Math Problem Statement

3

Solution

To solve the system of differential equations using the elimination method, we'll eliminate one variable to find the solution for the other. Here’s how we can proceed:

Given System:

\frac{dx}{dt} &= 3x - 2y, \quad \text{(1)} \\ \frac{dy}{dt} &= 2x - 3y. \quad \text{(2)} \end{aligned}$$ ### Step 1: Differentiate one of the equations Differentiate equation (1) with respect to $$t$$: $$\frac{d^2x}{dt^2} = 3\frac{dx}{dt} - 2\frac{dy}{dt}.$$ ### Step 2: Substitute from the other equation We can substitute $$\frac{dy}{dt}$$ from equation (2) into the differentiated equation: $$\frac{dy}{dt} = 2x - 3y.$$ Substitute this into the differentiated equation: $$\frac{d^2x}{dt^2} = 3\frac{dx}{dt} - 2(2x - 3y).$$ Simplify the equation: $$\frac{d^2x}{dt^2} = 3\frac{dx}{dt} - 4x + 6y. \quad \text{(3)}$$ Now, substitute $$y$$ from equation (1) into equation (3). From equation (1), solve for $$y$$: $$y = \frac{3x - \frac{dx}{dt}}{2}.$$ Substitute this into equation (3): $$\frac{d^2x}{dt^2} = 3\frac{dx}{dt} - 4x + 6\left(\frac{3x - \frac{dx}{dt}}{2}\right).$$ Simplify: $$\frac{d^2x}{dt^2} = 3\frac{dx}{dt} - 4x + 9x - 3\frac{dx}{dt},$$ $$\frac{d^2x}{dt^2} = 5x.$$ ### Step 3: Solve the resulting second-order differential equation The equation $$\frac{d^2x}{dt^2} = 5x$$ is a second-order linear homogeneous differential equation. Its characteristic equation is: $$r^2 - 5 = 0,$$ $$r = \pm \sqrt{5}.$$ So, the general solution for $$x(t)$$ is: $$x(t) = C_1 e^{\sqrt{5}t} + C_2 e^{-\sqrt{5}t}.$$ ### Step 4: Solve for $$y(t)$$ To find $$y(t)$$, substitute $$x(t)$$ back into equation (1): $$\frac{dx}{dt} = 3x - 2y.$$ Taking the derivative of $$x(t)$$ and then solving for $$y(t)$$ will give the corresponding solution for $$y$$. The derivative of $$x(t)$$ is: $$\frac{dx}{dt} = C_1\sqrt{5}e^{\sqrt{5}t} - C_2\sqrt{5}e^{-\sqrt{5}t}.$$ Substituting this into equation (1) and solving for $$y(t)$$: $$C_1\sqrt{5}e^{\sqrt{5}t} - C_2\sqrt{5}e^{-\sqrt{5}t} = 3(C_1 e^{\sqrt{5}t} + C_2 e^{-\sqrt{5}t}) - 2y(t).$$ This simplifies to: $$y(t) = C_1\left(\frac{\sqrt{5} - 3}{2}\right)e^{\sqrt{5}t} + C_2\left(\frac{-\sqrt{5} - 3}{2}\right)e^{-\sqrt{5}t}.$$ ### Final Solution: Thus, the solutions for $$x(t)$$ and $$y(t)$$ are: $$x(t) = C_1 e^{\sqrt{5}t} + C_2 e^{-\sqrt{5}t},$$ $$y(t) = C_1\left(\frac{\sqrt{5} - 3}{2}\right)e^{\sqrt{5}t} + C_2\left(\frac{-\sqrt{5} - 3}{2}\right)e^{-\sqrt{5}t}.$$ Would you like to see more details on any of these steps or have any questions? --- **Further questions for exploration:** 1. How does the solution behavior change if the eigenvalues are complex? 2. What does the solution look like if the initial conditions are given? 3. Can you find the specific solution given $$x(0) = 1$$ and $$y(0) = 0$$? 4. How would the phase portrait of this system look? 5. How do the constants $$C_1$$ and $$C_2$$ affect the general solution? **Tip:** When working with systems of differential equations, identifying and analyzing the characteristic equation is crucial for understanding the behavior of the solutions.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Second-Order Differential Equations
Linear Differential Equations

Formulas

Characteristic Equation

Theorems

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Suitable Grade Level

Advanced Undergraduate