Zeigen Sie: Die Differentialgleichung 2. Ordnung \(\vec{x}+2 \dot{x}+2 x=0\) besitzt die linear unabhängigen L?sungen $$ x_{1}=e^{-t} \cdot \sin t \quad \text { und } \quad x_{2}=e^{-t} \cdot \cos t $$ Wie lautet die allgemeine Lösung dieser Differentialgleichung?

A general solution of a differential equation encompasses all possible solutions, representing a family of curves. For the given second-order differential equation \( \frac{d^2 x}{dt^2} + 2\frac{dx}{dt} + 2x = 0 \), we aim to find a form of \( x(t) \) that includes all potential solutions.
We start by identifying individual solutions that satisfy the equation. In our case, these solutions are \( x_1 = e^{-t} \sin t \) and \( x_2 = e^{-t} \cos t \). With these specific solutions confirmed, the general solution can be expressed as a linear combination:
\[ x(t) = c_1 e^{-t} \sin t + c_2 e^{-t} \cos t \]
Here, \( c_1 \) and \( c_2 \) are arbitrary constants. These constants account for the diverse nature of initial conditions, ensuring every possible solution to the differential equation is covered by this general form.

Independent solutions of a differential equation are critical because they expand the solution space, allowing us to construct a general solution. For two solutions to be independent, their linear combination must produce no other trivial solutions.
In our exercise, we have \( x_{1}=e^{-t} \sin t \) and \( x_{2}=e^{-t} \cos t \). To verify their independence, we substitute these solutions into the original differential equation and confirm they satisfy it independently.
Specifically, independent solutions ensure their Wronskian is non-zero. The Wronskian, \( W(x_{1}, x_{2}) \), is given by:
\[ W(x_{1}, x_{2}) = x_{1}\frac{d}{dt}x_{2} - x_{2}\frac{d}{dt}x_{1} \]
When this Wronskian is non-zero, the solutions are guaranteed to be independent, thus suitable for constructing the general solution.

Differentiation is a fundamental tool used to verify whether given functions solve a differential equation. For our problem, we need first and second derivatives of the solutions \( x_1 \) and \( x_2 \).
For \( x_1 = e^{-t} \sin t \):
\( -d/dt = \frac{d}{dt} (e^{-t} \sin t) = -e^{-t} \sin t + e^{-t} \cos t \)
\( -d^2/dt^2 = \frac{d}{dt} (-e^{-t} \sin t + e^{-t} \cos t) = -2 e^{-t} \cos t \)
Substituting into the original equation confirms \( x_1 \) solves it.
Similarly for \( x_2 = e^{-t} \cos t \):
\( -d/dt = \frac{d}{dt} (e^{-t} \cos t) = -e^{-t} \cos t - e^{-t} \sin t \)
\( -d^2/dt^2 = \frac{d}{dt} (-e^{-t} \cos t - e^{-t} \sin t) = 2 e^{-t} \sin t \)
Substituting into the original equation confirms \( x_2 \) solves it.
This process of differentiation and substitution is crucial for validating candidate solutions.

Linear independence is essential in solving differential equations because it ensures that our solutions broaden the solution space. Two functions are linearly independent if no constant exists that can combine them into zero, except the trivial solution where all constants are zero.
In mathematical terms, for functions \( x_1 \) and \( x_2 \), if their Wronskian is non-zero, they are linearly independent.
For our exercise:
\[ x_1 = e^{-t} \sin t \quad \text{and} \quad x_2 = e^{-t} \cos t \]
Their Wronskian is:
\[ W(x_1, x_2) = x_1 \frac{d}{dt} x_2 - x_2 \frac{d}{dt} x_1 \]
Calculating the derivatives and substituting them, we find:
\[ W(e^{-t} \sin t, e^{-t} \cos t) = e^{-t} \sin t (-e^{-t} \cos t - e^{-t} \sin t) - e^{-t} \cos t (-e^{-t} \sin t + e^{-t} \cos t) \]
Simplifying this expression, we see that the Wronskian is indeed non-zero, thus confirming that \( x_1 \) and \( x_2 \) are linearly independent solutions, suitable for forming the general solution.