# Is this behaviour generalizable?

1. Mar 6, 2014

### Whovian

Notice that, when solving second-order linear homogeneous constant-coefficient ODEs, the classic linearly independent solutions are $y_1=e^{r_1\cdot t}$ and $y_2=e^{r_2\cdot t}$ for roots $r_1,r_2$ of our auxiliary equation. (Of course, we can pick any other basis for this two-dimensional vector space.)

If those are linearly dependent, the independent solutions chosen are $y_1=e^{r\cdot t}$ (no surprise) and $y_2=t\cdot y_1=\frac{\partial y_1}{\partial r}$.

Similarly, for homogeneous second-order (insert long list of adjectives here) Cauchy-Euler equations, we set up a quadratic equation and find $y_1=t^{r_1}$ and $y_2=t^{r_2}$ as our linearly independent solutions, or, if those are linearly dependent, $y_1=t^r$ and $y_2=t^r\cdot\log\left(t\right)=\frac{\partial y_1}{\partial r}$.

This is suggesting that, for nth-order linear homogeneous ODEs, if we can guess solutions of the form $f\left(r,t\right)$,

If n solutions $r_1,\ldots,r_n$ show up for $r$, we can use $\left\{f\left(r_1,t\right),\ldots,f\left(r_n,t\right)\right\}$ as a basis for the set of solutions.

If any repeated roots show up (slightly sketchy since it might not be a polynomial in terms of $r$,) we use $f\left(r_1,t\right),\ldots,f\left(r_n,t\right)$ along with the partial derivatives with respect to $r_k$, up to the $\mathrm{mult}\left(r_k\right)$th derivative, for all $k$, as our basis.

Now, for obvious reasons, this probably only holds in specific situations. Have I run into the specific couple for which our linearly independent solutions are generated by taking the partial derivative if a repeated root shows up, or is this generalizable in any way such as the near-unreadable example I suggested?

2. Mar 6, 2014

### lurflurf

How generalizable?
Cauchy-Euler equations are related to constant-coefficient by the chain rule.
$$\left( \dfrac{dx}{du} \dfrac{d}{dx} \right)^n=\left( \dfrac{d}{du} \right)^n$$
so if u=g(x)
$$\mathrm{y}_2(u)=u \, \mathrm{y}_1(u)$$
then
$$\mathrm{y}_2(\mathrm{g}(x))=\mathrm{g}(x) \, \mathrm{y}_1(\mathrm{g}(x))$$