# IVP and Boundary Value Problem

• Jan 28th 2010, 12:40 PM
IVP and Boundary Value Problem
Find an interval centered about x=0 for which the given IVP has a unique solution:
(x-2)y''+3y=x, y(0)=0, y'(0)=1

what exactly is the question asking?? Do I need to solve the IVP to find the boundaries?
• Jan 28th 2010, 05:49 PM
Jose27
Quote:

Find an interval centered about x=0 for which the given IVP has a unique solution:
(x-2)y''+3y=x, y(0)=0, y'(0)=1

what exactly is the question asking?? Do I need to solve the IVP to find the boundaries?

Well, you could solve it then prove that this solution is unique (not too difficult, I think, since it's a linear equation). Or you could use the existence and uniqueness theorem for linear equations that says that if $y''=P(x)+Q(x)y+R(x)y'$ where $P,Q,R$ are continous functions of $x$ on $a then there is a unique solution to this equation $y:(a,b) \rightarrow \mathbb{R}$ with the specified initial conditions on a point.
• Feb 1st 2010, 03:37 AM
chisigma
If we suppose that the solution we are searching is analitic in $x=0$ we can write...

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

$y^{'}(x)= \sum_{n=1}^{\infty} n\cdot a_{n}\cdot x^{n-1}$

$y^{''} (x)= \sum_{n=2}^{\infty} n\cdot (n-1)\cdot a_{n}\cdot x^{n-2}$ (1)

From (1) and the 'initial conditions' we derive that is...

$a_{0}=0$

$a_{1}=1$ (2)

... and the DE can be written as...

$3\cdot \sum_{n=2}^{\infty} a_{n}\cdot x^{n} + \sum_{n=2}^{\infty} n\cdot (n-1)\cdot a_{n}\cdot x^{n-1} -2\cdot \sum_{n=2}^{\infty} n\cdot (n-1)\cdot a_{n}\cdot x^{n-2} = -2\cdot x$ (3)

If we impose the identity (3) for all $n>1$ we obtain first...

$a_{2}=0$ (4)

... and for $n>2$ ...

$a_{n}= \frac{3\cdot a_{n-2} + (n-1)\cdot (n-2)\cdot a_{n-1}}{2\cdot n\cdot (n-1)}$ (5)

The conlusion is that the solution can be written as...

$y(x)= x + \sum_{n=3}^{\infty} \alpha_{n}\cdot (\frac{x}{2})^{n}$ (6)

... and because the sequence $\alpha_{n}$ in (6) is bounded the series in (6) will converge if $|x|<2$...

Kind regards

$\chi$ $\sigma$