# Can one prove that there are exactly $n$ arbitrary constants exist in the solution of a $n$th order differential equation?

I've heard that an $$n$$-th order differential equation will always have exactly $$n$$ arbitrary constants in its solution; that is, if $$y$$ satisfies the differential equation $$f(y(t),y^{(1)}(t),\cdots, y^{(n)}(t), t) = 0$$ then it will always be of the form $$y(t) = g(t; C_1, \cdots, C_n)$$ where $$C_1, \cdots C_n \in \mathbb{C}$$. This makes intuitive sense, as one might have to integrate $$n$$ times to get to the solution, which would mean $$n$$ constants of integration. It can even be proven quite easily in the constant-coefficient case; given that $$\sum _{i = 1} ^{n} a_i \hat{D}^i y(t) = f_0(t)$$, one can factor the differential operator $$\sum _{i = 1} ^{n} a_i \hat{D}^i$$ that acts on $$y$$ into $$n$$ first order derivatives according to solutions to the characteristic equation, which are all integrated to create $$n$$ arbitrary constants.

However, I am not satisfied with this argument. Consider, for example, the following second-order differential equation: \begin{align} \sin(y'') + (y')^2 = y\cos(t^2y''). \end{align} I cannot imagine how one would be able to apply either of the previous arguments (factorisation of the differential operator or $$n$$ integrations) to show that this has exactly two arbitrary constants. And even if one could show that there are two integrations in this case, the general case is still uncertain: Are there always going to be $$n$$ arbitrary constants? Is there perhaps a rigorous proof of this?

• Interesting question. Aug 14, 2020 at 13:24

This is, under some extra conditions on the functions defining the ODE, a consequence of the Picard-Lindelof Theorem for vector functions. Any n-th order ODE $$y^{(n)} = f(t,y,y',...,y^{(n-1)})$$ can be reduced to an ODE of the form $$v'=F(v,t)$$ where $$v: \mathbb{R} \rightarrow \mathbb{R}^{n}$$ is given by $$v_{1} = y, v_{2} = y',...,v_{n} = y^{(n-1)}$$ and so on. To see this take $$(F(v,t))_{i} = v_{i+1}$$ and $$(F(v,t))_{n}$$ to be the function defining the ODE, with $$y^{(k)}$$ replaced appropriately. If we take $$F$$ to be continuous in t and lipschitz in $$v$$ we get that any solution of the ODE is uniquely determined by the initial condition of $$v$$, which is n initial conditions on y and its' derivatives up to $$n-1$$.