Uniqueness of Solutions for 2nd Order Linear Homogeneous ODEs

[ognik]

Hi, please review my answer, I suspect I am missing some fine points...

y(x) is a solution to a 2nd order, linear, homogeneous ODE. Also y(x₀)=y₀ and dy/dz=y'₀
Show that y(x) is unique, in that no other solution passes through (x₀, y₀) with a slope of y'_0.

Expanding y(x) in a Taylor series, $ y(x) = \sum_{n=0}^{\infty} {y}^{n}({x}_{0}) \frac{{ (x -{x}_{0}) }^{n}}{n!} = y({x}_{0}) + {y}^{'}({x}_{0}) + \sum_{n=2}^{\infty} {y}^{n}({x}_{0}) \frac{{ (x -{x}_{0}) }^{n}}{n!} $

So I assume another solution is h(x) - but I am told that h(x₀) = y₀ and that the slope of h(x) is dy/dx, so when I expand h(x) in a Taylor series, I will get a series identical to y(x) - and because the 1st derivatives are equal, all higher derivatives will also be equal.

Therefore the function y(x) is the only unique solution.

 

[Fernando Revilla]

I will get a series identical to y(x) - and because the 1st derivatives are equal, all higher derivatives will also be equal.

You can't assure it. The question is that if $y(x)$ is a solution of $y^{\prime\prime}+ay^{\prime}+by=0$ with $y(0)=y_0$ and $y^{\prime}(0)=y^{\prime}_0,$ by a well-known theorem, $y(x)$ is unique and besides $y(x)$ is analytic. In the proof of the theorem we don't use series expansions.

You can verify the uniqueness, with the hypothesis of being $y(x)=\displaystyle\sum_{n\ge 0}a_n(x-x_0)^n,$ substituting this expansion in the differential equation and using the initial conditions. You'll obtain necessary conditions for the coefficients $a_n$ (i.e. uniqueness).

 

[ognik]

The book suggested the approach I used ("Hint. Assume a second solution satisfying these conditions and compare
the Taylor series expansions.") but you seem to have a good point so I looked at it as well.

But I'm not sure the problem is to expand about point x₀, I think y(x₀)=y₀ is just an initial condition?

Anyway, this is in a section on Frobenius's method, so I should use the substitution $y=\sum_{\lambda = 0}^{\infty}{a}_{\lambda}{x}^{k+\lambda} $, differentiate and substitute back into a standard ODE (y'' + Py' + Qy = 0). I get the expected indicial eqtn, with k=0 or 1. Following Frobenius from there, I don't see how to show the coefficiants I get mean only 1 coefficient, in fact I managed to get 2 coefficiants :-( ?