Solução_Calculo_Stewart_6e

F.
396 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS
TX.10
(b)
As k increases, the amplitude
increases, but the minimum value
stays the same.
As r increases, the amplitude and
the period decrease.
A change in φ produces slight
adjustments in the phase shift and
amplitude.
P (t) oscillates between P 0 e (k/r)(1+sin φ) and P 0 e (k/r)(−1+sin φ) (the extreme values are attained when rt − φ is an odd
multiple of π ), so lim P (t) does not exist.
2
t→∞
21. By Equation (7), P (t) =
1+tanhu =1+ eu − e −u
e u + e −u
K
1+Ae . By comparison, if c =(lnA)/k and u = 1 k(t − c), then
−kt 2
=
eu + e −u
e u + e + eu − e −u
−u e u + e = 2e u e−u
· −u e u + e−u e = 2
−u 1+e −2u
and e −2u = e −k(t−c) = e kc e −kt = e ln A e −kt = Ae −kt ,so
1
K 1+tanh 1
k(t − c) = K 2 2
2 [1 + tanh u] = K 2 · 2
1+e = K
−2u 1+e = K
= P (t).
−2u 1+Ae−kt 9.5 Linear Equations
1. y 0 +cosx = y ⇒ y 0 +(−1)y = − cos x is linear since it can be put into the standard linear form (1),
y 0 + P (x) y = Q(x).
3. yy 0 + xy = x 2 ⇒ y 0 + x = x 2 /y ⇒ y 0 − x 2 /y = −x is not linear since it cannot be put into the standard linear
form (1), y 0 + P (x) y = Q(x).
5. Comparing the given equation, y 0 +2y =2e x , with the general form, y 0 + P (x)y = Q(x),weseethatP (x) =2and the
integrating factor is I(x) =e P (x)dx = e 2 dx = e 2x . Multiplying the differential equation by I(x) gives
e 2x y 0 +2e 2x y =2e 3x ⇒ (e 2x y) 0 =2e 3x ⇒ e 2x y = 2e 3x dx ⇒ e 2x y = 2 3 e3x + C ⇒ y = 2 3 ex + Ce −2x .
7. xy 0 − 2y = x 2 [divide by x] ⇒ y 0 + − 2
y = x
x
().
I(x) =e P (x) dx = e (−2/x) dx = e −2ln|x| = e ln|x|−2 = e ln(1/x2) =1/x 2 . Multiplying the differential equation ()
by I(x) gives 1 x 2 y0 − 2 x 3 y = 1 x
⇒
1
x 2 y 0
= 1 x
⇒ 1 x 2 y =ln|x| + C ⇒
y = x 2 (ln |x| + C )=x 2 ln |x| + Cx 2 .