Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

350 0 CHAPTER 17 SECOND-ORDER DIFFERENTlAL EQUATIONS
o: 2o: 1 1 3x 1 -e 2 x
25. Yt = e , Y2 = e and Yt Y2 - Y2Y1 = e . So Ut = ( 1
) 3 =
· +e- o: e"'
e- x .
--- and
1 + e-x
U2 (x) = I 3 ex 2 dx = m(e"' + 1 ) -e-o: = ln(1 + e- "') - e-x. Hence
e"'+e"'
ex
yp(x) =ex 1nn .+e-x) + e 2 "'[ln(1 +e-x)- e-"'J and the general solution is
y(x) = (c1 + ln(1 + e- "')]e"' + [c2 - e- x + ln(1 + e- ")]e 2 "'.
27. r 2 - 2r + 1 = (r -1) 2 = 0 => r = 1 so Yc(x) = Ct e"' + c2xe"'. Thus Yt =ex, Y2 = xe"' and
xe'c · e"' /(1 + x 2 )
ex
x
+ x
Y I Y~- y2y~ = e"'(x + 1)ex- xe"'e"' = e 2 "'. Sou~=-
2
= - - 1
-- 2
~
Ut = -
I x 1 ( 2 ) 1 e"' · e"' /(1 + x2 ) 1 I 1 d -I d
1 + x 2 dx = - 2
In 1 + x , u2 = e 2 "' = 1
+ x 2 ~ U2 =
1 + x 2 x = tan x an
yp(x) = - 4e"' ln(1 +x 2 ) +xe"'tan- 1 x. Hence the general solution isy(x) = e"' [c1 + c2x- ~ ln(1 + x 2 ) +xtan- 1 x].
17.3 Applications of Second-Order Differential Equations
1. By Hooke's Law k(0.25) = 25 so k = 100 is the spring constant and the differential equation is 5x" + 100x = 0.
The auxiliary equation is 5r 2 + 100 = 0 with roots r = ±2 J5 i , so the general solution to the differential equation is
x(t) = c 1 cos(2 V5 t) + c2 sin(2 V5 t). We are given that x(O) = 0.35 ~ c1 = 0.35 and x 1 (0) = 0 ~
2 J5 c2 = 0 => c2 = 0, so the position of the mass after t seconds is x(t) = 0.35 cos(2 J5 t).
3. k(0.5) = 6 or k = 12 is the ·spring constant, so the initial-value problem is 2x" + 14x 1 + 12x = 0, x(O) = 1, x 1 (0) = 0.
The general solution is x(t ) = c1e-Gt + c2e-t. But 1 = x(O) = c1 + c2 and 0 = x 1 (0) = -6c1 - c2. Thus the position is
given by :i:(t) = - ~e- 6 t
+ ~e-t.
5. For critical damping we need c 2 - 4mk = 0 or m = c 2 /(4k) = ·14 2 /(4 · 12) = ~ kg.
2 . d
7. We are given m = 1, k = 100, x(O) = -0.1 and x 1 (0) = 0. From (3), the differential equation is ~t~ + c d; + 100x = 0
with auxiliary equation r 2 + cr + 100 = 0.
If c = 10, we have two complex roots r = - 5 ± 5 v'3 i, so the motion is underdamped and the solution is
x =e-st [ct cos(5 v'3t) + c 2 sin(5 v'3 t)J. Then -0.1 = x(O) = c1 and 0 = x 1 (0) = 5 v'3c2 - 5ct
. - 1
~ C2 - "i:ii"7s,
sox=e-st [--;0.1cos(5v'3t)- 10 J:isin(5v'3t)].
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