]$$ Multiplying \(P\), \(e^{Dt}\), and \(P^{-1}\), we obtain: $$e^{At} = \left[\begin{array}{ccc} 1+2e^{t}-2e^{2t} & -3e^{t}+3e^{2t} & 3e^{t}-3e^{2t} \\ 3e^{t}-6e^{2t} & 1+9e^{t}-9e^{2t} & -9e^{t}+9e^{2t} \\ 3e^{t}-6e^{2t} & -e^{t}+e^{2t} & 1+5e^{t}-5e^{2t} \end{array}\right]$$ Now that we have the matrix exponential \(e^{At}\), we can obtain the general solution of the system as \(\mathbf{y}(t) = e^{At}\mathbf{y}(0)\).

Apply the initial condition

Applying the given initial condition $\mathbf{y}(0)=\left[\begin{array}{c} 15 \\ 17 \\ 10 \end{array}\right]$, we have: $$\mathbf{y}(t) = e^{At}\mathbf{y}(0)=\left[\begin{array}{ccc} 1+2e^{t}-2e^{2t} & -3e^{t}+3e^{2t} & 3e^{t}-3e^{2t} \\ 3e^{t}-6e^{2t} & 1+9e^{t}-9e^{2t} & -9e^{t}+9e^{2t} \\ 3e^{t}-6e^{2t} & -e^{t}+e^{2t} & 1+5e^{t}-5e^{2t} \end{array}\right] \left[\begin{array}{c} 15 \\ 17 \\ 10 \end{array}\right]$$ Computing the matrix-vector product, we get the following expression for the solution: $$\mathbf{y}(t)=\left[\begin{array}{c} (1+5e^{t}-15e^{2t}) \\ (1+19e^{t}-45e^{2t}) \\ 11e^{t}-30e^{2t} \end{array}\right]$$

Key concepts

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations that relate functions of one variable to their derivatives. They describe how a function changes, making them essential tools in modeling dynamic systems in fields like physics, biology, and engineering. For example, ODEs can describe the motion of planets, the decay of radioactive materials, or the growth of populations. In the context of this exercise, we are dealing with a system of ODEs, which involves multiple interdependent differential equations that have to be solved simultaneously.

• In our problem, we have a system described by the matrix equation \( \mathbf{y}^{\prime} = A \mathbf{y} \)
• Here, \( \mathbf{y} \) is a vector of functions we want to determine, and \( A \) is a constant matrix known as the coefficient matrix.
• Such systems are typically solved using techniques like diagonalization or the matrix exponential method.

This particular exercise asks us to apply these techniques to solve the initial value problem, providing the solution vector as a function of time \( t \).

Matrix Exponential

The matrix exponential is a way to generalize the idea of exponentiation from numbers to matrices. It plays a crucial role when solving linear systems of differential equations. For a square matrix \( A \), the matrix exponential is defined by the power series:

• \(e^{At} = I + At + \frac{A^2t^2}{2!} + \frac{A^3t^3}{3!} + \cdots \), where \( I \) is the identity matrix.

This infinite series doesn't just depend on the matrix itself but also on its eigenvalues and eigenvectors.

• When the matrix can be diagonalized as \( A = PDP^{-1} \), calculating the matrix exponential simplifies significantly.
• We compute \( e^{Dt} \), a diagonal matrix where each diagonal entry is an exponential function, and then find \( e^{At} = Pe^{Dt}P^{-1} \).

Through this method, we acquire a tool to find solutions to systems governed by linear ODEs, showing the state of the system at any time \( t \).

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are central to understanding linear transformations and systems of linear equations. An eigenvalue \( \lambda \) of a matrix \( A \) is a scalar such that, when \( A \) is multiplied by its corresponding eigenvector \( v \), the result is a vector scaled by \( \lambda \):

• \( Av = \lambda v \)

Finding eigenvalues involves solving the characteristic equation, \( \det(A - \lambda I) = 0 \), which is a polynomial equation in \( \lambda \). Each eigenvalue corresponds to one or more eigenvectors, which are found by solving \( (A - \lambda I)x = 0 \).

• They provide a basis for diagonalization, making calculations like matrix exponentials more manageable.
• In our exercise, the matrix \( A \) was found to have eigenvalues \( \lambda_1 = 1 \), and \( \lambda_2 = \lambda_3 = 2 \), with corresponding eigenvectors.

This knowledge allows the matrix to be expressed in a diagonalizable form, simplifying the computation of the matrix exponential and helping solve the system of ODEs.

Initial Value Problem

An initial value problem involves solving a differential equation under a given set of conditions at a particular point, usually at time \( t = 0 \). It is a critical concept because it specifies a unique solution to a differential equation based on the initial conditions provided.

• The initial condition in our problem is \( \mathbf{y}(0) = \left[\begin{array}{c} 15 \ 17 \ 10 \end{array}\right] \).
• This tells us where the system state is at time zero, allowing us to determine the system's evolution over time.

To solve an initial value problem for a system of ODEs, we find the general solution (using tools like the matrix exponential) and substitute the initial conditions to determine any unknown constants. The result is a specific solution that satisfies both the differential equation and the initial condition constraints.

• For our problem, after computing the matrix exponential, we multiply it by the initial condition vector to get the actual solution \( \mathbf{y}(t) \), which describes the system behavior for all \( t \geq 0 \).

Understanding this concept ensures that solutions to differential equations are not just general but tailored to specific scenarios dictated by initial data.