{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# ODEHW2 - Math 135 HW2 solutions to selected exercises Mike...

This preview shows pages 1–2. Sign up to view the full content.

Math 135, HW2 solutions to selected exercises Mike Hall p. 384 5d) Find a function whose Laplace transform is equal to 1 p ( p +1) . Using partial fractions decomposition we see that 1 p ( p + 1) = 1 p - 1 p + 1 = L [1] - L [ e - x ] = L [1 - e - x ] so f ( x ) = 1 - e - x is a solution (which is unique, up to some restrictions we don’t care that much about, by some theorem we take for granted for now). p. 394 1b) Find the Laplace transform of f ( x ) = (1 - x 2 ) e - x . By the properties of the Laplace transform, we have L [ f ( x )]( p ) = L [(1 - x 2 ) e - x ]( p ) = L [1 - x 2 ]( p + 1) = ± 1 s - 2 s 3 ²³ ³ ³ ³ s = p +1 = 1 p + 1 - 2 ( p + 1) 3 . 3d) Solve: y 00 + y 0 = 3 x 2 , y (0) = 0 , y 0 (0) = 1. Taking Laplace transforms of both sides, we get ( p 2 L [ y ] - py (0) - y 0 (0)) + ( pL [ y ] - y (0)) = 3 · 2 p 3 and plugging in for y (0) and y 0 (0), we get L [ y ]( p 2 + p ) - 1 = 6 p 3 L [ y ] = ± 6 p 3 + 1 ² 1 p ( p + 1) We use partial fractions decomposition to break up the right hand side into recognizable Laplace transforms. One thing to note is that it’s not always best to combine everything into one big fraction, as it may only

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

ODEHW2 - Math 135 HW2 solutions to selected exercises Mike...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online