# hw1_s - Homework 1 | Due on January 22(Thursday | Solution...

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

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.

Unformatted text preview: Homework 1 | Due on January 22 (Thursday) | Solution 1 1. Find the general solutions of the following 1st order ODE: (a) xy + 2 y = sin x First, find the integrating factor μ ( x ): xμ ( x ) y + 2 μ ( x ) y = μ ( x )sin x . [ xμ ( x )] = 2 μ ( x ) ⇒ μ ( x ) + xμ ( x ) = 2 μ ( x ) ⇒ xμ ( x ) = μ ( x ) ⇒ μ ( x ) /μ ( x ) = 1 /x ⇒ [ln | μ ( x ) | ] = 1 /x ⇒ ln | μ ( x ) | = ln | x | ⇒ μ ( x ) = x Thus we have x 2 y + 2 xy = [ x 2 y ( x )] = x sin x . Hence, integrating by parts, we get x 2 y ( x ) = Z x sin xdx = x (- cos x )- Z (- cos x ) dx =- x cos x +sin x + c. So the general solution is y ( x ) =- cos x x + sin x x 2 + c x 2 . (b) (1 + x 2 ) y + 4 xy = 1 1 + x 2 First, find the integrating factor μ ( x ): (1 + x 2 ) μ ( x ) y + 4 xμ ( x ) y = μ ( x ) 1 + x 2 . £ (1 + x 2 ) μ ( x ) / = 4 xμ ( x ) ⇒ 2 xμ ( x ) + (1 + x 2 ) μ ( x ) = 4 xμ ( x ) ⇒ (1 + x 2 ) μ ( x ) = 2 xμ ( x ) ⇒ μ ( x ) /μ ( x ) = 2 x/ (1 + x 2 ) ⇒ [ln | μ ( x ) | ] = [ln | 1 + x 2 | ] ⇒ ln | μ ( x ) | = ln | 1 + x 2 | ⇒ μ ( x ) = 1 + x 2 Thus we have (1 + x 2 ) 2 y + 4 x (1 + x 2 ) y = [(1 + x 2 ) 2 y ] = 1. Hence, by integrating, we get (1 + x 2 ) 2 y ( x ) = x + c . So the general solution is y ( x ) = x + c (1 + x 2 ) 2 . MATH-2400 | Introduction to Differential Equations | Spring 2004 | Jeong-Rock Yoon Homework 1 | Due on January 22 (Thursday) | Solution 2 2. Solve the following initial value problems:2....
View Full Document

## This homework help was uploaded on 02/24/2008 for the course MATH 2400 taught by Professor Yoon during the Spring '04 term at Rensselaer Polytechnic Institute.

### Page1 / 6

hw1_s - Homework 1 | Due on January 22(Thursday | Solution...

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

View Full Document
Ask a homework question - tutors are online