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 \(t \geq 0 .\) \(\begin{array}{ll}{x_{1}^{\prime}=2 x_{2},} & {x_{1}(0)=3} \\\ {x_{2}^{\prime}=-2 x_{1},} & {x_{2}(0)=4}\end{array}\)

Short answer

Question: Transform the given system of first-order differential equations into a single second-order differential equation, solve it, and sketch the graph of the solution in the \(x_1x_2\) plane for \(t \geq 0\). System: \(\begin{cases} x_1' = 2x_2 \\ x_2' = -2x_1 \end{cases}\) Initial conditions: \(x_1(0) = 3\) \(x_2(0) = 4\) Provide the forms of \(x_1(t)\) and \(x_2(t)\), and describe the appearance of the graph in the \(x_1x_2\) plane.

Step-by-step solution

Transform the System into a Second-Order Differential Equation

We are given a system of first-order differential equations: $$\begin{cases} x_1' = 2x_2 \\ x_2' = -2x_1 \end{cases}$$ To obtain a single second-order differential equation, we will differentiating one of the initial equations concerning the variable \(x_2\) and eliminate \(x_2\) using the other equation. Differentiate the first equation with respect to time, we get: $$x_1'' = 2x_2'$$ Now, using the second equation, we can substitute \(x_2' = -2x_1\): $$x_1'' = 2(-2x_1)$$ This results in a second-order differential equation: $$x_1'' = -4x_1$$

Solve the Second-Order Differential Equation

Now, we'll solve the second-order differential equation. The general solution to this equation is: $$x_1(t) = C_1 \cos{(2t)} + C_2 \sin{(2t)}$$ From the first initial condition \(x_1(0) = 3\), we can find \(C_1\) and \(C_2\): $$x_1(0) = C_1 \cos{(0)} + C_2 \sin{(0)} = 3$$ $$C_1 = 3$$ Now, we need to find \(x_2(t)\). We know that \(x_1' = 2x_2\), and we can find \(x_1'\) by differentiating \(x_1(t)\): $$x_1' = -6\sin{(2t)} + 2C_2\cos{(2t)} = 2x_2$$ So, we get: $$x_2(t) = -3\sin{(2t)} + C_2\cos{(2t)}$$

Use Initial Conditions to Find Specific Solutions

We now use the second initial condition \(x_2(0) = 4\) to find \(C_2\): $$x_2(0) = -3\sin{(0)} + C_2\cos{(0)} = 4$$ $$C_2 = 4$$ So our final expressions for \(x_1(t)\) and \(x_2(t)\) are: $$x_1(t) = 3\cos{(2t)} + 4\sin{(2t)}$$ $$x_2(t) = -3\sin{(2t)} + 4\cos{(2t)}$$

Sketch the Graph of the Solution

To sketch the graph of the solution in the \(x_1x_2\) plane, we use the parametric equation representation for \(x_1\) and \(x_2\): $$x_1(t) = 3\cos{(2t)} + 4\sin{(2t)}$$ $$x_2(t) = -3\sin{(2t)} + 4\cos{(2t)}$$ The graph will be an ellipse centered around the origin with a horizontal axis length of \(2\cdot3\) and a vertical axis length of \(2\cdot4\). You can plot the parametric equation in software like Desmos, GeoGebra, or Mathematica to see the graph for \(0 \leq t \leq 2\pi\).