# How to find the number of arbitrary constants in solution of a given differential equation? [closed]

How to find the number of arbitrary constants in solution of a given differential equation ?

• An $n$th order differential equation will have $n$ arbitrary constants in its solution. Aug 15, 2014 at 18:03
• I think you will need to elaborate a little on the context and scope of your question before you can get a good answer. But briefly: there are as many constants as there are derivatives. Aug 15, 2014 at 18:03

e.g. $$y\color{red}'=f(x)$$ has $\color{red}{\text{one}}$ constant of integration, $$y\color{red}{''}+y'=f(x)$$ has $\color{red}{\text{two}}$, $$y\color{red}{'''}+y''=f(x)$$ has $\color{red}{\text{three}},$ etc..
To generalise, suppose we've an ODE of the form $$a_ny^{(\color{green}{n})}+a_{n-1}y^{^{(n-1)}}+\cdots+a_2y''+a_1y'+a_0y=f(x),$$ where $a_i$ is a function of $x$.
Then the number of arbitrary constants in the general solution to this equation is $\color{green}{n}.$
Notation: $y^{(n)}=\frac{d^ny}{dx^n}$
• Can we show this fact from an algebraic point of view. I mean, since the solution of a linear (homogeneous) ODE of order $n$ is a vector space whose dimension is $n$, if $y_i$ is each of this element, then any linear combination of these $y_i$, say, $y = \sum_i \alpha_i y_i$ where $\alpha_i$ are arbitrary constants $\neq 0$, spans the set of all possible solutions of the ODE. Am I write? Aug 15, 2014 at 18:39