Solução_Calculo_Stewart_6e

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
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