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My Problem is: i have a given differential equation: $$y^{\prime\prime\prime}-\frac{x^2}{x^2-2x+2}\cdot y^{\prime\prime}+\frac{2x}{x^2-2x+2}\cdot y^{\prime}-\frac{2}{x^2-2x+2}\cdot y= 0$$and the given functions : $$y_{1}=x \quad y_{2}=x^2 \quad y_{3}=e^{x}$$

A fundamental system is defined as a:

$\{y_1,\ldots,y_n\} \quad $ with: $\quad \mathcal{L} := \{y \in C^1([a,b]; \mathbb{R}^n)\ |\ y = \sum_{k=1}^na_ky_k\ ,\ a_1, \ldots, a_n \in \mathbb{R}\}$

How can i show, that the given functions are a fundamental system for the given differential equation?

My Approach was: i managed to show, that the given functions are solutions for the differential equation.

for $y_{1} = x$ $$y^{\prime}=1 \quad y^{\prime\prime}=0 \quad y^{\prime\prime\prime}=0$$ that's why: $$0 - 0 + \frac{2x}{x^2-2x+2}\cdot 1 -\frac{2}{x^2-2x+2}\cdot x =0$$ $$\frac{2x}{x^2-2x+2} -\frac{2x}{x^2-2x+2}=0$$ $$0=0$$

for $y_{2} = x^2$ $$y^{\prime}=2x \quad y^{\prime\prime}=2 \quad y^{\prime\prime\prime}=0$$ that's why: $$0 - \frac{x^2}{x^2-2x+2}\cdot 2 +\frac{2x}{x^2-2x+2}\cdot 2x -\frac{2}{x^2-2x+2}\cdot x^2 =0$$ $$0 - \frac{2x^2}{x^2-2x+2} +\frac{4x^2}{x^2-2x+2} -\frac{2x^2}{x^2-2x+2} =0$$ $$0=0$$

for $y_{3} = e^x$ $$y^{\prime}= e^x \quad y^{\prime\prime}= e^x \quad y^{\prime\prime\prime}= e^x$$ that's why: $$ e^x - \frac{x^2 e^x}{x^2-2x+2} +\frac{2x e^x}{x^2-2x+2} -\frac{2 e^x}{x^2-2x+2} =0$$ $$e^x +\frac{ -e^x(x^2-2x+2)}{x^2-2x+2}=e^x-e^x=0$$

But now i am stuck, i don't think, that this is the proof i am looking for. i dont know how to show it... maybe you can help?

P.S.: edits were made to improve language and latex

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up vote 1 down vote accepted

Your proof is missing two parts:

  1. That every function of the form $\sum a_i y_i$ is also a solution.
  2. That every solution is of the form $\sum a_i y_i$.

$1$ is satisfied since this is a linear ODE. $2$ is satisfied since the solutions are linearly independent and the order of the ODE is equal to the number of solutions.

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