17a_ISMCHAP17_COSV6 - INSTRUCTOR’S SOLUTIONS MANUAL...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: INSTRUCTOR’S SOLUTIONS MANUAL SECTION 17.2 (PAGE 913) CHAPTER 17. ORDINARY DIFFEREN- TIAL EQUATIONS NOTE: SECTIONS 17.2 AND 17.5 AND THE REVIEW EXERCISES FOR CHAPTER 17 IN CALCULUS OF SEVERAL VARIABLES HAVE MORE EXERCISES THAN THE CORRESPONDING VERSIONS IN CALCULUS: A COMPLETE COURSE AND SINGLE-VARIABLE CALCULUS. ONLY THE SOLUTIONS FOR THOSE UNITS ARE GIVEN HERE; FOR THE OTHERS SEE CHAPTER 17. Section 17.2 Solving First-Order Equations (page 913) 1. dy dx = y 2 x 2 dy y = dx x 2 ln y = ln x + C 1 ⇒ y 2 = Cx 2. dy dx = 3 y − 1 x dy 3 y − 1 = dx x 1 3 ln | 3 y − 1 | = ln | x | + 1 3 ln C 3 y − 1 x 3 = C ⇒ y = 1 3 ( 1 + Cx 3 ). 3. dy dx = x 2 y 2 ⇒ y 2 dy = x 2 dx y 3 3 = x 3 3 + C 1 , or x 3 − y 3 = C 4. dy dx = x 2 y 2 dy y 2 = x 2 dx − 1 y = 1 3 x 3 + 1 3 C ⇒ y = − 3 x 3 + C . 5. dY dt = tY ⇒ dY Y = t dt ln Y = t 2 2 + C 1 , or Y = Ce t 2 / 2 6. dx dt = e x sin t e − x dx = sin t dt − e − x = − cos t − C ⇒ x = − ln ( cos t + C ). 7. dy dx = 1 − y 2 ⇒ dy 1 − y 2 = dx 1 2 1 1 + y + 1 1 − y dy = dx 1 2 ln 1 + y 1 − y = x + C 1 1 + y 1 − y = Ce 2 x or y = Ce 2 x − 1 Ce 2 x + 1 8. dy dx = 1 + y 2 dy 1 + y 2 = dx tan − 1 y = x + C ⇒ y = tan ( x + C ). 9. dy dt = 2 + e y ⇒ dy 2 + e y = dt e − y dy 2 e − y + 1 = dt − 1 2 ln ( 2 e − y + 1 ) = t + C 1 2 e − y + 1 = C 2 e − 2 t , or y = − ln Ce − 2 t − 1 2 10. We have dy dx = y 2 ( 1 − y ) dy y 2 ( 1 − y ) = dx = x + K . Expand the left side in partial fractions: 1 y 2 ( 1 − y ) = A y + B y 2 + C 1 − y = A ( y − y 2 ) + B ( 1 − y ) + Cy 2 y 2 ( 1 − y ) ⇒ − A + C = ; A − B = ; B = 1 . ⇒ A = B = C = 1 . Hence, dy y 2 ( 1 − y ) = 1 y + 1 y 2 + 1 1 − y dy = ln | y | − 1 y − ln | 1 − y | . Therefore, ln y 1 − y − 1 y = x + K . 647 SECTION 17.2 (PAGE 913) R. A. ADAMS: CALCULUS 11. dy dx = x + y x − y Let y = v x v + x d v dx = x ( 1 + v) x ( 1 − v) x d v dx = 1 + v 1 − v − v = 1 + v 2 1 − v 1 − v 1 + v 2 d v = dx x tan − 1 v − 1 2 ln ( 1 + v 2 ) = ln | x | + C 1 tan − 1 ( y / x ) − 1 2 ln x 2 + y 2 x 2 = ln | x | + C 1 2 tan − 1 ( y / x ) − ln ( x 2 + y 2 ) = C . 12. dy dx = xy x 2 + 2 y 2 Let y = v x v + x d v dx = v x 2 ( 1 + 2 v 2 ) x 2 x d v dx = v 1 + 2 v 2 − v = − 2 v 3 1 + 2 v 2 1 + 2 v 2 v 3 d v = − 2 dx x − 1 2 v 2 + 2 ln | v | = − 2 ln | x | + C 1 − x 2 2 y 2 + 2 ln | y | = C 1 x 2 − 4 y 2 ln | y | = Cy 2 . 13. dy dx = x 2 + xy + y 2 x 2 Let y = v x v + x d v dx = x 2 ( 1 + v + v 2 ) x 2 d v 1 + v 2 = dx x tan − 1 v = ln | x | + C y x = tan ln | x | + C y = x tan ln | x | + C . 14. dy dx = x 3 + 3 xy 2 3 x 2 y + y 3 Let y = v x v + x d v dx = x 3 ( 1 + 3 v 2 ) x 3 ( 3 v + v 3 ) x d v dx = 1 + 3 v 2 3 v + v 3 − v = 1 − v 4 v( 3 + v 2 ) ( 3 + v 2 )v d v 1 − v 4 = dx x Let u = v 2 du = 2 v d v 1 2 3 + u 1 − u 2 du = ln | x | + C 1 3 4 ln u + 1 u − 1 − 1 4 ln | 1 − u 2 | = ln | x | + C 1 3 ln y 2 + x 2 y 2 − x 2 − ln x 4 − y 4 x 4 = 4 ln...
View Full Document

This note was uploaded on 01/29/2012 for the course ECONOMICS 3400 taught by Professor Kroger during the Spring '11 term at Georgia State.

Page1 / 14

17a_ISMCHAP17_COSV6 - INSTRUCTOR’S SOLUTIONS MANUAL...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online