chapter_09 - 11:46 L24-ch09 Sheet number 1 Page number 395...

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

View Full Document Right Arrow Icon
January 27, 2005 11:46 L24-ch09 Sheet number 1 Page number 395 black 395 CHAPTER 9 Mathematical Modeling with Differential Equations EXERCISE SET 9.1 1. y 0 =9 x 2 e x 3 =3 x 2 y and y (0) = 3 by inspection. 2. y 0 = x 3 2 sin x, y (0) = 3 by inspection. 3. (a) frst order; dy dx = c ;(1+ x ) dy dx =(1+ x ) c = y (b) second order; y 0 = c 1 cos t c 2 sin t, y 0 + y = c 1 sin t c 2 cos t +( c 1 sin t + c 2 cos t )=0 4. (a) frst order; 2 dy dx + y =2 ³ c 2 e x/ 2 +1 ´ + ce x/ 2 + x 3= x 1 (b) second order; y 0 = c 1 e t c 2 e t ,y 0 y = c 1 e t + c 2 e t ( c 1 e t + c 2 e t ) =0 5. 1 y dy dx = y 2 +2 xy dy dx , dy dx (1 2 xy 2 )= y 3 , dy dx = y 3 1 2 xy 2 6. 2 x + y 2 xy dy dx = 0, by inspection. 7. (a) IF: µ = e 3 R dx = e 3 x , d dx £ ye 3 x ¤ ,ye 3 x = C,y = Ce 3 x separation o± variables: dy y = 3 dx, ln | y | = 3 x + C 1 = ± e 3 x e C 1 = 3 x including C = 0 by inspection (b) IF: µ = e 2 R dt = e 2 t , d dt [ 2 t ]=0 2 t = = 2 t separation o± variables: dy y dt, ln | y | t + C 1 = ± e C 1 e 2 t = 2 t including C = 0 by inspection 8. (a) IF: µ = e 4 R xdx = e 2 x 2 , d dx h 2 x 2 i = 2 x 2 separation o± variables: dy y =4 x dx, ln | y | x 2 + C 1 = ± e C 1 e 2 x 2 = 2 x 2 including C = 0 by inspection (b) IF: µ = e R dt = e t , d dt £ t ¤ = t separation o± variables: dy y = dt, ln | y | = t + C 1 = ± e C 1 e t = t including C = 0 by inspection 9. µ = e R 4 dx = e 4 x , e 4 x y = ± e x dx = e x + C , y = e 3 x + 4 x 10. µ = e 2 R = e x 2 , d dx h x 2 i = xe x 2 x 2 = 1 2 e x 2 + = 1 2 + x 2
Background image of page 1

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

View Full DocumentRight Arrow Icon
January 27, 2005 11:46 L24-ch09 Sheet number 2 Page number 396 black 396 Chapter 9 11. µ = e R dx = e x , e x y = ± e x cos( e x ) dx = sin( e x )+ C , y = e x sin( e x Ce x 12. dy dx +2 y = 1 2 , µ = e R 2 dx = e 2 x , e 2 x y = ± 1 2 e 2 x dx = 1 4 e 2 x + C , y = 1 4 + 2 x 13. dy dx + x x 2 +1 y =0 = e R ( x/ ( x 2 +1)) dx = e 1 2 ln( x 2 +1) = p x 2 , d dx h y p x 2 i ,y p x 2 +1= C, y = C x 2 14. dy dx + y = 1 1 e x , µ = e R dx = e x , e x y = ± e x 1 e x dx =ln | 1 e x | + C , y = e x ln | 1 e x | + x 15. 1 y dy = 1 x dx ,ln | y | | x | + C 1 ¯ ¯ ¯ y x ¯ ¯ ¯ = C 1 , y x = ± e C 1 = C , y = Cx including C = 0 by inspection 16. dy 1+ y 2 =2 x dx, tan 1 y = x 2 + C,y = tan ( x 2 + C ) 17. dy y = x x 2 dx, ln | y | = p x 2 + C 1 , y = ± e 1+ x 2 e C 1 = 1+ x 2 , y = 1+ x 2 1 ,C 6 18. ydy = x 3 dx x 4 , y 2 2 = 1 4 ln(1 + x 4 C 1 , 2 y 2 = ln(1 + x 4 C, y = ± p [ln(1 + x 4 C ] / 2 19. ² 2(1 + y 2 ) y dy = e x dx, 2ln | y | + y 2 = e x + C ; by inspection, y = 0 is also a solution 20. dy y = x dx, ln | y | = x 2 / 2+ C 1 = ± e C 1 e x 2 / 2 = x 2 / 2 , including C = 0 by inspection 21. e y dy = sin x cos 2 x dx = sec x tan xdx , e y = sec x + C , y = ln(sec x + C ) 22. dy y 2 =(1+ x ) dx, tan 1 y = x + x 2 2 + C, y = tan( x + x 2 / C ) 23. dy y 2 y = dx sin x , ± · 1 y + 1 y 1 ¸ dy = ± csc x dx, ln ¯ ¯ ¯ ¯ y 1 y ¯ ¯ ¯ ¯ | csc x cot x | + C 1 , y 1 y = ± e C 1 (csc x cot x )= C (csc x cot x ) = 1 1 C (csc x cot x ) 6 =0; by inspection, y = 0 is also a solution, as is y =1.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/10/2010 for the course MATH 200-177 taught by Professor Richardwhite during the Spring '10 term at Drexel.

Page1 / 29

chapter_09 - 11:46 L24-ch09 Sheet number 1 Page number 395...

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