Chem Differential Eq HW Solutions Fall 2011 184

# Chem Differential - A184 Appendix A Ordinary Diﬀerential Equations Review of Concepts and Methods Wronskian cos x sin x − sin x cos x W x = = 1

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: A184 Appendix A Ordinary Diﬀerential Equations: Review of Concepts and Methods Wronskian: cos x sin x − sin x cos x W ( x) = = 1. We now apply the variation of parameters formula with g ( x) = yp sec x = 1 ; cos x −y2 g(x) dx + y2 W ( x) y1 g(x) dx W ( x) = y1 = − cos x = cos x · ln (| cos x|) + x sin x. sin x dx + sin x cos x dx Thus the general solution is y = c1 cos x + c2 sin x + cos x · ln (| cos x|) + x sin x. 29. x2y + 3 xy + y = indicial equation is √ x. The homogeneous equation is an Euler equation. The r2 + 2r + 1 = 0 ⇒ (r + 1)2 = 0. We have one double indicial root r = −1. Hence the solutions of the homogenous equation y1 = x−1 and y2 = x−1 ln x. Wronskian: 1 x W ( x) = −1 x2 1 x ln x 1−ln x x2 = 1 − ln x ln x 1 + 3 = 3. 3 x x x We now apply the variation of parameters formula with √ x 3 g ( x) = = x− 2 ; 2 x yp −y2 g(x) dx + y2 W ( x) = y1 = − = 1 − x = − = y1 g(x) dx W ( x) 1 2 2 3/2 4 1/2 =x. x x33 9 1 x ln x ln x 3 − 3 x x 2 dx + x x dv u ln x 1 3 −3 x x 2 dx x √ x dx + ln x x 1 2 3/2 x ln x − x3 1 x 2 dx 2 3/2 1 ln x 2 3/2 x dx + x 3 x x3 Thus the general solution is 4 y = c1 x−1 + c2x−1 ln x + x1/2. 9 33. x2y + 3xy + y = 0. See Exercise 29. 37. x2y + 7xy + 13 y = 0. Euler equation with α = 7, β = 13, indicial equation r2 + 6 r + 13 = 0; indicial roots: √ r = −3 ± −4; r1 = −3 − 2i, r2 = −3 + 2i. ...
View Full Document

## This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

Ask a homework question - tutors are online