17a_ISMCHAP17_COSV6

# 17a_ISMCHAP17_COSV6 - INSTRUCTOR’S SOLUTIONS MANUAL...

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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...
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## 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.

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17a_ISMCHAP17_COSV6 - INSTRUCTOR’S SOLUTIONS MANUAL...

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