I am trying to find how many solutions of a second order ODE satisfy given initial conditions. For example how many solutions of $3xy'' + y' = 0$ satisfy $y(1)=y'(1)=1$, $y''(1)=-3$?
Is there an existence and uniqueness theorem I can apply for second order equations to show there is one solution? Or shall I solve the equation and see how many solutions I end up with?
First, note that for a single first-order equation, we have the standard existence and uniqueness theorem: if we have $$ u' = F(x, u), \qquad u(a) = b, $$ where $F$ and $\partial F/\partial u$ are continuous at $(a,b)$, then there exists a finite-sized interval centered at $a$ on which a unique solution to this ODE exists. This theorem can also be extended to coupled systems of ODEs: $$ u_i' = F_i(x, u_j), \qquad u_i(a) = b_i, $$ where $i$ & $j$ run over the number of functions we're trying to solve for, and we have extended the derivative conditions on the functions $F_i$ in the obvious way.
Second, note that any $n$th-order equation can be put into first-order form: if we have the equations and initial conditions $$ y^{(n)} = F(x, y, y', ... y^{(n-1)}), \quad y(a) = b_1, y'(a) = b_2, \dots, y^{(n-1)} = b_{n}, $$ then we can define $u_i = y^{(i-1)}$, and the system becomes \begin{align*} u_1' &= u_2 & u_1(a) &= b_1 \\ u_2' &= u_3 & u_2(a) &= b_2 \\ &\vdots && \vdots& \\ u'_{n-1} &= u_n & u_{n-1}(a) &= b_{n-1} \\ u'_n &= F(x, u_1, u_2, \dots, u_n) & u_n(a) &= b_n \end{align*} which is in the first-order form from above. Thus, a unique solution to this system (and therefore to our original ODE) exists in some interval centered about $a$.
Finally, note that the interval of validity can be found explicitly for a linear ODE of any order in the same way; in this case we have $$ u_i' + \sum_j f_{ij}(x) u_j = g_i(x) $$ and the interval of validity of our solution will be the largest interval that contains $a$ and does not contain any singularities of the functions $f_{ij}(x)$ or $g_i(x)$.