Ein Massenpunkt bewege sich in der \(x, y\)-Ebene so, daB seine kartesischen Koordinaten \(x\) und \(y\) den folgenden Differentialgleichungen genügen: $$ \vec{x}=\dot{y}, \quad \ddot{y}=-\dot{x} $$ Bestimmen Sie die Bahnkurve für die Anfangswerte $$ x(0)=y(0)=0, \quad \dot{x}(0)=0 . \quad \dot{y}(0)=1 $$ Hinweis: Das Differentialgleichungssystem l?Bt sich mit Hilfe der Substitutionen \(u=\dot{x}\) und \(v=\dot{y}\) auf ein System linearer Differentialgleichungen \(t\). Orinung zurückf?hren. L?sen Sie zunächst dieses System. Durch Ricksubstitution und anschlieBende Integration erhalten Sie dann die gesuchten zeitabhängigen Koordinaten \(x=x(t)\) und \(y=y(t)\) des Massenpunktes.

A harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force proportional to the displacement. In mathematical terms, it is often described by a second-order differential equation. For our example, we have the equation: \[ \frac{d^2 u}{dt^2} = -u \].
This equation is typical of simple harmonic motion, where the solution involves sine and cosine functions. The presence of these functions suggests that the motion is periodic. In our case, the solutions are \(u(t) = A \cos(t) + B \sin(t)\), demonstrating the cyclic nature of oscillating systems.

A system of linear differential equations consists of multiple equations involving derivatives of several dependent variables. These equations are linear, meaning each term is either a constant or the product of a constant and the first power of one variable.
In our exercise, we transformed the original system using substitutions \(u = \dot{x}\) and \(v = \dot{y}\), leading to the following system: \[ \begin{align*} \dot{u} &= v, \ \dot{v} &= -u. \ \end{align*} \].
Systems like this are prevalent in modeling various physical phenomena, such as coupled oscillators and electric circuits. Solving these systems often requires techniques like elimination or matrix algebra, though simpler substitutions can also work.

Initial conditions specify the state of a system at the beginning of the observation, often at \(t = 0\). They are crucial for solving differential equations as they allow the determination of unknown constants in the general solution.
For our problem, the initial conditions were given as \(x(0) = y(0) = 0\), \(\dot{x}(0) = 0\), and \(\dot{y}(0) = 1\).
These conditions help us find specific solutions fitting our exact scenario. For example, after obtaining the general solution for \(u(t) = A \cos(t) + B \sin(t)\), applying \(u(0) = 0\) and \(\dot{u}(0) = 0\) revealed that \(A = 0\) and \(B = 1\). This specificity is crucial in real-world applications where precise initial states are known and solutions need to reflect that.

Integration is the process of finding the integral of a function, which can be thought of as the 'reverse' of differentiation. This is crucial for solving differential equations when given the rate of change and needing to find the original function.
In our exercise, after determining \(\dot{x}(t) = \sin(t)\) and integrating, we obtained \(x(t) = -\cos(t) + C\).
We use initial conditions to solve for constants of integration. Here, since \(x(0) = 0\), we found \(C = 1\), hence \(x(t) = 1 - \cos(t)\). Integration is fundamental for solving many physical problems, converting differential equations back into functions describing real-world phenomena.

The substitution method involves replacing variables or expressions with new variables to simplify solving differential equations. It is particularly useful for transforming non-linear systems into linear ones or reducing the order of differential equations.
In our problem, we used the substitutions \(u = \dot{x}\) and \(v = \dot{y}\), transforming the system into: \[ \begin{align*} \dot{u} &= v, \ \dot{v} &= -u \].
These new equations are simpler to solve, leading us to: \( \ddot{u} = -u \), a classic harmonic oscillator equation. Substitution is a powerful method in differential equations, allowing us to leverage known solutions and techniques for simpler forms to solve more complex problems.