Motivation: It is a well-known fact that $ay''+by'+cy=0$ has solutions which are found from substituting the ansatz $y=e^{\lambda t}$ into the DEqn. It turns out that we replace the calculus problem $ay''+by'+cy=0$ with the algebra problem of solving the characteristic equation $a\lambda^2+b\lambda+c=0$. When the solution is a conjugate pair of complex numbers or distinct pair of real numbers the solutions arise from $e^{\lambda t}$. On the other hand, when the solution is real and repeated then the ansatz solution $y=e^{\lambda t}$ only covers half of the general solution.
Suppose that $a\lambda^2+b\lambda+c=0$ has double root solution $\lambda = r$ then we form the general solution of $ay''+by'+cy=0$ as $$ y(t) = c_1e^{rt}+c_2te^{rt}. $$ The inclusion of the $t$ in the solution is surprising to many students. I think many have asked "where'd the $t$ come from?". Of course, we could just as well ask "where the $e^{\lambda t}$ come from?". I know of several ways to derive the $t$. In particular:
$y''=0$ integrates twice to $y=c_1+tc_2$ and $e^{0t}=1$ so this is an example of the double root. A simple change of coordinates allows this derivation to be extended to an arbitrary double-root.
reduction of order to a system of ODEs in normal form. We'll obtain a $2 \times 2$ matrix which is not diagonalizable. However, the matrix exponential gives a solution and the generalized e-vector piece generates the $t$ in the second solution.
you can use the second linearly independent solution formula from the theory of ODEs. This formula is found by making a reduction of order based on the fact $y=e^{rt}$ is a solution. After a bit the problem reduces to a linear ODE which integrates to give a lovely formula with nested integrals. This formula also will derive the $t$ in the double root solution.
Laplace transforms. We can transform the given ODE in $t$ to obtain an algebra equation with $(s-r)^2Y$ which gives $\frac{F(s)}{(s-r)^2}$ and upon inverse transform the appearance of the $(s-r)^2$ in the denominator gives us the $te^{rt}$ solution
Inverse operators. By writing the given ODE as $(D-r)^2[y]=0$ we can integrate in a certain way and again derive the $te^{rt}$ solution.
Series solution techniques.
added 10/6: start with the distinct root solution $y=c_1e^{\lambda_1 t}+c_2e^{\lambda_2t}$ and consider the limit $\lambda_1 \rightarrow \lambda_2$ to derive the second solution.
These are the methods which seem fairly obvious in view of the introductory course (up to notation, several of these are the same method). My question is this:
Question: What is the history of the solution $y=te^{rt}$? Who studied the problem $ay''+by'+cy=0$ and found this solution?
I'm also interested in the particular sub-histories of the other methods I mention above.
Thanks in advance for any insights!