Using arbitrary initial conditions to solve an IVP Suppose that we have a second order linear differential equation $$ay'' + by' + cy = 0.$$
In order to solve that differential equation we need two additional pieces of information, namely the initial conditions (or boundary conditions depending on the problem). Now, typically the initial conditions given in an example like this are $y(0)$ and $y'(0)$. But wouldn't we be able to solve the equation if we were given say $y(0)$ and $y''(0)$, or any pair $y^{(m)}(0)$ and $y^{(n)}(0)$ where $m$ and $n$ need not be less than two?
Thank you in advance.
 A: Solutions to the equation you provide are generally in the form $$y=C_1e^{r_1t}+C_2e^{r_2t}$$
when the characteristic equation has real distinct roots. Then differentiating gives us
$$y'=r_1C_1e^{r_1t}+r_2C_2e^{r_2t}$$
$$y''=r_1^2C_1e^{r_1t}+r_2^2e^{r_2t}$$
$$y^{(n)}=r_1^nC_1e^{r_1t}+r_2^ne^{r_2t}$$
Say you were given initial conditions $y^{(n)}(0)=k_1$ and $y^{(m)}(0)=k_2$. Then we have the system of equations $$\begin{cases}k_1=r_1^nC_1+r_2^nC_2\\k_2=r_1^mC_1+r_2^mC_2\end{cases}$$
which will generally give a unique solution for $C_1,C_2$. But things can go wrong. What if $r_1=0$ or $r_2=0$? Then we don't necessarily have a solution at all! And even if we do, it won't be unique. We've found a way things can go wrong, even by assuming the roots were real and distinct.
There is actually a very good reason the initial conditions are usually given for $y$ and $y'$: The Existence and Uniqueness Theorem for Second Order Linear Differential Equations. This theorem assumes the initial conditions are given in that form. Otherwise, things can go wrong as I've shown above.
