Zeigen Sie, daB die Funktionen \(y_{1}=\mathrm{e}^{x}, y_{2}=\mathrm{e}^{-x}\) und \(y_{3}=\mathrm{e}^{-2 x}\) eine Fundamentalbasis der homogenen linearen Differentialgleichung 3. Ordnung $$ y^{\prime \prime \prime}+2 y^{\prime \prime}-y^{\prime}-2 y=0 $$ bilden.

A homogeneous linear differential equation is a type of differential equation where the function and all its derivatives appear linearly. This means there are no products or powers of the function or its derivatives.
The general form of a homogeneous linear differential equation of order 3 is given as:
$$ y^{(3)} + a_{2}y^{(2)} + a_{1}y^{\text{'}} + a_{0}y = 0. $$
Each term involves either the function \(y\) or its derivatives multiplied by constants \(a_{0}\), \(a_{1}\), and \(a_{2}\). The term 'homogeneous' signifies that the right side of the equation is zero.
Our specific equation is:
$$ y^{(3)} + 2y^{(2)} - y^{\text{'}} - 2y = 0. $$
This means we'll be examining solutions where \(y = e^{x}\), \(y = e^{-x}\), and \(y = e^{-2x}\) to verify if they satisfy this equation.

The Wronskian-Determinant is a tool used to determine whether a set of functions are linearly independent. It's named after the Polish mathematician Józef Maria Hoene-Wroński.
To calculate the Wronskian for functions \(y_{1}\), \(y_{2}\), and \(y_{3}\), we form a determinant from the functions and their derivatives:
$$ W(y_{1}, y_{2}, y_{3}) = \begin{vmatrix} y_{1} & y_{2} & y_{3} \ y_{1}^{'} & y_{2}^{'} & y_{3}^{'} \ y_{1}^{''} & y_{2}^{''} & y_{3}^{''} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ end{vmatrix}. $$
Evaluate this determinant to check if it's non-zero. If it is, the functions are linearly independent.
For our functions \(y_{1} = e^{x}\), \(y_{2} = e^{-x}\), and \(y_{3} = e^{-2x}\), we follow these steps:
1. Compute first and second derivatives.
2. Form the determinant matrix.
3. Evaluate at a specific point such as \(x = 0\). If this determinant is non-zero, we conclude that our functions are linearly independent.

Linear independence is crucial for understanding if a set of solutions forms a fundamental set.
If the functions \(y_{1}\), \(y_{2}\), and \(y_{3}\) are linearly independent, it means none of these functions can be written as a combination of the others.
In mathematical terms, this means the only solution to:
$$ c_{1}y_{1} + c_{2}y_{2} + c_{3}y_{3} = 0 $$
where \(c_1\), \(c_2\), and \(c_3\) are constants, is \(c_{1} = c_{2} = c_{3} = 0\).
Verifying this involves computing the Wronskian determinant as discussed earlier. If the Wronskian is non-zero at any point, it implies linear independence.
Once we determine linear independence, we can be confident that our set of functions makes up a fundamental set of solutions to the differential equation.

Third-order differential equations involve derivatives up to the third degree. These equations can describe more complex systems and phenomena than lower-order differential equations.
Our specific third-order equation is:
$$ y^{'''} + 2y^{''} - y^{'} - 2y = 0. $$
To solve it, we need three independent solutions since it is a third-order differential equation.
We look for solutions of the form \(e^{kx}\) and then find the necessary \(k\) by substituting back into the equation.
This leads to solving characteristic equations and finding values for \(k\) that help form the general solution. For our equation, we verify solutions \(y_{1} = e^{x}\), \(y_{2} = e^{-x}\), and \(y_{3} = e^{-2x}\). Substituting each and showing they work helps us confirm these as valid solutions.

The fundamental set of solutions for a differential equation is a set of linearly independent solutions that can span the entire solution space.
For a third-order linear homogeneous differential equation, this set will consist of exactly three linearly independent solutions.
Given the Wronskian determinant of our solutions \(y_{1} = e^{x}\), \(y_{2} = e^{-x}\), and \(y_{3} = e^{-2x}\), if it is non-zero, we can confirm they form a fundamental set of solutions.
This completeness implies that any solution to the differential equation can be expressed as:
< $$ y = c_{1}y_{1} + c_{2}y_{2} + c_{3}y_{3}, $$ < < \br> for any constants \(c_{1}\), \(c_{2}\), and \(c_{3}\).
This is why calculating the Wronskian and verifying linear independence is so important – it ensures our solutions can cover the entire solution space.