Math 280: Final Study Guide, Spring 2017 Answers
1.2: 15-16, 17-24, 25-28, 29-34 1) Find all (x0 , y0 ) such that the initial value problem with y(x0 ) = y0 has a unique
solution y(x) on a region with x near x0 . Do not solve.
(a) y0 6= 0, x0 6=
π
2
+ kπ.
(b) y0 6= 2, x0 6= ±1.
2.1: 1-4, 19-20, 21-27, 29-30, 38-40 2) (a) See pictures.
(b) See pictures.
(c) See pictures.
3) (a) See pictures.
(b) See pictures.
4) (a) See pictures.
(b) See pictures.
2.2: 1-11, 15-18, 21-22, 23-26, 29-30, 31-33 5) (a) y = tan(ln(x) − π4 ).
(b) y + ln(y) = 12 (ln(x))2 + 12 .
6) (a) y =
1
.
3−x
Solution defined on (−∞, 3).
(b) xy 0 = −2y, y(2) = 1. y =
Rx
7) (a) y = 1 1t e2t dt + 3.
Rx 2
(b) 31 y 3 = 2 et dt + 9.
2.3: 1-24, 25-36, 43-48 8) (a) y = − 12 + Cex
2
(b) y = x2 + Cx2 e−x
(c) y = 1 +
√C
x2 +1
4
.
x2
Solution defined on (0, ∞).
2.2, 2.3: Be able to recognize if an equation is linear or separable. 9) (a) y = 1/(3 − (2x + 2)e−x )
(b) y = sin(x) + 2 cos(x)
2.4: 1-18, 21-26 10) (a) x2 (y + 1) + y + xe−y = 2.
(b) x ln(y) −
1
y
+ 3x = 5.
3.1: 1-8, 11-12, 13-18, 29-30, 35-38 11) (a) T (t) = 325 − 300ekt .
1
(b) k = 20
ln 23 .
(c) See pictures.
12) (a) i(t) = 3e−5t + 10t − 2.
(b) See pictures.
13) (a) k = 98.
(b) y = 45e−.98t + 10
(c) See pictures.
3.2: 1-4, 11-12, 15-17, 22 14) (a) y(t) =
200
.
1+e−.2t
(b) y(∞) = 200. (There are two ways to see this. With the explicit solution, or
looking at when dP
= 0.)
dt
(c) See pictures.
15) A rock dropped into a pool of liquid satisfies m dv
= mg − kv 2 .
dt
p
(a) v∞ = mg
.
k
4t (b) v(t) = 2 ee4t −1
.
+1
4.1: 1-6, 9-10, 15-19, 21, 23-28 16) (a) W (x) = 2e4x . Linearly independent.
(b) W (x) = −2x3 . Linearly independent.
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(c) −1(1) + 1 cos(2x) + 2 sin2 (x) = 0. Linearly dependent.
17) (a) Plug both solutions in and show that you get 0. W (x) = x5 . W (x) > 0 on (0, ∞).
(b) Plug both solutions in and show that you get 0. W (x) = −ex (x − 2)2 . W (x) < 0
on (−∞, 2).
4.2: 1-4, 6, 9-12, 16 18) (a) y = c1 x3 + c2 x5 .
(b) y = c1 ex + c2 (2x + 1)e−x .
4.3: 1-11, 15-24, 29-31, 35-36, 49-58 19) (a) y = c1 e−2x cos(x) + c2 e−2x sin(x) + c3 .
(b) y = c1 e−x + c2 xe−x + c3 x2 e−x .
(c) y = c1 cos(x) + c2 sin(x) + c3 x cos(x) + c4 x sin(x).
20) (a) y 00 + 6y 0 + 10y = 0.
(b) y 000 − 6y 00 + 12y 0 − 8y = 0.
(c) y 0000 + 25y 00 = 0.
4.4: 1-14, 19, 21, 27-34; 4.5: 35-50, 55-58 21) (a) y = 2e2x − 2 − x2 − x + e−x .
(b) y = ex + 2xex + x2 ex − e2x .
22) (a) y = − ω(ω2γ−γ 2 ) sin(ωx) +
(b) y =
1
2ω 2
sin(ωx) −
1
ω 2 −γ 2
sin(γx)
1
x cos(ωx).
2ω
4.6: 1-6, 9, 11-12, 15, 23-26, 27-28 23) yp = 41 e3x .
24) (a) yp = − 12 sec(x) + tan(x) sin(x).
(b) y = c1 cos(x) + c2 sin(x) − 12 sec(x) + tan(x) sin(x).
(c) y = 12 cos(x) − sin(x) + 12 sec(x) + tan(x) sin(x).
25) y = c1 x + c2 x3 + 18 x5
26) y = c1 x + c2 x ln(x) + 16 x(ln(x))3 .
Rx t
R
1 x x e−t
27) y = 12 e−x 0 t−e
dt.
4 +1 dt + 2 e
0 t4 +1
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4.4, 4.6: Know when to use undetermined coefficients and when to use variation of parameters. 28) Find the general solutions.
(a) y = c1 e−x + c2 xe−x +
1 −x
e .
2x
−x
(b) y = c1 e−x cos(x) + c2 e
sin(x) + 12 xe2x + 12 e2x .
5.1: 1-6, 8, 21-24, 25-31, 33-37, 41-42 29) (a) x(t) = −e−t cos(2t).
(b) See pictures.
(c) Underdamped.
30) (a) y = 3te−3t .
(b) Critically damped.
(c) y 0 (t) = 0 when t = 13 . y
1
3
= 1e .
(d) See pictures.
31) (a) y(t) = 4e−2t − 5e−t + 3 sin(t) + cos(t)
(b) 4e−2t − 5e−t is transient, 3 sin(t) + cos(t) is steady-state.
7.1: 1-16, 19-36, 37-38 32) Find the Laplace transforms directly from the definition.
(a) F (s) =
1
.
(s−2)2
(b) F (s) =
1
(e−3 − e−3s ) + 1s e−3s .
s−1
1
(4s2 − 2 + (4s + 2)e−2s ) .
s3
(c) F (s) =
7.2: 1-30, 35-44, 45-46 33) Find L−1 {F (s)}.
(a) f (t) = t2 + 2t − 1 + e−2t .
(b) f (t) = 2 sin(2t) − e−2t + e2t .
(c) f (t) = cos(t) − 3 sin(t) − cos(3t) + sin(3t).
34) (a) y = e−3x + e3x − 2.
(b) y = sin(t) − cos(t) + e2t .
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(c) y = e−2t + 1 + 2 sin(t) − cos(t).
7.3: 1-8, 11-18, 21-30, 37-48, 49-54, 55-62, 63-68 35 Solve the initial value problems by Laplace transforms.
(a)
(b)
(c)
(d)
y = e−t cos(2t) + 2e−t sin(2t) − 2e−2t .
y = 2e2t − e−t − 2t + 1.
y(t) = 2et − 2e−t − te−t .
y(t) = −2 cos(2t) − 23 sin(2t) + 5te−t + 2e−t
36) Write in terms of unit step functions. Then find the Laplace transform.
(a) f (t) = 1 + u(t − 2) + 2u(t − 3). F (s) = 1s + 1s e−2s + 2s e−3s .
(b) f (t) = u(t − 3) − u(t − 5). F (s) = 1s e−3s − 1s e−5s .
(c) f (t) = t − tu(t − 3) = t − (t − 3)u(t − 3) − 3u(t − 3). F (s) =
1
s2
−
1 −3s
e
s2
− 3s e−3s .
8.1: 1-6, 7-10, 11-16, 17-20, 21-23 ~ 1 (t) and X
~ 2 (t) solve the matrix equation. The determinant of the
37) (a) Check that X
−2t
solution matrix is e
which is never 0.
~ 1 (t) and X
~ 2 (t) solve the matrix equation. The determinant of the
(b) Check that X
solution matrix is −6 which is never 0.
8.2: 1-6, 13, 21-24, 31, 35-40, 48 1 −1 ~
0
~
X(t).
38) (a) X (t) =
4 5
1
1
0
3t
3t
~
(b) X(t)
= c1 e
+ c2 e
t
+
.
−2
−2
−1
(c) x(t) = c1 e3t + c2 te3t , y(t) = −2c1 e3t − (2t + 1)c2 e3t .
8
6 ~
0
~
X(t).
39) (a) X (t) =
−3 −1
1
2
2t
5t
~
(b) X(t)
= c1 e
+ c2 e
−1
−1
1
2
2t
5t
= −3e
+e
.
−1
−1
(c) x(t) = c1 e2t + 2c2 e5t = −3e2t + 2e5t , y(t) = −c1 e2t − c2 e5t = 3e2t − e5t .
1
−2
0
~ (t) =
~
40) (a) X
X(t).
5 −1
1
−1
~
cos(3t) +
sin(3t).
(b) X(t)
=
2
1
(c) x(t) = cos(3t) − sin(3t) , y(t) = 2 cos(3t) + sin(3t).
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