I have heard that the solution of nth order linear homogeneous DE of constant coefficient is expressed as e^mx, xe^mx, (x^2)e^mx, ... , (x^(k-1))e^mx when the auxiliary equation have repeated roots of k multiplicity.
But how can we prove it? I searched it for 4 hours...But I can't find the proof
First, checking these are solutions: for $n\ge 1$
$$\frac{d}{dx}x^ne^{mx}=nx^{n-1}e^{mx}+mx^ne^{mx}$$
So,
$$\left(\frac{d}{dx}-m\right)x^ne^{mx}=nx^{n-1}e^{mx}$$
$$\implies\left(\frac{d}{dx}-m\right)^nx^ne^{mx}=n!e^{mx}$$ Therefore, $\left(\frac{d}{dx}-m\right)^kx^ne^{mx}=0$ for $k>n$, so $e^{mx},xe^{mx},...,x^{k-1}e^{mx}$ are all solutions to $\left(\frac{d}{dx}-m\right)^kf=0$. Moreover, they are linearly independent as the monomials $x^i$ are themselves linearly independent.
Now, the solution set of an order $k$ linear ode has dimension $k$, so our set of $k$ linearly independent solutions must in fact be a basis.