- Thread starter
- #1

#### find_the_fun

##### Active member

- Feb 1, 2012

- 166

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

\(\displaystyle (x-2)y''+3y=x\) where \(\displaystyle y(0)=0\) and \(\displaystyle y'(0)=1\)

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 text book we're presented with Existence of Unique Solution theorem which I kind of don't get. It states

Let \(\displaystyle a_n(x),a_{n-1}(x),...,a_1(x),a_0(x)\) and \(\displaystyle g(x)\) be continuous on an interval I and let \(\displaystyle a_n(x) \neq 0\) for every \(\displaystyle x\) in this interval. If \(\displaystyle x=x_0\) is any point in this interval then a solution \(\displaystyle y(x) \)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 \(\displaystyle (x-2)y''+(x-3)y=x\) where \(\displaystyle y(0)=0\) and \(\displaystyle y'(0)=1\)?

\(\displaystyle (x-2)y''+3y=x\) where \(\displaystyle y(0)=0\) and \(\displaystyle y'(0)=1\)

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 text book we're presented with Existence of Unique Solution theorem which I kind of don't get. It states

Let \(\displaystyle a_n(x),a_{n-1}(x),...,a_1(x),a_0(x)\) and \(\displaystyle g(x)\) be continuous on an interval I and let \(\displaystyle a_n(x) \neq 0\) for every \(\displaystyle x\) in this interval. If \(\displaystyle x=x_0\) is any point in this interval then a solution \(\displaystyle y(x) \)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 \(\displaystyle (x-2)y''+(x-3)y=x\) where \(\displaystyle y(0)=0\) and \(\displaystyle y'(0)=1\)?

Last edited: