Calculus: Early Transcendentals, by Anton, 7th Edition,ch09

Calculus - Early Transcendentals

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372 CHAPTER 9 Mathematical Modeling with Differential Equations EXERCISE SET 9.1 1. y 0 =2 x 2 e x 3 / 3 = x 2 y and y (0) = 2 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 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 = x dy dx + y, dy dx (1 xy )= y 2 , dy dx = y 2 1 xy 6. 2 x + y 2 +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 xdx, 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 3 dx = e 3 x , e 3 x y = ± e x dx = e x + C , y = e 2 x + 3 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
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Exercise Set 9.1 373 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 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 | =ln | x | + C 1 ¯ ¯ ¯ y x ¯ ¯ ¯ = C 1 , y x = ± e C 1 = C , y = Cx including C = 0 by inspection 16. dy y 2 = x 2 dx, tan 1 y = 1 3 x 3 + C,y = tan ² 1 3 x 3 + 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 1 y + y dy = e x dx, ln | y | + y 2 / 2= e x + C ; by inspection, y = 0 is also a solution 20. dy y = xdx, 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 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. 24. 1 tan y dy = 3 sec x dx , cos y sin y dy = 3 cos | sin y | = 3 sin x + C 1 , sin y = ± e 3 sin x + C 1 = ± e C 1 e 3 sin x = 3 sin x 6 =0, y = sin 1 ( 3 sin x ) ,asis y = 0 by inspection
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374 Chapter 9 25.
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Calculus: Early Transcendentals, by Anton, 7th Edition,ch09...

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