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.)(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Is this behaviour generalizable?

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