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 = 2 x 2 e x 3 / 3 = x 2 y and y (0) = 2 by inspection. 2. y = x 3 2 sin x, y (0) = 3 by inspection. 3. (a) first order; dy dx = c ; (1 + x ) dy dx = (1 + x ) c = y (b) second order; y = c 1 cos t c 2 sin t, y + y = c 1 sin t c 2 cos t + ( c 1 sin t + c 2 cos t ) = 0 4. (a) first order; 2 dy dx + y = 2 c 2 e x/ 2 + 1 + ce x/ 2 + x 3 = x 1 (b) second order; y = c 1 e t c 2 e t , y 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 dx = e 3 x , d dx ye 3 x = 0 , ye 3 x = C, y = Ce 3 x separation of variables: dy y = 3 dx, ln | y | = 3 x + C 1 , y = ± e 3 x e C 1 = Ce 3 x including C = 0 by inspection (b) IF: µ = e 2 dt = e 2 t , d dt [ ye 2 t ] = 0 , ye 2 t = C, y = Ce 2 t separation of variables: dy y = 2 dt, ln | y | = 2 t + C 1 , y = ± e C 1 e 2 t = Ce 2 t including C = 0 by inspection 8. (a) IF: µ = e 4 x dx = e 2 x 2 , d dx ye 2 x 2 = 0 , y = Ce 2 x 2 separation of variables: dy y = 4 x dx, ln | y | = 2 x 2 + C 1 , y = ± e C 1 e 2 x 2 = Ce 2 x 2 including C = 0 by inspection (b) IF: µ = e dt = e t , d dt ye t = 0 , y = Ce t separation of variables: dy y = dt, ln | y | = t + C 1 , y = ± e C 1 e t = Ce t including C = 0 by inspection 9. µ = e 3 dx = e 3 x , e 3 x y = e x dx = e x + C , y = e 2 x + Ce 3 x 10. µ = e 2 x dx = e x 2 , d dx ye x 2 = xe x 2 , ye x 2 = 1 2 e x 2 + C, y = 1 2 + Ce x 2
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Exercise Set 9.1 373 11. µ = e 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 2 dx = e 2 x , e 2 x y = 1 2 e 2 x dx = 1 4 e 2 x + C , y = 1 4 + Ce 2 x 13. dy dx + x x 2 + 1 y = 0 , µ = e ( x/ ( x 2 +1)) dx = e 1 2 ln( x 2 +1) = x 2 + 1 , d dx y x 2 + 1 = 0 , y x 2 + 1 = C, y = C x 2 + 1 14. dy dx + y = 1 1 + e x , µ = e dx = e x , e x y = e x 1 + e x dx = ln(1 + e x ) + C , y = e x ln(1 + e x ) + Ce x 15. 1 y dy = 1 x dx , ln | y | = ln | x | + C 1 , ln y x = C 1 , y x = ± e C 1 = C , y = Cx including C = 0 by inspection 16. dy 1 + y 2 = x 2 dx, tan 1 y = 1 3 x 3 + C, y = tan 1 3 x 3 + C 17. dy 1 + y = x 1 + x 2 dx, ln | 1 + y | = 1 + x 2 + C 1 , 1 + y = ± e 1+ x 2 e C 1 = Ce 1+ x 2 , y = Ce 1+ x 2 1 , C = 0 18. y dy = x 3 dx 1 + x 4 , y 2 2 = 1 4 ln(1 + x 4 ) + C 1 , 2 y 2 = ln(1 + x 4 ) + C, y = ± [ln(1 + x 4 ) + C ] / 2 19. 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 = x dx, ln | y | = x 2 / 2 + C 1 , y = ± e C 1 e x 2 / 2 = Ce x 2 / 2 , including C = 0 by inspection 21. e y dy = sin x cos 2 x dx = sec x tan x dx , e y = sec x + C , y = ln(sec x + C ) 22. dy 1 + y 2 = (1 + x ) dx, tan 1 y = x + x 2 2 + C, y = tan( x + x 2 / 2 + C ) 23. dy y 2 y = dx sin x , 1 y + 1 y 1 dy = csc x dx, ln y 1 y = ln | csc x cot x | + C 1 , y 1 y = ± e C 1 (csc x cot x ) = C (csc x cot x ) , y = 1 1 C (csc x cot x ) , C = 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 x dx , ln | sin y | = 3 sin x + C 1 , sin y = ± e 3 sin x + C 1 = ± e C 1 e 3 sin x = Ce 3 sin x , C = 0, y = sin 1 ( Ce 3 sin x ) , as is y = 0 by inspection
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374 Chapter 9 25.
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