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 \(t \geq 0 .\) \(\begin{array}{ll}{x_{1}^{\prime}=1.25 x_{1}+0.75 x_{2},} & {x_{1}(0)=-2} \\\ {x_{2}^{\prime}=0.75 x_{1}+1.25 x_{2},} & {x_{2}(0)=1}\end{array}\)

Short answer

Answer: The equation for \(x_2'\) is \(x_2' = 1.75e^{-1.25t} + 0.75x_1 + 0.45x_2\).

Step-by-step solution

Rewrite the system as a single second-order equation

First, we will differentiate the second equation with respect to \(t\), to eliminate \(x_2\). \(x_{2}^{\prime\prime}=0.75 x_{1}^{\prime} + 1.25 x_{2}^{\prime}\) Now, we can substitute \(x_{1}^{\prime}=1.25 x_{1}+0.75 x_{2}\) in place of \(x_1'\): \(x_{2}^{\prime\prime}=0.75 (1.25x_{1}+0.75x_{2}) + 1.25 x_{2}^{\prime}\) This now gives us the second-order equation: \(x_{2}^{\prime\prime} - 1.25x_{2}^{\prime} = 0.9375x_1 + 0.5625x_2\)

Find \(x_1\) and \(x_2\) that satisfy the initial conditions

To do this, we need to plug in the initial conditions \(x_1(0)=-2\) and \(x_2(0)=1\). From the first given equation: \(x_{1}^{\prime}(0)=1.25 (-2) + 0.75 (1)\). Now, using the second equation: \(x_{2}^{\prime}(0)=0.75 (-2) + 1.25 (1)\). Hence, we get \(x_{1}^{\prime}(0)=-1.75\) and \(x_{2}^{\prime}(0)=0.25\).

Plot the graph of the solution in the \(x_1 x_2\)-plane for \(t \geq 0\)

To plot this graph, we must find the general solution of the single second-order equation from Step 1. It looks like a first-order linear inhomogeneous equation in \(x_{2}^{\prime}\). Let the integrating factor be \(I(t) = e^{\int 1.25 dt} = e^{1.25t}\). Multiplying the equation by \(I(t)\), we get: \(e^{1.25t} x_{2}^{\prime\prime} - 1.25e^{1.25t} x_{2}^{\prime} = 0.9375e^{1.25t} x_1 + 0.5625e^{1.25t} x_2\) This equation is now exact. So, we can integrate it w.r.t \(t\): \(x_2'= ce^{-1.25t} + 0.75x_1 + 0.45x_2, c\in \mathbb{R}\) Substitute the initial values of \(x_1\), \(x_2\), and \(x_2'\) to find the constant \(c\): \(0.25 = -1.5 + c\), therefore \(c = 1.75\). Now, we have: \(x_2' = 1.75e^{-1.25t} + 0.75x_1 + 0.45x_2\) Finally, integrate \(x_2'\) w.r.t \(t\) to get \(x_2\): \(x_2 = -\frac{7}{3}e^{-1.25t} + 3x_1 + \frac{2}{3}x_2 + b\) We can now plot the graph of the solution within the \(x_1 x_2\)-plane for \(t\geq 0\). Keep in mind that this is not a complete solution, as it does not include the unknown value of \(b\). However, this representation gives you an idea of how the solution evolves over time.

Key concepts

Initial Conditions

Initial conditions play a crucial role in solving differential equations, as they dictate the specific solution among many possible solutions of a differential equation. In essence, initial conditions are the values of the function and its derivatives at a starting point, generally at time zero, which the solution must satisfy.

For example, in our exercise, we are given the initial conditions:

• \(x_1(0) = -2\)
• \(x_2(0) = 1\)

These conditions specify the initial state of our system of equations. They ensure that the solution curve passes through the point specified by these initial values. Without initial conditions, a differential equation has an infinite number of solutions, as it could pass through any point on the plane. By applying initial conditions, however, we find the unique solution relevant to our specific problem.

For systems of differential equations like ours, both initial values and derivatives (such as \(x_1'(0)\) and \(x_2'(0)\)) are necessary to fully describe the starting behavior of both functions involved.

System of Differential Equations

A system of differential equations consists of multiple equations that involve derivatives of several interdependent variables. These systems often describe complex phenomena, where one equation alone isn't enough. For instance, in our exercise, we have the following system:

• \(x_1' = 1.25x_1 + 0.75x_2\)
• \(x_2' = 0.75x_1 + 1.25x_2\)

These equations are coupled; that is, the derivative of one variable, for instance, \(x_1'\), depends on the current values of both \(x_1\) and \(x_2\).

The goal in such problems is often to decouple these into a single equation or to solve them using numerical methods that consider the system as a whole. In our solution process, we converted the system into a single second-order equation to simplify solving it.

Systems of differential equations are common in fields such as physics, engineering, and economics, where multiple interconnected factors must be analyzed simultaneously.

Graphical Solution

Upon solving the system of differential equations, the next step is often visual analysis through a graphical solution. This involves plotting the trajectories of the solutions in a chosen plane, such as the \(x_1 x_2\)\-plane in our exercise.

By graphing the equations, you can visually interpret the behavior of the solutions over time and analyze how changes in initial conditions can affect these trajectories. In the context of our example:

• The graph will start at the initial conditions given, namely the point \((x_1(0), x_2(0)) = (-2, 1)\)\.
• As \(t\) increases, tracking the intersection of solution curves helps illustrate their evolution and response dynamics.

Graphical solutions offer an intuitive way to grasp concepts that might be less obvious through algebraic solutions alone. They reveal attractions, stability, or spiraling behaviors in systems, which are fundamental in systems' dynamics.

Inhomogeneous Equations

An inhomogeneous equation is one that includes a non-zero function on one side, distinguishing it from a homogeneous equation, which equals zero. These equations often express that external forces or inputs affect the system.

For the equation obtained from our system,\[x_{2}^{\prime\prime} - 1.25x_{2}^{\prime} = 0.9375x_1 + 0.5625x_2\] the right-hand side is not zero, making it inhomogeneous.

In solving such equations, you typically find two parts:

• The complementary solution, derived from the associated homogeneous equation, where the right side is set to zero.
• The particular solution, which accounts for the non-zero right-hand side and represents the system's steady response to external inputs.

Combining these solutions yields the general solution to the inhomogeneous differential equation. For our problem, integrating the inhomogeneous part necessitated finding an integrating factor, ensuring we accurately represented how the system's behavior responds over time to changes in \(x_1\) and \(x_2\).