Math 2201
Test 3 practice problems
Test 3, on 4/23, covers §’s 4.1, 4.3, 4.4, 4.6, 4.7, 8.1, 8.2
1. Solve the initial value problem: 4y
′′ − y = xex/2
, y(0) = 1, y′
(0) = 0. Answer: y(x) = 3e
x/2+e
−x/2
4 +
e
x/2
8
(x
2 − 2x).
2. Solve the initial value problem: y
′′ − y = 1
x
, x ∈ (0, ∞), y(1) = 0, y′
(1) = 1.
Answer: y(x) = 1
2
h
e
x−1 − e
1−x + e
x
R x
1
e
−r
r
dr − e
−x
R x
1
e
r
r
dri
3. Find the general solution of y
′′ + 3y
′ + 2y = sin(e
x
). Answer: y(x) = c1e
−2x + c2e
−x − e
−2x
sin(e
x
).
4. Find the general solution of y
(4) + 2y
′′ + y = 0.
5. What’s wrong with the following initial value problem
(x − 1)y
′′ − 2y
′ = x, x ∈ (0, ∞)
y(1) = 2
y
′
(1) = 1
?????
6. If we use the method of undetermined coefficents to find a particular solution yp to the nonhomogeneous DE
y
′′′ − 4y
′′ + 13y
′ = e
2x
(cos(3x) + 1) − cos x − 3,
what will be the final form of our yp? Answer: yp(x) = Axe2x
cos(3x) + Bxe2x
sin(3x) + Ce2x + D cos x + E sin x + F x
7. Same question for
L[y] = 2e
x+1 sin x − 3 sin x − 4e
x−1
but now L[y] is a 5th order, linear operator with real, constant coefficients, such that 3 of the roots of its characteristic
polynomial are r = 1, i, i. Again, what’s the final form of yp(x)?
Answer: yp(x) = Aex
cos x + Bex
sin x + Cx2
cos x + Dx2
sin x + Exex
8. Find a 4th order, homogeneous, linear DE that has y(x) = 2 cos(3x) − x
5
as a solution. Answer: y
(4) + 9y
′′ = 0.
9. Find the general solutions of following Cauchy-Euler equations defined on (0, ∞)
(a) x
2y
′′ + 2xy′ − 2y = 0. Answer: y(x) = c1x + c2x
−2
, where c1, c2 are arbitrary.
(b) x
3y
′′′ + 5x
2y
′′ + 7xy′ + 8y = 0.
(c) 2x
2y
′′ + 5xy′ + y = x
2 − x
(d) Solve the initial value problem on (0, ∞): x
2y
′′+xy′−y = ln x, y(1) = 4, y′
(1) = 5. Answer: y(x) = 5x−x
−1−ln x.
10. Find a 3rd order homogeneous Cauchy-Euler equation on (0, ∞) that has y(x) = (ln x)
2
3x
as a solution.
Answer: x
3y
′′′ + 6x
2y
′′ + 7xy′ + y = 0.
11. Find the simplest 2nd order nonhomogeneous Cauchy-Euler equation on (0, ∞) that has y(x) = (ln x)
2
3x
as a
solution. Hint: Simply differentiate twice, then multiply by x
2
: x
2y
′′ =
2
3
1−3 ln x+(ln x)
2
x
.
12. Find the simplest 2nd order nonhomogeneous linear DE with constant coefficients that has y(x) = x
R x
2
e
t
√
t
dt as
a solution. Answer: y
′′ = e
x
3
2
√
x +
√
x
.
13. (a) Let g(x) be continuous. Solve the initial value problem
xy′′ + y
′ =
g(x)
x
, x ∈ (0, ∞)
y(3) = ln 3
y
′
(3) = 1
3
(b) For your solution y(x), what’s y
′
(e), y′′(1)?
Answer: (a) y(x) = ln x −
R x
3
ln t·g(t)
t
dt + ln x
R x
3
g(t)
t
dt, (b) y
′
(e) = 1
e +
1
e
R e
3
g(t)
t
dt, (c) y
′′(1) = −1 −
R 1
3
g(t)
t
dt + g(1).
1
2
14. §8.1:#s 3,9,13,19,23,25
15. Find the general solutions: (a) X′ =
0 1 1
0 1 1
0 0 2
X, (b) X′ =
1 1 1
1 1 1
1 1 1
X, (c) X′ =
1 −1 2
−1 1 0
−1 0 1
X
Answer to (a): X(t) = c1
1
0
0
+ c2e
t
1
1
0
+ c3e
2t
1
1
1
, where c1, c2, c3 are arbitrary.
16. (Bad Repeats) Find the general solution: X′ =
2 1 6
0 2 5
0 0 2
X.
17. Solve the initial value problem
dx
dt = z, x(0) = 1
dy
dt = −z, y(0) = 2
dz
dt = y, z(0) = 3.
Answer: X(t) = 3 1
0
0
+ 3 sin t
− sin t
cos t
+ 2 − cos t
cos t
sin t