{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

3 - init_cond

# 3 - init_cond - Series sol near ordinary points w given...

This preview shows pages 1–3. Sign up to view the full content.

Series sol. near ordinary points w/ given initial conds. Let’s revisit the 2 problems that we have solved in class – but let’s solve them with some initial conditions. We will solve them two using two different methods. 1. Using the results we obtained in ordinary_ps.pdf . 2. Using a Taylor series expansion of the solution about x 0 . Before doing so, recall that the Talor series expansion of a function about point x 0 is given by y ( x ) = y ( x 0 ) + y 0 ( x 0 )( x - x 0 ) + y 00 ( x 0 ) 2! ( x - x 0 ) 2 + y 000 ( x 0 ) 3! ( x - x 0 ) 3 + y (4) ( x 0 ) 4! ( x - x 0 ) 4 + · · · . (1) For a given second order initial value problem (IVP), we already know y ( x 0 ) and y 0 ( x 0 ) . What we need to find are y 00 ( x 0 ) , y 000 ( x 0 ) , · · · . We will find these using the given differential equation, as illustrated in the problems below. 1 c HF, 2011

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Series sol. near ordinary points w/ given initial conds. Problem 1 (1 + x ) y 00 + y = 0 , x 0 = 0 . y (0) = 1 , y 0 (0) = 1 . (2) Solution – Method A : Recall that we previous found the general solution to the problem to be (see ordinary_ps.pdf ) y = a 0 1 - 1 2 x 2 + 1 6 x 3 - 1 24 x 4 + · · · + a 1 x - 1 6 x 3 + 1 12 x 4 + · · · , (3) where a 0 and a 1 are two arbitrary constants. Using the first initial conditions, we find y (0) = 1 = a 0 1 - 1 2 0 2 + 1 6 0 3 - 1 24 0 4 + · · · + a 1 0 - 1 6 0 3 + 1 12 0 4 + · · · a 0 = 1 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

3 - init_cond - Series sol near ordinary points w given...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online