MHB How Does the Interval Affect the Unique Solution of an Initial Value Problem?

[find_the_fun]

Find an interval centered about x=0 for which the given initial value problem has a unique solution.

[math](x-2)y''+3y=x[/math] where [math]y(0)=0[/math] and [math]y'(0)=1[/math]

I've seen the answer on a few different sites (1)(2) and still don't get it

The question speaks of a unique solution with regards to an interval: what does interval have to do with unique solutions? Well in my textbook we're presented with Existence of Unique Solution theorem which I kind of don't get. It states

Let [math]a_n(x),a_{n-1}(x),...,a_1(x),a_0(x)[/math] and $$g(x)$$ be continuous on an interval I and let [math]a_n(x) \neq 0[/math] for every $$x$$ in this interval. If [math]x=x_0[/math] is any point in this interval then a solution $$y(x) $$of the initial value problem exists on the interval and is unique.
​

All this theorem says is lists some random terms that are continuous and says there's a unique solution, there are no action items.

EDIT: why divide the question by (x-2) and not 3? Why is it we're trying to make y'' have 1 as it's coefficient and not y?

What would the answer be to [math](x-2)y''+(x-3)y=x[/math] where [math]y(0)=0[/math] and [math]y'(0)=1[/math]?

 

[chisigma]

Find an interval centered about x=0 for which the given initial value problem has a unique solution.

[math](x-2)y''+3y=x[/math] where [math]y(0)=0[/math] and [math]y'(0)=1[/math]

I've seen the answer on a few different sites (1)(2) and still don't get it

The question speaks of a unique solution with regards to an interval: what does interval have to do with unique solutions? Well in my textbook we're presented with Existence of Unique Solution theorem which I kind of don't get. It states

Let [math]a_n(x),a_{n-1}(x),...,a_1(x),a_0(x)[/math] and $$g(x)$$ be continuous on an interval I and let [math]a_n(x) \neq 0[/math] for every $$x$$ in this interval. If [math]x=x_0[/math] is any point in this interval then a solution $$y(x) $$of the initial value problem exists on the interval and is unique.
​

All this theorem says is lists some random terms that are continuous and says there's a unique solution, there are no action items.

EDIT: why divide the question by (x-2) and not 3? Why is it we're trying to make y'' have 1 as it's coefficient and not y?

What would the answer be to [math](x-2)y''+(x-3)y=x[/math] where [math]y(0)=0[/math] and [math]y'(0)=1[/math]?

Let's write the ODE as...

$\displaystyle y^{\ ''} + \frac{3}{x-2}\ y = \frac{x}{x-2}\ ,\ y(0)=0\ ,\ y^{\ '}(0)=1\ (1)$

This ODE has a [regular...] singular point in x=2, so that in x=0 it has an analytic solution...

$\displaystyle y(x) = \sum_{n=0}^{\infty} a_{n}\ x^{n}\ (2)$

A first question: how to find the $a_{n}$?... the initial conditions permit us to find $a_{0}=0$ and $a_{1}=1$. From the (1) we derive...

$\displaystyle y^{\ ''} = \frac{x - 3\ y}{x-2} \implies y^{\ ''} (0) = 0 \implies a_{2}=0\ (3)$

$\displaystyle y^{\ '''} = \frac{1 - 3\ y^{\ '}}{x-2} - \frac{x - 3\ y}{(x-2)^{2}} \implies y^{\ '''} (0)=1 \implies a_{3} = \frac{1}{6}\ (4)$

... and so one. A second question: what is the convergence interval of (2)?... the fact that x=2 is a singular point means that the series converges for $-2 < x < 2$. For saying what happens outside this interval we have to analyse better the solution...

Kind regards

$\chi$ $\sigma$