# Does any $n$-th order ODE have to have $n$ arbitrary contants in its general solution? [duplicate]

In the lecture, my proffesor stated that

Any $n$-th order ODE has $n$ arbitrary constants in its general solution.

However, we have later seen that some ODEs have some singularities, so the so called "the general solution" is not so general at all in many cases, and this lead me to question that is there any execeptions to the rule that I have quoted.

For any ordinary DE, does the general solution of this ODE has to have $n$ arbitrary constants ? If so, how ? if not, can you provide an example ?

Edit:

As I have pointed out, I'm asking this question for ODEs, and not PDEs.

If your ODE is of the following form $$x^{(n)}(t)=f(t,x^{(0)}(t),\ldots, x^{(n-1)}(t))$$ and $f:\mathbb{R}\times\mathbb{R}^n\rightarrow\mathbb{R}$ is locally Lipschitz, then the general solution has to include $n$ arbitrary constants. It boils down to an application of the Picard-Lindelöf theorem. You can transform your ODE into the following form: $$y'(t)=F(t,y(t))$$ with $F:\mathbb{R}\times\mathbb{R}^{n}\rightarrow\mathbb{R}^n$ locally Lipschitz. Then you can impose an arbitrary initial condition $y_0=y(\hat{t})\in\mathbb{R}^n$ and Picard-Lindelöf gives you a unique solution. Therefore you can impose exactly $n$ arbitrary constants, which have to be reflected in the general solution.

A counterexample can be found by looking at nonunique solutions of ODE's, for example $x'=\sqrt{x}$ with $x(0)=0$. it has at least two solutions, $x_1(t)=0$ and $x_2(t)=\frac{t^2}{4}$ but you can combine these two, to obtain a more general set of solutions like this: $$x(t)=\left\{\begin{array}{cc}\frac{(t-t_0)^2}{4},& t\geq t_0\\ 0,& t<t_0\end{array}\right.$$ with $t_0\geq 0$. Hence you can impose an initial condition and at least another parameter $t_0$.

To the PDE Question of things: For PDE's there is no general uniqueness statement, therefore you cannot expect to find a general solution which allows you to prescribe a given number of parameters. Even if you have uniqueness the number of prescribable parameters usually becomes infinite. Let me explain what I mean in an example: Imagine you have the following PDE $\Delta u= f$ on a compact smooth domain $\Omega\subset \mathbb{R}^n$, with a given $f:\Omega\rightarrow\mathbb{R}$ at least integrable. Then you can prescribe so called boundary conditions, which will make the solution of the problem unique. What I mean is the following: You can prescribe $u|_{\partial\Omega}=g$ with $g:\partial\Omega\rightarrow\mathbb{R}$ for example continuous. Since $\partial\Omega$ has infinitely many points, you have prescribed infinitely many parameters.

As an additional comment: You might want to check out Green's functions, they give you a kind of general solution to some linear PDEs. Here is a wiki link: https://en.wikipedia.org/wiki/Green%27s_function

• Shouldn't those be $\mathbb{R} \times \mathbb{R}^{n-1}$ ?
– Our
Feb 15, 2018 at 6:20
• @onurcanbektas Actually no, I made a different mistake, when I was writing down the ODE $x^{(n)} = f(t,x^{(1)},\ldots,x^{(n-1)})$ but it has to be $x^{(0)}$ instead of $x^{(1)}$. I corrected the answer. Feb 15, 2018 at 9:51