Question

Systems of first order equations can sometimes be transformed into a single equation of higher order. Consider the system $$ x_{1}^{\prime}=-2 x_{1}+x_{2}, \quad x_{2}^{\prime}=x_{1}-2 x_{2} $$ (a) Solve the first equation for \(x_{2}\) and substitute into the second equation, thereby obtaining a second order equation for \(x_{1}\). Solve this equation for \(x_{1}\) and then determine \(x_{2}\) also. (b) Find the solution of the given system that also satisfies the initial conditions \(x_{1}(0)=2,\) \(x_{2}(0)=3 .\) (c) Sketch the curve, for \(t \geq 0,\) given parametrically by the expressions for \(x_{1}\) and \(x_{2}\) obtained in part (b).

Short answer

In summary, we have determined the solution to the given system of differential equations with the given initial conditions. The expressions for x1(t) and x2(t) are: $$ x_{1}(t) = e^t \left(2\cos(\sqrt{3} t) + \frac{1}{4}\sin(\sqrt{3} t)\right) $$ $$ x_{2}(t) = e^t \left(5\cos(\sqrt{3} t) + 2\sqrt{3}\sin(\sqrt{3} t) \right) $$ The sketch of the curve given parametrically by these expressions illustrates a spiral that starts at the point (2, 3) and grows away from the origin due to the exponential factor \(e^t\).

Step-by-step solution

Solving x1′ equation for x2

First, solve the x1′ equation for x2: $$ x_{1}^{\prime} = -2 x_{1} + x_{2} \implies x_{2} = x_{1}^{\prime} + 2x_{1} $$

Substituting x2 into x2′ equation

Now replace x2 in the second equation with the expression obtained in Step 1: $$ x_{2}^{\prime} = x_{1} - 2x_{2} \implies (x_{1}^{\prime} + 2x_{1})' = x_{1} - 2(x_{1}^{\prime} + 2x_{1}) $$

Solving the second-order equation for x1

Now we have a second-order linear ODE for x1: $$ x_{1}^{\prime\prime} - 2x_{1}^{\prime} + 4x_{1} = 0 $$ The characteristic equation of this ODE is: $$ r^2 - 2r + 4 = 0 $$ which has the complex roots \(r = 1 \pm i\sqrt{3}\). So the general solution for x1 is: $$ x_{1}(t) = e^t\left(A\cos(\sqrt{3} t) + B\sin(\sqrt{3} t)\right) $$ where A and B are constants to be determined by the initial conditions.

Determining x2

We now determine x2 using the expression we found in Step 1: $$ x_{2}(t) = x_{1}^{\prime}(t) + 2x_{1}(t) $$ Taking the derivative of x1(t) with respect to t: $$ x_{1}^{\prime}(t) = e^t \left((A - \sqrt{3} B)\cos(\sqrt{3} t) + (A + \sqrt{3} B)\sin(\sqrt{3} t)\right) $$ Hence, the general solution for x2 is: $$ x_{2}(t) = e^t \left((2A + 2B)\cos(\sqrt{3} t) + (4A + 2\sqrt{3} B)\sin(\sqrt{3} t)\right) $$

Applying the initial conditions

We are given the initial conditions \(x_{1}(0) = 2\) and \(x_{2}(0) = 3\). Plugging these into our general solutions, we have: $$ 2 = x_{1}(0) = e^0\left(A\cos(0) + B\sin(0)\right) \implies A = 2 $$ $$ 3 = x_{2}(0) = e^0\left((2 \cdot 2 + 2B)\cos(0) + (4 \cdot 2 + 2\sqrt{3} B)\sin(0)\right) \implies 4B = 1 \implies B=\frac{1}{4} $$ The solution given by the initial conditions, x1(t) and x2(t), is $$ x_{1}(t) = e^t \left(2\cos(\sqrt{3} t) + \frac{1}{4}\sin(\sqrt{3} t)\right) $$ $$ x_{2}(t) = e^t \left(5\cos(\sqrt{3} t) + 2\sqrt{3}\sin(\sqrt{3} t) \right) $$

Sketching the curve

Now, we will sketch the curve of the parametric system, where \(x_{1}(t)\) represents the horizontal axis and \(x_{2}(t)\) represents the vertical axis. We'll make use of the solution for the initial conditions £(0) = 2\( and \)x_{2}(0) = 3$: 1. Plot the point (2, 3) as our starting point. 2. Consider the growth exponential factor \(e^t\), which means that the curve will get farther from the origin for positive values of t. 3. Observe that when t increases, the expressions for \(x_{1}(t)\) and \(x_{2}(t)\) will oscillate with the sine and cosine components, creating a spiral-like shape. Sketching the curve, we get a spiral that starts at the point (2, 3) and grows away from the origin due to the exponential factor \(e^t\).

Key concepts

Second-Order Linear ODE

A second-order linear ordinary differential equation (ODE) is a type of differential equation characterized by derivatives up to the second degree and linear terms. In other words, an equation of the form \( y'' + p(t)y' + q(t)y = g(t) \), where \( p(t) \), \( q(t) \), and \( g(t) \) are functions of \( t \) and \( y'' \) represents the second derivative of \( y \) with respect to \( t \). These ODEs are fundamental in modeling a myriad of physical phenomena, from simple harmonic oscillators to RLC circuits in electrical engineering.

The challenge often lies in finding the general solution to such equations, which is where the method of the characteristic equation comes in handy, greatly simplifying the process of solving these equations, especially with constant coefficients. The transformation from a system of first-order equations to a single second-order equation, as seen in our exercise, is a powerful technique employed when dealing with interconnected dynamic systems, allowing us to deduce complex motion or behavior from seemingly simple relationships.

Characteristic Equation

The characteristic equation unlocks the general solution to constant-coefficient second-order linear ODEs. Formed by substituting a trial solution of the form \( y = e^{rt} \) into the homogeneous version of the ODE, the characteristic equation resembles an algebraic equation. In our exercise, the characteristic equation \( r^2 - 2r + 4 = 0 \) corresponds to the ODE \( x_1'' - 2x_1' + 4x_1 = 0 \).

When roots are complex, as they are in this case, the solution involves sines and cosines, indicative of oscillatory behavior. The real part of the complex roots, if one exists, will affect the growth or decay rate of the oscillations. As such, the characteristic equation provides the basis for understanding the qualitative features of a system's dynamics, and is a crucial step in the solution process.

Initial Conditions

Initial conditions specify the state of a system at a particular starting time, usually \( t = 0 \). They are vital for finding the specific solution to an ODE that applies to a given physical scenario. In the context of a second-order ODE, typically two initial conditions are required, one for the function itself and one for its first derivative.

In our exercise, the initial conditions \( x_1(0) = 2 \) and \( x_2(0) = 3 \) serve to pinpoint the constants \( A \) and \( B \) within the general solution. They transform the general solution into a unique solution that describes the behavior of the system for all future times. Properly applying the initial conditions often involves plugging the conditions into both the function and its derived functions, as was done to solve for \( A \) and \( B \) in our example. This further highlights how initial conditions are instrumental in accurately predicting the evolution of a system over time.