Question

Proceed as in Problem 7 to transform the given system into a single equation of second order. Then find \(x_{1}\) and \(x_{2}\) that also satisfy the given initial conditions. Finally, sketch the graph of the solution in the \(x_{1} x_{2}\) -plane for \(t \geq 0 .\) \(\begin{array}{ll}{x_{1}^{\prime}=3 x_{1}-2 x_{2},} & {x_{1}(0)=3} \\\ {x_{2}^{\prime}=2 x_{1}-2 x_{2},} & {x_{2}(0)=\frac{1}{2}}\end{array}\)

Short answer

Answer: x2(t) = x1(t) * ((1/2 + t) / 3)

Step-by-step solution

Transform the given system into a single second-order equation

To transform the given system into a single second-order equation, we differentiate one of the equations with respect to \(t\) and substitute the other equation to eliminate one of the variables. Let's differentiate the first equation: \(x_1^{\prime\prime} = 3x_1^{\prime} - 2x_2^{\prime}\) Now, substitute the values of \(x_1^{\prime}\) and \(x_2^{\prime}\) from the given equations: \(x_1^{\prime\prime} = 3(3x_1 - 2x_2) - 2(2x_1 - 2x_2)\) Simplify the equation: \(x_1^{\prime\prime} = 7x_1 - 6x_2\)

Solve the second-order equation

To solve the second-order equation \(x_1^{\prime\prime} = 7x_1 - 6x_2\), we will consider \(x_1^{\prime} = x_3\). This way, we can rewrite the equations as follows: \(x_1^{\prime} = x_3\) \(x_3^{\prime} = 7x_1 - 6x_2\) Now, solve for \(x_1\) and \(x_2\) using the initial conditions: \(x_1(0) = 3\) \(x_1^{\prime}(0) = x_3(0) = 3x_1(0) - 2x_2(0) = 3 \cdot 3 - 2 \cdot \frac{1}{2} = 8\)

Find \(x_1\) and \(x_2\) satisfying the initial conditions

With the initial conditions known, we can integrate the second-order equation to find \(x_1\) and \(x_2\): \(x_1(t) = 3e^{2t}\) \(x_2(t) = \frac{1}{2}e^{2t} + te^{2t}\)

Sketch the graph of the solution in the \(x_1 x_2\)-plane

To sketch the graph of the solution \(x_1(t) = 3e^{2t}\) and \(x_2(t) = \frac{1}{2}e^{2t} + te^{2t}\) in the \(x_1 x_2\)-plane, we can eliminate \(t\) by taking \(x_2\) as a function of \(x_1\). Divide the second equation by the first equation: \(\frac{x_2(t)}{x_1(t)} = \frac{\frac{1}{2}e^{2t} + te^{2t}}{3e^{2t}} = \frac{\frac{1}{2} + t}{3}\) Now, solving for \(x_2(t)\) as a function of \(x_1(t)\): \(x_2(t) = x_1(t) \cdot \left(\frac{\frac{1}{2} + t}{3}\right)\) To sketch the graph, plot the function \(x_2(t)\) as a function of \(x_1(t)\). The graph should show an exponentially increasing curve that starts at the initial conditions \((3, \frac{1}{2})\) and has an increasing slope as the time \(t\) increases.

Key concepts

Systems of Equations

Systems of equations are sets of two or more equations that have common variables. In mathematics, these can often show us relationships between different quantities. When solving systems of equations, our goal is to find values for the variables that satisfy all the given conditions simultaneously. This is often achieved by substituting one equation into another or by using methods such as substitution or elimination.

In the original exercise, we deal with a system of equations given by:

• \(x_1'(t) = 3x_1 - 2x_2\)
• \(x_2'(t) = 2x_1 - 2x_2\)

Here, \(x_1\) and \(x_2\) are functions of \(t\), creating interdependency between the two equations. To simplify and solve the equations, the solution requires mixing and matching terms until you get a single equation in one variable. This allows us to reduce complexity and solve for one variable at a time.

Second-Order Equations

Second-order equations are differential equations where the highest derivative is the second derivative. These equations often describe systems in physics and engineering, such as oscillations or other dynamic systems.

For the given system of equations, transforming them into a second-order equation involves differentiating one of the equations and substituting into another. Take the first equation: \(x_1^{\prime} = 3x_1 - 2x_2\). By differentiating it with respect to \(t\), we get: \(x_1^{\prime\prime} = 3x_1^{\prime} - 2x_2^{\prime}\).

Then, substitute \(x_1^{\prime}\) and \(x_2^{\prime}\) from the original system to eliminate one of the variables. After simplification, this yields the second-order equation \(x_1^{\prime\prime} = 7x_1 - 6x_2\). Such transformations provide a valuable tool for solving through standard methods for ordinary differential equations.

Initial Conditions

Initial conditions are crucial in finding a unique solution to differential equations. They specify the state of the system at the beginning (usually at \(t = 0\)), which then guides the evolution of the solutions over time.

In our problem, the initial conditions are given as \(x_1(0) = 3\) and \(x_2(0) = \frac{1}{2}\). By applying these conditions, we determine specific constants or coefficients when solving the differential equations.

For instance, when solving the second-order equation derived from the system, using these initial values allows the determination of specific expressions for \(x_1(t)\) and \(x_2(t)\). With \(x_1(t) = 3e^{2t}\) and \(x_2(t) = \frac{1}{2}e^{2t} + te^{2t}\), the initial conditions tell us where exactly the functions \(x_1(t)\) and \(x_2(t)\) begin on the \(x_1 - x_2\) plane.

Exponential Functions

Exponential functions have the general form \(y = ae^{kt}\), where \(a\) and \(k\) are constants. These functions model processes that grow or decay at rates proportional to their current value. They are common in physics, biology, and economics.

In the current exercise, solutions involving exponential functions include the time-dependent behavior given by \(x_1(t) = 3e^{2t}\) and \(x_2(t) = \frac{1}{2}e^{2t} + te^{2t}\). The term \(e^{2t}\) indicates exponential growth, affected by the constant rate \(2\) that defines the system's dynamics.

Such functions illustrate how solutions can exponentially increase, aligning with phenomena where change becomes faster as the value grows. When modeling such types of systems, understanding the nature of the exponential solution gives insights into long-term behavior. Plots of these functions often take the shape of curves rapidly moving upward or downward, depending on the parameters.