Name:
Exam 3 Solutions – MATH 230, Spring 2011
Instructor: Dr. Zachary Kilpatrick
No calculators. Show all your work. Simplify as much as possible.
1. (20pts) (a) Compute (or just write) the Maclaurin series for f (x) = cos x.
cos x = 1 −
x2 x4 x6 x8 x10
+
−
+
−
+···
2!
4!
6!
8!
10!
(b) Use part (a) to compute the Taylor series for f (x) = cos x3 centered at zero.
cos x3 = 1 −
x6 x12 x18 x24 x30
+
−
+
−
+···
2!
4!
6!
8!
10!
(c) Use part (b) to compute the following integral using infinite series:
Z
3
x cos x dx =
=
=
=
Z
x cos x3 dx
x6 x12 x18 x24 x30
+
−
+
−
+ · · · dx
x 1−
2!
4!
6!
8!
10!
Z x7 x13 x19 x25 x31
+
−
+
−
+ · · · dx
x−
2!
4!
6!
8!
10!
x8
x14
x20
x26
x32
x2
−
+
−
+
−
+···
2
8(2!) 14(4!) 20(8!) 26(8!) 32(10!)
∞
X
x6n+2
(−1)n
(6n + 2)(2n)!
n=0
Z
1
2. (20pts)
f (x) = ln(4 + x2 )
(a) Compute f ′ (x) and f ′′ (x).
2x
4 + x2
2(4 + x2 ) − 2x(2x)
8 − 2x2
f ′′ (x) =
=
(4 + x2 )2
(4 + x2 )2
f ′ (x) =
(b) Use part (a) to write down the first two nonzero terms of the Taylor series for f (x)
centered at zero (Maclaurin series).
note from (a): f (0) = ln 4; f ′ (0) = 0; f ′′ (0) = 12 , so
f (x) ≈ f (0) +
f ′ (0)
f ′′ (0) 2
x2
x+
x = ln 4 +
1!
2!
2(2!)
2
(c) Work out the geometric series for
2x
4 + x2
x2 x4 x6
2x
x
x8
1
2x
1−
=
=
+
−
+
−···
4 + x2
4 1 − (−x2 /4)
2
4
16 64 256
x7
x9
x x3 x5
−
+
−
+
−···
=
2
4
16 64 256
(d) Use part (c) to find a power series for f (making sure f (0) is correct).
x x3 x5 x7
x9
ln(4 + x ) =
−
+
−
+
− · · · dx
2
4
16 64 256
x4
x6
x8
x10
x2
−
+
−
+
−···
= C+
2 · 2 4 · 4 6 · 16 8 · 64 10 · 256
∞
X
x2n+2
= ln 4 +
(−1)n
(2n + 2)4n
n=0
2
Z (since when x = 0, f should be ln 4)
(e) Find the interval of convergence for the power series in part (d).
(−1)n+1 x2(n+1)+2 (2n + 2)4n (2n + 2)x2 2n + 2 x2
an+1 x2
= =
=
→
(2(n + 1) + 2)4n+1 (−1)n x2n+2 (2n + 4)4 2n + 4 4
an 4
thus x2 /4 < 1 or x2 < 4 meaning −2 < x < 2 ensures convergence. when x = ±2
∞
X
(−1)n
converges
2n
+
2
n=0
since 1/(2n + 2) is decreasing and the series alternates.
thus the interval of convergence is x ∈ [−2, 2].
3
as n → ∞
′
3. (20pts) For the first order linear differential equation: y +
cos x
y = −3 cos2 x
sin x
(a) Find the associated integrating factor I(x).
Z
cos x
dx = ln(sin x)
sin x
thus
I(x) = eln(sin x) = sin x
(b) Use the integrating factor to find the general solution.
multiplying the differential equation by the integrating factor
(sin x)y ′ + (cos x)y = −3 cos2 x sin x
(sin xy)′ = −3 cos2 x sin x
Z
sin xy =
(−3 cos2 x sin x)dx = cos3 x + C
y =
C
cos3 x
+
sin x
sin x
(c) Specify the free constant of the solution with the initial condition y( π2 ) = 0.
y( π2 ) = C = 0
thus
y(x) =
cos3 x
sin x
4
4. (20pts) (a) Solve y ′′ − 4y ′ + 3y = 0
use the characteristic equation: r 2 − 4r + 3 = 0, note (r − 1)(r − 3) = 0 so r = 1, 3. thus
y(x) = c1 ex + c2 e3x
(b) Solve using method of undetermined coefficients: y ′′ − 4y ′ + 3y = xex − ex
use the trial solution yp = (Ax2 + Bx)ex , so yp′ = (2Ax + B + Ax2 + Bx)ex
and yp′′ = (2A + 2Ax + B + 2Ax + B + Ax2 + Bx)ex . thus
(2A + 2B + 4Ax + Bx + Ax2 − 4B − 8Ax − 4Bx − 4Ax2 + 3Bx + 3Ax2 )ex = xex − ex
2A − 2B − 4Ax = x − 1
so A = −1/4 and B = 1/4, so
yp (x) =
x − x2 x
e
4
adding this to the complementary solution from part (a)
y(x) = c1 ex + c2 e3x +
x − x2 x
e
4
5
(c) Find constants of solution in part (b) using the initial conditions y(0) = 0 and y ′ (0) = 0.
y(0) = c1 + c2 = 0
→
y ′(x) = c1 + 3c2 +
1
=0
4
c1 = −c2
and
→
−2c2 +
1
4
→ c2 =
1
8
and c1 = −1/8, meaning
y(x) =
e3x − ex x − x2 x
+
e
8
4
5. (20pts) A spring with a 2 kg mass and damping constant 2 has natural length 3 m and a
force of 4 N is required to stretch it to a length of 4 m.
(a) Compute the spring constant k.
k = 4/(4 − 3) = 4/1 = 4
(b) Find the general solution for the position x(t) of the mass at time t.
2x′′ + 2x′ + 4x = 0
which has characteristic equation: 2r 2 + 2r + 4 = 0, which means
√
√
−2 ± 4 − 32
7
1
r=
=− ±
i
4
2
2
since the root is complex, we have an underdamped spring, so the solution involves products
of exponential and trigonometric functions
√ !!
√ !
7
7
t + c2 sin
t
x(t) = e−t/2 c1 cos
2
2
6
(c) Find constants in the case the spring is started in the equilibrium position with velocity
3 m/s (at t = 0). Plot the solution x(t) versus t.
starts at equilibrium position: x(0) = 0, so
x(0) = c1 = 0
then, using
′
x (t) = c2
√
√
√ !
1 −t/2
7t
3t
−t/2 7
− e
sin
+e
cos
2
2
2
2
and the initial velocity 3m/s: x′ (0) = 3 means
√
7
6
′
x (0) = c2
=3
→
c2 = √
2
7
thus
√
6 −t/2
7t
x(t) = √ e
sin
2
7
(d) Find the damping constant that would produce critical damping in the spring with a 2
kg mass and spring constant k from part (a).
for
2x′′ + cx′ + 4x = 0
we want the characteristic equation 2r 2 + cr + 4 = 0 to have a double root and
p
√
−c ± c2 − 4(2)(4)
−c ± c2 − 32
=
r=
4
4
√
so we want c2 − 32 = 0 meaning c2 = 32 or c = 4 2 (since c must be positive)
7
6. (Extra credit: 5pts) Solve y ′′ − y ′ = 1
characteristic equation: r 2 − r = 0 so r = 0, 1 meaning the complementary solution is
yc = c1 + c2 ex
notice, this means 1 solves the complementary equation, so we must pick yp = Ax as a trial
solution. plugging yp = Ax into the differential equation: −A = 1 or A = −1, so
y = c1 + c2 ex − x
1
7. (Extra credit: 5pts) Compute lim
h→0 3h
Z
3h
cos(x3 )dx
h
plug in the infinite series from problem 1(b) for cos x3
Z 3h
Z 3h 1
x6 x12
1
3
cos(x )dx = lim
1−
+
− · · · dx
lim
h→0 3h h
h→0 3h h
2!
4!
(3h)7 − h7 (3h)13 − h13
1
2h −
+
−···
= lim
h→0 3h
7(2!)
13(4!)
2 37 h6 − h6 313 h12 − h12
−
+
−···
= lim
h→0 3
21(2!)
39(4!)
2
=
3
since all terms except the first will go to zero as h → 0
8