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$$e^x, e^{2x},...,e^{nx}$$ - fundamental system of solutions of Linear differential equation. How to find form of Linear differential equation?


I've read that we need to do next (Wronskian): $$A_{n}=\begin{bmatrix} y(x)&e^x&e^{2x}& \cdots &e^{nx}\\ y'(x)&e^x&2e^{2x}& \cdots &ne^{nx}\\ \vdots & \vdots & \ddots & \vdots \\ y^{(n)}(x)&e^x&2^{n}e^{2x}& \cdots &n^{n}e^{nx}\end{bmatrix}=0$$ and then make Laplace expansion along the first column yields. We will have $$y(x)*det_1+y'(x)*det_2+...+y^{(n)}(x)*det_n=0$$ But how to find det1, det2,..,detn ?

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The differential equation $y^{(n)} + a_{n-1} y^{(n-1)} + \ldots + a_0 y = 0$ has solution $e^{r x}$ if $r$ is a root of the polynomial $r^n + a_{n-1} r^{n-1} + \ldots + a_0$. So we just need to find a polynomial whose roots are $1, \ldots, n$: namely $(r - 1)(r-2) \ldots (r-n)$.

The coefficients turn out to be signed Stirling numbers of the first kind.

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Hint: Your eigenvalues are

$\lambda_k=k$. The characteristic polynomial is given by

$$P(\lambda)=(\lambda-\lambda_1)\cdot \ldots \cdot (\lambda-\lambda_n).$$

Use the theorem of Vieta to obtain the coefficients of

$$P(\lambda)=\lambda^n+a_{n-1}\lambda^{n-1}+\ldots+a_0.$$

The differential equation is given as:

$$y^{(n)}+a_{n-1}y^{(n-1)}+\ldots+a_0=0.$$

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