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


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.