Zeigen Sie: Die Funktionen $$ y_{1}(x)=\mathrm{e}^{2 x} \quad \text { und } \quad y_{2}(x)=x \cdot \mathrm{e}^{2 x} $$ bilden eine Fundamentalbasis der homogenen Differentialgleichung 2 . Ordnung \(y^{\prime \prime}-4 y^{\prime}+4 y=0\)

In the context of differential equations, a **fundamental basis** refers to a set of solutions to a homogeneous differential equation that can be used to express any solution of the equation. For a second-order homogeneous linear differential equation, this means finding two solutions whose linear combinations represent the general solution.
For example, the functions \(y_1(x) = e^{2x}\) and \(y_2(x) = xe^{2x}\) can form a fundamental basis for the differential equation \(y'' - 4y' + 4y = 0\). The key lies in showing that both functions are valid solutions and are linearly independent.
This makes it possible to write any solution \(y(x)\) of this differential equation in the form:
\[ y(x) = c_1 y_1(x) + c_2 y_2(x) \]
where \(c_1\) and \(c_2\) are constants.

For a set of solutions to form a fundamental basis, the solutions must not only satisfy the differential equation but also be **linearly independent**.
**Linear independence** means that no solution in the set can be written as a linear combination of the others. Mathematically, for two functions \(y_1(x)\) and \(y_2(x)\), they are linearly independent if there are no constants \(c_1\) and \(c_2\), not both zero, such that:
\[ c_1 y_1(x) + c_2 y_2(x) = 0 \text{ for all } x \]
In our example problem, to prove the functions \(y_1(x) = e^{2x}\) and \(y_2(x) = xe^{2x}\) are linearly independent, we can use the Wronskian determinant.

The **Wronskian Determinant** provides a robust way to test if two functions are linearly independent. For two functions \(y_1(x)\) and \(y_2(x)\), the Wronskian determinant \(W(y_1, y_2)\) is given by:
\[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \ \end{vmatrix} \]
In our example, we calculate \(W(y_1(x), y_2(x))\) as:
\[ W(y_1, y_2) = e^{2x}(e^{2x} + 2xe^{2x}) - xe^{2x}(2e^{2x}) = e^{4x} + 2xe^{4x} - 2xe^{4x}\]
Simplifying, we get:
\[ W(y_1, y_2) = e^{4x}\]
Since \( W(y_1, y_2) eq 0\), \(y_1(x) = e^{2x}\) and \(y_2(x) = xe^{2x}\) are linearly independent and thus form a fundamental basis.

A **second-order differential equation** involves the second derivative of the unknown function. It can generally be written as:
\[ a(x)y'' + b(x)y' + c(x)y = 0 \]
where \(a(x), b(x),\) and \(c(x)\) are given functions of \(x\). For homogeneous linear differential equations, the coefficients are functions of the independent variable, and the equation equals zero.
In our solved example, the differential equation is:
\[ y'' - 4y' + 4y = 0 \]
To solve it, we found two solutions, \(y_1(x) = e^{2x}\) and \(y_2(x) = xe^{2x}\), and showed they are linearly independent to form the fundamental basis.
These steps ensure that any solution to the differential equation can be written as a combination of these two functions. Thus, the general solution to the equation is:
\[ y(x) = c_1 e^{2x} + c_2 xe^{2x} \]
with \(c_1\) and \(c_2\) as arbitrary constants.