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As a consequence of uniqueness and existence for ODE's we know that a ODE of nth order can have no more than n linear independent solutions. This question sums it up pretty well.

Unfortunately Qiaochu Yuan restricts itself to ODE with constant coefficients. Under what circumstances do we get a ODE with less than n linear independent solutions?

2 Answers2

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The argument using the existence and uniqueness theorem (plus Grönwall's inequality to show that existence is global) shows that there are exactly $n$ for any $n$'th order homogeneous linear differential equation on an interval where the coefficients are continuous and the leading coefficient is nonzero.

Where you can have fewer than $n$ is when the leading coefficient is $0$ at some point in the interval. Thus $x^2 y'' - 2 y = 0$ has only one linearly independent solution (namely $x^2$) defined on an interval containing $0$: the general solution is $a x^2 + b/x$ which is not defined at $x=0$ if $b \ne 0$.

Robert Israel
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  • So assuming continuous coefficients we can, probably a bit sloppy speaking, only get less than n solution in certain points, not on a whole interval? As in your example, I can just restrict myself to an interval without {0} to get 2 lin. ind. solutions again. – Felix Crazzolara Jun 28 '17 at 16:04
  • Or I could just multiply by x. – Felix Crazzolara Jun 28 '17 at 16:19
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A linear ODE of nth order has always $n$ linear independent solutions. This fact follows, for example, from the general theory of first order systems of linear equations.

Rigel
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