- #1

ognik

- 643

- 2

y(x) is a solution to a 2nd order, linear, homogeneous ODE. Also y(x

_{0})=y

_{0}and dy/dz=y'

_{0}

Show that y(x) is unique, in that no other solution passes through (x

_{0}, y

_{0}) 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(x

_{0}) = y

_{0}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.