Solving System Of Differential Equations Find X(ln 3) + Y(ln 3)

When faced with a system of differential equations like this, our goal is to find functions x(t) and y(t) that satisfy both equations simultaneously, while also adhering to the given initial conditions. This problem delves into the fascinating world of differential equations, which are fundamental tools for modeling dynamic systems in various fields, including physics, engineering, and economics. The specific system we're tackling here is a linear system of first-order differential equations, characterized by the fact that the derivatives of the unknown functions (x and y) are expressed as linear combinations of the functions themselves. To effectively solve such systems, we often turn to the powerful technique of eigenvalue analysis. This method allows us to decouple the equations, transforming the system into a set of simpler, independent equations that are much easier to solve. Let's embark on a step-by-step journey to unravel this problem.

To begin, we will transform the system of differential equations into matrix form, which is a standard approach for simplifying linear systems. This involves representing the derivatives and the functions as vectors and the coefficients as a matrix. The matrix representation provides a compact and organized way to work with the system, making it easier to apply linear algebra techniques. The system

$\dot{x} = 2y$,

$\dot{y} = 2x + 3y$,

can be written in matrix form as:

$\begin{bmatrix} \dot{x} \\ \dot{y} \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$.

Let's denote the matrix as $A = \begin{bmatrix} 0 & 2 \\ 2 & 3 \end{bmatrix}$ and the vector function as $\mathbf{v}(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix}$. Then, the system can be written as $\dot{\mathbf{v}} = A \mathbf{v}$. This compact notation will allow us to leverage the tools of linear algebra to solve the system. The next crucial step is to find the eigenvalues and eigenvectors of the matrix A. These values hold the key to understanding the behavior of the system and constructing the general solution. The eigenvalues represent the rates of growth or decay of the solutions, while the eigenvectors define the directions along which these changes occur. By determining these values, we can effectively decompose the system into simpler components and find the general solution. This process is a cornerstone of solving linear systems of differential equations.

Finding Eigenvalues and Eigenvectors

The eigenvalues $\lambda$ of the matrix $A$ are the solutions to the characteristic equation, which is given by $\text{det}(A - \lambda I) = 0$, where $I$ is the identity matrix. In our case, the characteristic equation is:

$\text{det}\left(\begin{bmatrix} 0 & 2 \\ 2 & 3 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) = 0$.

This simplifies to:

$\text{det}\begin{bmatrix} -\lambda & 2 \\ 2 & 3 - \lambda \end{bmatrix} = 0$.

Expanding the determinant, we get:

$(-\lambda)(3 - \lambda) - (2)(2) = 0$,

$\lambda^2 - 3\lambda - 4 = 0$.

This quadratic equation factors nicely as:

$(\lambda - 4)(\lambda + 1) = 0$.

Thus, the eigenvalues are $\lambda_1 = 4$ and $\lambda_2 = -1$. Eigenvalues are crucial because they dictate the fundamental modes of behavior of the system. A positive eigenvalue indicates exponential growth, while a negative eigenvalue signifies exponential decay. The magnitude of the eigenvalue determines the rate of this growth or decay. Therefore, knowing the eigenvalues provides a qualitative understanding of the system's long-term behavior. Now that we have the eigenvalues, the next step is to find the corresponding eigenvectors. The eigenvectors provide the directions in which these growth or decay modes operate. Finding these eigenvectors will allow us to construct the general solution of the system.

For each eigenvalue, we need to find the corresponding eigenvector. An eigenvector $\mathbf{v}$ satisfies the equation $(A - \lambda I)\mathbf{v} = \mathbf{0}$.

For $\lambda_1 = 4$, we have:

$\begin{bmatrix} -4 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$.

This gives us the equation $-4v_1 + 2v_2 = 0$, which simplifies to $v_2 = 2v_1$. We can choose $v_1 = 1$, so the eigenvector corresponding to $\lambda_1 = 4$ is $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$. The eigenvector provides the direction in which the solution will grow exponentially with a rate determined by the eigenvalue. The eigenvector we just found tells us that solutions corresponding to the eigenvalue 4 will move in a direction that is a combination of 1 unit in the x-direction and 2 units in the y-direction.

For $\lambda_2 = -1$, we have:

$\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$.

This gives us the equation $v_1 + 2v_2 = 0$, which simplifies to $v_1 = -2v_2$. We can choose $v_2 = 1$, so the eigenvector corresponding to $\lambda_2 = -1$ is $\mathbf{v}_2 = \begin{bmatrix} -2 \\ 1 \end{bmatrix}$. This eigenvector, associated with the negative eigenvalue -1, signifies a direction of exponential decay. The solution components aligned with this vector will diminish over time. The eigenvector $\mathbf{v}_2 = \begin{bmatrix} -2 \\ 1 \end{bmatrix}$ shows that solutions associated with the eigenvalue -1 will move in a direction that is a combination of -2 units in the x-direction and 1 unit in the y-direction. The eigenvectors we have found are linearly independent, which is a crucial requirement for constructing the general solution of the system.

General Solution and Applying Initial Conditions

The general solution of the system is a linear combination of the solutions corresponding to each eigenvalue and eigenvector pair. This means that the solution can be expressed as a sum of exponential functions, each multiplied by a constant coefficient and the corresponding eigenvector. The general solution of the system is given by:

$\mathbf{v}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2$,

where $c_1$ and $c_2$ are constants determined by the initial conditions. Substituting the eigenvalues and eigenvectors we found, we get:

$\begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = c_1 e^{4t} \begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} -2 \\ 1 \end{bmatrix}$.

This gives us the following expressions for $x(t)$ and $y(t)$:

$x(t) = c_1 e^{4t} - 2c_2 e^{-t}$,

$y(t) = 2c_1 e^{4t} + c_2 e^{-t}$.

To find the constants $c_1$ and $c_2$, we use the initial conditions $x(0) = 10$ and $y(0) = -5$. Plugging in $t = 0$, we get:

$x(0) = c_1 - 2c_2 = 10$,

$y(0) = 2c_1 + c_2 = -5$.

This is a system of linear equations in $c_1$ and $c_2$. We can solve this system using substitution or elimination. Multiplying the second equation by 2, we get $4c_1 + 2c_2 = -10$. Adding this to the first equation, we get:

$5c_1 = 0$,

so $c_1 = 0$. Substituting $c_1 = 0$ into the equation $2c_1 + c_2 = -5$, we get $c_2 = -5$. The initial conditions play a crucial role in selecting a particular solution from the family of general solutions. They represent the state of the system at a specific time, allowing us to pinpoint the unique solution that matches that initial state. In our case, the initial conditions $x(0) = 10$ and $y(0) = -5$ provide the necessary constraints to determine the constants $c_1$ and $c_2$. With these constants in hand, we can fully define the solution functions $x(t)$ and $y(t)$.

Final Calculation of x(ln 3) + y(ln 3)

Now that we have found the constants $c_1 = 0$ and $c_2 = -5$, we can write the particular solution as:

$x(t) = -2(-5)e^{-t} = 10e^{-t}$,

$y(t) = -5e^{-t}$.

We want to find $x(\ln 3) + y(\ln 3)$. Plugging in $t = \ln 3$, we get:

$x(\ln 3) = 10e^{-\ln 3} = 10e^{\ln (3^{-1})} = 10 \cdot \frac{1}{3} = \frac{10}{3}$,

$y(\ln 3) = -5e^{-\ln 3} = -5e^{\ln (3^{-1})} = -5 \cdot \frac{1}{3} = -\frac{5}{3}$.

Therefore,

$x(\ln 3) + y(\ln 3) = \frac{10}{3} - \frac{5}{3} = \frac{5}{3}$.

The final step in solving the problem is to evaluate the expression $x(\ln 3) + y(\ln 3)$. This involves substituting $t = \ln 3$ into the particular solutions we derived for $x(t)$ and $y(t)$. By performing this substitution, we obtain numerical values for $x(\ln 3)$ and $y(\ln 3)$, which we can then add together to arrive at the final answer. This step demonstrates how the mathematical solution translates into a concrete prediction about the system's state at a specific time.

Thus, the final answer is $\frac{5}{3}$.