Question

In Exercises $17-24$ solve the initial value problem. $${\mathbf{y}}^{\prime }=\left[\begin{array}{ll}4& -6\\ 3& -2\end{array}\right]\mathbf{y},\mathbf{y}(0)=\left[\begin{array}{l}5\\ 2\end{array}\right]$$

Short answer

In this initial value problem, the particular solution for the first-order linear system of ordinary differential equations is given by: $$\mathbf{y}(t)=\left[\begin{array}{l}2\\ 1\end{array}\right]e^{2t}+\left[\begin{array}{l}2\\ 3\end{array}\right].$$

Step-by-step solution

Find the eigenvalues and eigenvectors of the matrix

First, we want to find the eigenvalues of the matrix. For this, we will set up the characteristic equation by calculating the determinant of $(A-\lambda I)$, where $A$ is the given matrix and $\lambda$ is the eigenvalue. Then, we'll solve for the eigenvalues $\lambda$. For our matrix, $$A=\left[\begin{array}{ll}4& -6\\ 3& -2\end{array}\right]$$ The characteristic equation is given by $$\text{det}(A-\lambda I)=\left|\begin{array}{ll}4-\lambda & -6\\ 3& -2-\lambda \end{array}\right|=(4-\lambda )(-2-\lambda )-(-6)(3)=0$$ Solving the quadratic equation, we get the eigenvalues: $${\lambda }_1=2,{\lambda }_2=0.$$ Now, we will find the eigenvectors corresponding to each of these eigenvalues by solving $(A-\lambda I)\mathbf{v}=0.$ For ${\lambda }_1=2,$ $$(A-{\lambda }_{1}I)\mathbf{v}=\left[\begin{array}{ll}2& -6\\ 3& -4\end{array}\right]\mathbf{v}=0$$ We can see the eigenvector is $${\mathbf{v}}_1=\left[\begin{array}{l}2\\ 1\end{array}\right].$$ For ${\lambda }_2=0,$ $$(A-{\lambda }_{2}I)\mathbf{v}=\left[\begin{array}{ll}4& -6\\ 3& -2\end{array}\right]\mathbf{v}=0$$ We can see the eigenvector is $${\mathbf{v}}_2=\left[\begin{array}{l}2\\ 3\end{array}\right].$$

Write the general solution for the system of ODEs

Using the eigenvalues and eigenvectors, we can write the general solution for the system of ODEs as: $$\mathbf{y}(t)=c_1{\mathbf{v}}_1{e}^{{\lambda }_{1}t}+c_2{\mathbf{v}}_2{e}^{{\lambda }_{2}t}=c_1\left[\begin{array}{l}2\\ 1\end{array}\right]e^{2t}+c_2\left[\begin{array}{l}2\\ 3\end{array}\right].$$

Apply the initial condition to find the particular solution

Now, we apply the initial condition $\mathbf{y}(0) = \left[$$\begin{array}{l}5\\ 2\end{array}$$\right]$tofindthecoefficients$c_1$and$c_2$. From the general solution, we have: $$\mathbf{y}(0)=c_1\left[\begin{array}{l}2\\ 1\end{array}\right]e^0+c_2\left[\begin{array}{l}2\\ 3\end{array}\right]=\left[\begin{array}{l}5\\ 2\end{array}\right].$$ Solving this system of linear equations, we get the coefficients: $$c_1=1,c_2=1.$$

Final answer

Therefore, the particular solution for the initial value problem is: $$\mathbf{y}(t)=\left[\begin{array}{l}2\\ 1\end{array}\right]e^{2t}+\left[\begin{array}{l}2\\ 3\end{array}\right].$$

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra that play a crucial role in a wide variety of mathematical applications. An eigenvalue is a scalar that indicates how a linear transformation changes the size of a vector, while an eigenvector is a non-zero vector that only changes in magnitude, not direction, under that linear transformation.

When dealing with a square matrix, like the one in our example problem, finding the eigenvalues involves solving the characteristic equation. The matrix in question represents a linear transformation, and finding its eigenvalues tells us about the invariant points of that transformation - where a vector is scaled rather than rotated. The eigenvectors associated with those eigenvalues are the directions along which these scaling effects occur.

An important property to remember is that different eigenvalues always have linearly independent eigenvectors. This property is essential when solving systems of differential equations, as it allows us to express the general solution as a linear combination of these eigenvectors scaled by exponential functions of time and their corresponding eigenvalues. In the system of differential equations, each eigenvalue-eigenvector pair provides a piece of the puzzle, revealing the system's behavior over time.

Characteristic Equation

The characteristic equation is a polynomial whose roots are the eigenvalues of a matrix. It is derived from the determinant of the matrix subtracted by an unknown scalar, $\lambda$, times the identity matrix. This equation, in the context of the given matrix A, is written as $\text{det}(A-\lambda I)=0$.

In our initial value problem, we found the characteristic equation of a 2x2 matrix. We used the property of determinants to set up a quadratic equation that, when solved, gave us the eigenvalues of the matrix. It's worth noting that the degree of the characteristic equation corresponds to the size of the matrix, and hence a 2x2 matrix gives us a quadratic characteristic equation. Solving the characteristic equation is a critical step because it leads us straight to the eigenvalues, which are key to understanding the system's dynamics and forming the general solution.

System of Differential Equations

A system of differential equations consists of multiple equations involving derivatives of several interrelated variables. These systems frequently appear in modeling real-world phenomena where variables change with respect to one another over time.

In our example, the system was represented in matrix form, and the solution involved finding a general solution that satisfied the differential equations set by that matrix. After obtaining the eigenvalues and corresponding eigenvectors, we used them to construct the general solution. Specifically, the general solution to a linear system of differential equations is a linear combination of terms, each of which is an eigenvector multiplied by the exponential of its corresponding eigenvalue and time. Each term is then multiplied by an arbitrary constant, which is determined by the initial conditions of the problem.

Finally, applying the initial conditions helps us find these constants and refine the general solution into a particular solution that exactly satisfies the given conditions. This step completes the process and provides a concrete solution that can predict the system's behavior at any given time, making the initial value problem completely solvable.