Question

In Exercises $17-24$ solve the initial value problem. $${\mathbf{y}}^{\prime }=\left[\begin{array}{rrr}4& 4& 0\\ 8& 10& -20\\ 2& 3& -2\end{array}\right]\mathbf{y},\mathbf{y}(0)=\left[\begin{array}{l}8\\ 6\\ 5\end{array}\right]$$

Short answer

To solve the given initial-value problem, we first found the eigenvalues and eigenvectors of the given matrix. Then, we used them to form the general solution, which consists of three terms with different factors of eigenvectors and exponential functions. Finally, we used the initial condition to find the particular solution, which is a linear combination of these three terms with specific constants. The particular solution we found is: $$\mathbf{y}(t)=\frac{7}{5}{e}^{8t}\left[\begin{array}{r}5\\ 2\\ 1\end{array}\right]+\frac{17}{5}{e}^{2t}\left[\begin{array}{r}-1\\ 1\\ 0\end{array}\right]+\frac{3}{5}{e}^{2t}\left[\begin{array}{r}5\\ 0\\ 1\end{array}\right].$$

Step-by-step solution

Find the eigenvalues and eigenvectors of the given matrix

By examining the given matrix, we find that it is a $3\times 3$ matrix: $$A=\left[\begin{array}{rrr}4& 4& 0\\ 8& 10& -20\\ 2& 3& -2\end{array}\right].$$ To find its eigenvalues, set the determinant of $A-\lambda I$ equal to zero and solve for $\lambda$. Here $I$ is the $3\times 3$ identity matrix. $$det(A-\lambda I)=det\left(\left[\begin{array}{ccc}4-\lambda & 4& 0\\ 8& 10-\lambda & -20\\ 2& 3& -2-\lambda \end{array}\right]\right)$$ By calculating the determinant, we obtain the characteristic equation: $$(\lambda -2{)}^2(\lambda -8)=0$$ So, we find three eigenvalues: ${\lambda }_1=8$, ${\lambda }_2=2$, and ${\lambda }_3=2$. Now, we should find the corresponding eigenvectors for each eigenvalue. For ${\lambda }_1=8$, set up the augmented matrix for the system $(A-8I)\mathbf{v}=\mathbf{0}$: $$\left[\begin{array}{rrrr}-4& 4& 0& 0\\ 8& 2& -20& 0\\ 2& 3& -10& 0\end{array}\right].$$ By row operations, we get to the reduced row-echelon form: $$\left[\begin{array}{rrrr}1& 0& -5& 0\\ 0& 1& -2& 0\\ 0& 0& 0& 0\end{array}\right].$$ The eigenvector for ${\lambda }_1=8$ is: $${\mathbf{v}}_1=\left[\begin{array}{r}5\\ 2\\ 1\end{array}\right].$$ For ${\lambda }_2=2$, and since this eigenvalue has multiplicity 2, we should find at least two linearly independent eigenvectors by solving the system $(A-2I)\mathbf{v}=\mathbf{0}$. $$\left[\begin{array}{rrrr}2& 4& 0& 0\\ 8& 8& -20& 0\\ 2& 3& -4& 0\end{array}\right].$$ Again, by row operations, we reach the reduced row-echelon form: $$\left[\begin{array}{rrrr}1& 1& -5& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right].$$ From this, we can find two linearly independent eigenvectors for ${\lambda }_2=2$: $${\mathbf{v}}_2=\left[\begin{array}{r}-1\\ 1\\ 0\end{array}\right],{\mathbf{v}}_3=\left[\begin{array}{r}5\\ 0\\ 1\end{array}\right].$$

Form the general solution

Now that we have found the eigenvalues and eigenvectors, we can use them to express the general solution of the system as follows: $$\mathbf{y}(t)=c_1{e}^{8t}{\mathbf{v}}_1+c_2{e}^{2t}{\mathbf{v}}_2+c_3{e}^{2t}{\mathbf{v}}_3,$$ where $c_1,c_2,$ and $c_3$ are constants we need to determine from the initial condition.

Find the particular solution using the initial condition

The initial condition is given as: $\mathbf{y}(0)=\left[\begin{array}{l}8\\ 6\\ 5\end{array}\right]$. Plugging $t=0$ into the general solution, we get: $$\mathbf{y}(0)=c_1{e}^0{\mathbf{v}}_1+c_2{e}^0{\mathbf{v}}_2+c_3{e}^0{\mathbf{v}}_3=c_1{\mathbf{v}}_1+c_2{\mathbf{v}}_2+c_3{\mathbf{v}}_3,$$ Substituting the initial condition and the eigenvectors yields: $$\left[\begin{array}{l}8\\ 6\\ 5\end{array}\right]=c_1\left[\begin{array}{r}5\\ 2\\ 1\end{array}\right]+c_2\left[\begin{array}{r}-1\\ 1\\ 0\end{array}\right]+c_3\left[\begin{array}{r}5\\ 0\\ 1\end{array}\right].$$ Solving the system of equations, we obtain $c_1=\frac{7}{5},c_2=\frac{17}{5},$ and $c_3=\frac{3}{5}$.

Write down the particular solution

Using the constants $c_1,c_2,$ and $c_3$, we can write the particular solution of the given system as: $$\mathbf{y}(t)=\frac{7}{5}{e}^{8t}\left[\begin{array}{r}5\\ 2\\ 1\end{array}\right]+\frac{17}{5}{e}^{2t}\left[\begin{array}{r}-1\\ 1\\ 0\end{array}\right]+\frac{3}{5}{e}^{2t}\left[\begin{array}{r}5\\ 0\\ 1\end{array}\right].$$ This is the solution to the given initial-value problem.

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors play a crucial role in solving systems of differential equations, especially when dealing with linear transformations represented by matrices. Eigenvalues are scalars associated with a matrix, and they provide essential insights into the matrix's properties.

An eigenvalue, often denoted by $\lambda$, is a value for which a non-zero eigenvector exists to satisfy the equation $A\mathbf{v}=\lambda \mathbf{v}$, where $\mathbf{\text{v}}$ is the eigenvector and $\mathbf{\text{A}}$ is the matrix in question. The said eigenvector is then a vector that remains parallel to its original direction under the corresponding linear transformation.

To identify these values, we solve the characteristic equation, which arises from setting the determinant of $A-\lambda I$ to zero. Here, $I$ represents the identity matrix of the same order as $A$. Eigenvalues are indicative of the behavior of the system of differential equations, and their corresponding eigenvectors lay the groundwork for the system's solution.

Characteristic Equation

The characteristic equation is a polynomial equation derived from the matrix representing a system of linear equations. This equation is central to understanding the system's dynamics.

To form the characteristic equation of a matrix $A$, you calculate the determinant of $A-\lambda I$, and set it equal to zero. In the case of a $3\times 3$ matrix, solving the resulting cubic equation yields the eigenvalues of the matrix. As demonstrated in the exercise, the characteristic equation for the given matrix is $(\lambda -2{)}^2(\lambda -8)=0$. It reveals the eigenvalues ${\lambda }_1=8$, and ${\lambda }_2=2$ (with a multiplicity of 2).

The roots of the characteristic equation, which are the eigenvalues, inform us about the nature of the system's equilibria and its long-term behavior, which is key to predicting the system's responses over time. It's the foundation from which the behavior of the system can be extrapolated.

Systems of Differential Equations

Systems of differential equations are sets of related functions and their derivatives. They often arise when multiple quantities and their rates of change are interdependent. In the physical sciences and engineering, systems of differential equations are indispensable tools for modeling dynamic processes.

When faced with a linear system of differential equations, one approach to finding a solution involves the use of eigenvalues and eigenvectors. Given a set of eigenvalues and their corresponding eigenvectors, we can express the system's general solution as a linear combination of terms that each involve an exponential function of time, multiplied by an eigenvector. In mathematical terms, for a system represented by matrix $A$, the general solution takes the form $\mathbf{y}(t)=c_1{e}^{{\lambda }_{1}t}{\mathbf{v}}_1+c_2{e}^{{\lambda }_{2}t}{\mathbf{v}}_2+\cdots +c_n{e}^{{\lambda }_{n}t}{\mathbf{v}}_n$, where $c_i$ are constants to be determined from initial conditions, ${\lambda }_i$ are the eigenvalues, and ${\mathbf{v}}_i$ are the respective eigenvectors.

Applying initial conditions allows us to solve the system for a particular solution, specifying how the system evolves over time from a defined starting point. This is well-illustrated in the exercise, where the initial values are given, and the particular solution is found by fitting those values into the general solution equation.