Solução_Calculo_Stewart_6e

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F. TX.10 9 DIFFERENTIAL EQUATIONS 9.1 Modeling with Differential Equations 1. y = x − x −1 ⇒ y 0 =1+x −2 . To show that y is a solution of the differential equation, we will substitute the expressions for y and y 0 in the left-hand side of the equation and show that the left-hand side is equal to the right-hand side. LHS= xy 0 + y = x(1 + x −2 )+(x − x −1 )=x + x −1 + x − x −1 =2x =RHS 3. (a) y = e rx ⇒ y 0 = re rx ⇒ y 00 = r 2 e rx . Substituting these expressions into the differential equation 2y 00 + y 0 − y =0,weget2r 2 e rx + re rx − e rx =0 ⇒ (2r 2 + r − 1)e rx =0 ⇒ (2r − 1)(r +1)=0 [since e rx is never zero] ⇒ r = 1 or −1. 2 (b) Let r 1 = 1 2 and r 2 = −1, so we need to show that every member of the family of functions y = ae x/2 + be −x is a solution of the differential equation 2y 00 + y 0 − y =0. y = ae x/2 + be −x ⇒ y 0 = 1 2 aex/2 − be −x ⇒ y 00 = 1 4 aex/2 + be −x . LHS = 2y 00 + y 0 1 − y =2 4 aex/2 + be −x 1 + 2 aex/2 − be −x − (ae x/2 + be −x ) = 1 2 aex/2 +2be −x + 1 2 aex/2 − be −x − ae x/2 − be −x = 1 2 a + 1 2 a − a e x/2 +(2b − b − b)e −x =0=RHS 5. (a) y =sinx ⇒ y 0 =cosx ⇒ y 00 = − sin x. LHS = y 00 + y = − sin x +sinx =06= sinx,soy =sinx is not a solution of the differential equation. (b) y =cosx ⇒ y 0 = − sin x ⇒ y 00 = − cos x. LHS = y 00 + y = − cos x +cosx =06= sinx,soy =cosx is not a solution of the differential equation. (c) y = 1 x sin x ⇒ 2 y0 = 1 (x cos x +sinx) ⇒ 2 y00 = 1 (−x sin x +cosx +cosx). 2 LHS = y 00 + y = 1 (−x sin x +2cosx)+ 1 x sin x =cosx 6= sinx, soy = 1 x sin x is not a solution of the 2 2 2 differential equation. (d) y = − 1 x cos x ⇒ 2 y0 = − 1 (−x sin x +cosx) ⇒ 2 y00 = − 1 2 (−x cos x − sin x − sin x). LHS = y 00 + y = − 1 (−x cos x − 2sinx)+ 2 − 1 x cos x 2 =sinx =RHS,soy = − 1 2 x cos x is a solution of the differential equation. 7. (a) Since the derivative y 0 = −y 2 is always negative (or 0 if y =0), the function y must be decreasing (or equal to 0)onany interval on which it is defined. (b) y = 1 2 x + C ⇒ 1 y0 = − (x + C) 2 . LHS = 1 1 y0 = − (x + C) 2 = − = −y 2 =RHS x + C 381
F. 382 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS TX.10 (c) y =0is a solution of y 0 = −y 2 that is not a member of the family in part (b). (d) If y(x) = 1 x + C ,theny(0) = 1 0+C = 1 C .Sincey(0) = 0.5, 1 C = 1 2 ⇒ C =2,soy = 1 x +2 . 9. (a) dP dt =1.2P 1 − P . Now dP 4200 dt > 0 ⇒ 1 − P P > 0 [assuming that P>0] ⇒ 4200 4200 < 1 ⇒ P
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Solução_Calculo_Stewart_6e

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