MAT 275 TEST 2 PRACTICE PROBLEMS
I. Existence and uniqueness. Fundamental sets.
1. Determine the longest interval on which the given initial value problem is certain to have a unique twice
differentiable solution.
(a)
x−3y ' ' x
x−3y '
x−1y=0, y2=0, y' 2=1
(b)
t−1y ' ' t y 'y=sect, y0=1, y ' 0=3
(c)
tt−4y ' ' 3y ' lnty=sin t, y1=1, y ' 1=1
2. Which of the following is a true statement?
I. Two functions defined on an open interval I are said to be linearly independent on I provided that one is a
constant multiple of the other on I.
II. Two functions defined on an open interval I are said to be linearly dependent on I provided that one is a
constant multiple of the other on I.
III. Two functions defined on an open interval I are said to be linearly independent on I provided that neither is a
constant multiple of the other on I.
IV. Two functions defined on an open interval I are said to be linearly dependent on I provided that neither is a
constant multiple of the other on I.
3. Which of the following pairs of functions is linearly independent on the entire real line?
A.
{
sin x , cos x
}
B.
{
ex, x ex
}
C.
{
x ,
e
3
x
}
D.
{
x , 3x
}
E.
{
1, e−t
}
F.
{
cos t , sin t/2
}
G.
{
e−2tcos 2t , e−2t sin 2t
}
H.
{
2e−t,4e−t3
}
I.
{
e2t, e2t−6
}
J.
{
x ,
∣
x
∣
}
4. Which of the following is NOT a fundamental set of solutions for
y ' '−y=0?
A.
{
et, e−t
}
B.
{
2et,2e−t
}
C.
{
t et,e−t
}
D.
{
ete−t,1
2ete−t
}
E.
{
1
2ete−t,1
2et−e−t
}
F.
{
ete−t, et
}
5. Suppose
y1t=t
and
y2t=t2
are both solutions of the second order linear equation
y ' ' pty 'qty=0.
Which of the functions below are guaranteed to also be solutions of the same equation?
A.
y=t2−1
B.
y=5t2
C.
y=−9t217t
D. y = 0
6. Consider the ODE
t2y ' ' 3t y ' y=0
with the initial conditions y(1) = 1, y'(1) = 1.
(i) What is the maximum interval of validity, I, of the solution?
(ii) Verify that the functions
y1t=t−1
and
y2t=t−1lnt
satisfy the ODE for t in the interval I.
(iii) Use the Wronskian to show that the functions
y1
and
y2
from ii. form a fundamental set of solutions.
(iv) Solve the initial value problem.
II. HODEs/IVP with constant coefficients.
1. Find a real valued solution to the following initial value problems. Sketch a graph of the solution.
a.
y ' ' −6y ' 13 y=0
with
y0=1, y' 0=1.
b.
y ' ' 4y ' 4y=0,
with
y0=1, y ' 0=−4.
c.
y ' ' 3y '2y=0,
with
y0=3, y' 0=0.
2. For which values of
(if any) are all solutions of
y ' ' −2−1y ' −1y=0
unbounded as
t∞
?
3. The characteristic equation of a homogeneous 9th order linear Differential Equation with constant coefficients has roots
r=0
with multiplicity three,
r=−2
with multiplicity two,
r=−3±2i
with multiplicity two.
Write the general solution of the Differential Equation.
4. One solution of the DE
6y45y325 y ' ' 20 y ' 4y=0
is
y=cos2x.
Find the general solution.
III. Reduction of order:
1. The ODE
t2y ' ' 3t y ' y=0
has a solution
y1t=t−1
for t > 0. Find the general solution.
2. The ODE
t2y ' '−tt2y 't2y=0
has a solution
y1t=t
for t > 0. Find the general solution.
IV. Undetermined coefficients
1. Find the general solution of the ODE
y ' ' 2y ' y=e−t
2. Solve the IVP:
y ' ' −y ' −2y=6x6e−x, y0=1, y' 0=0
3. Solve the IVP:
y ' ' −y ' −2y=6t e2t , y 0=0, y ' 0=1
4. Determine a suitable form for the particular solution Y(t) if the method of undetermined coefficients is to be used.
You do not need to determine the values of the coefficients.
(i)
y ' ' 3y ' =2t2t2e−3tsin 3t
(ii)
y ' ' y=t1sin t
(iii)
y ' ' −5y '6y=etcos 2te2t 3t4sin t
(iv)
y ' ' 2y '2y=3e−t2e−tcos t4e−tt2sin t
(v)
y ' ' −4y'4y=2t24t e2ttsin2t
V. Mass-Spring system
1. Consider the IVP:
y ' ' 4y=0
with
y0=−3
and
y ' 0=6.
. Write the solution as
yt=Rcos0t−.
2. A mass of 2 kilograms stretches a spring 0.5 meters. If the mass is set in motion from its equilibrium with a downward
velocity of 10 cm/s, and there is no damping, write an IVP for the position u (in meters) of the mass at any time t ( in
seconds). Use g=9.8 m/s2 for the acceleration due to gravity.
3. For the following, choose the best description of the system from the following:
Simple Harmonic Motion (SHM) Overdamped (OD) Underdamped (UD) Critically Damped (CD)
Beating (B) Resonant (R) Steady-State plus Transient (SST)
a.
y ' ' 4y=0
b.
y ' ' 1.82y=cos2t
c.
y ' ' 4y=cos2t
d.
y ' 'y 'y=0
e.
y ' ' y 'y=cost
f.
y ' ' 2y ' y=0
4. The motion of a force mass-spring system is described by the following IVP:
u ' '9u=cos3t, u0=0, u ' 0=0
(a) Explain why you expect resonance to occur.
(b) solve this IVP and sketch the graph of the solution.
5. The motion of a force mass-spring system is described by the following IVP:
u ' ' 2.82u=cos3t, u 0=0,u ' 0=0.
(a) Explain why you expect the beats phenomenon to occur.
(b) Solve the IVP and write your solution in the form
Asin tsint
(c) Determine the length of the beats and the period of the oscillation.
6. A mass m =1 is attached to a spring with constant k = 2 and damping constant γ. Determine the value of γ so that the
motion is critically damped.
7. The position function of a mass-spring system satisfies the differential equation
m x' ' x'k x=cos t,
x0=x ' 0=0
.
Assume m = 1 and k = 9.
If
≠0,
the amplitude of the forced oscillation is given by
C=1
0
2−2222.
Assume
=1
.
Differentiate C to find the value of ω at which practical resonance occurs. Determine the corresponding value of C.
VI. Introduction to systems:
1. Transform the given IVP into an initial value problem for two first order equations.
u ' '4u '5t u=7−sin 2t, u0=−2, u ' 0=1
2. (a) Write the following IVP for a system of 2 linear ODEs as an IVP for a single second-order linear ODE
x'=− y , x0=1
y '=10 x−7y , y0=−7
(1)
(b) Find the solution of the IVP (1)
3. Match the description of the phase portrait with the corresponding system (one description will not match)
I
x '=y , y' =−x
II
x '=y , y' =x
III
x '=−2y , y' =x
A. circles B. Ellipses C. hyperbolas D. parallel lines
TEST 2 ANSWERS TO PRACTICE PROBLEMS
I.
1. (a) 1 < x < 3 (b)
−
2t1
(c) 0 < t < 4.
2. II and III
3. A, B, E, G, I, J
4. C:
tet
is not a solution; D: the two functions are not linearly independent.
5. By the principle of superposition: B, C, D
6. (i) t > 0 (iii)
Wy1, y2=t−3
Since the Wronskian is nonzero on I,
y1
and
y2
form a fundamental set of solutions.
(iv)
y=12ln t
t
II.
1. (a)
y=−e3tsin 2te3tcos2t
(b)
yt=1−2te−2t
(c)
yt=−3e−2t6e−t
2. The general solution is
y=c1etc2e−1t.
Thus all solutions are unbounded if
1
3.
c1c2tc3t2c4e−2tc5t e−2te−3tc6cos2tc7sin 2tte−3t c8cos2tc9sin2t
4. Since
y=cos2x
is a solution, 2i and –2i must be roots of the characteristic equation and
r24
must be a factor.
Using long division, another factor is
6r25r1
. Thus the characteristic equation can be written as
r243r12r1
and the general solution is
y=c1cos2xc2sin2xc3e−x/3c4e−x/2
III.
1.
yt=c1t−1c2t−1ln t
2.
yt=c1tc2t et
IV.
1.
y=c1e−tc2t e−t1
2t2e−t
2.
y=3
2e2x−2e−x
3
2−3x
−2xe−x
3.
y=5
9e2t −5
9e−tt2−2
3te2t
4.
(i)
Yt=tA0t2A1tA2tB0t2B1tB2e−3tEsin3tFcos3t
(ii)
Yt= A0tA1tB0tB1sinttC0tC1cost
(iii)
Yt=A0etcos2t A1etsin 2te2tB0tB1sin te2tC0tC1cos t
(iv)
Yt= A0e−tt e−tB0t2B1tB2sin tt e−tC0t2C1tC2cos t
(v)
Yt= A0t2A1tA2t2B0tB1e2tC0tC1sin2tD0tD1cos2t
V.
1.
yt=−3 cos2t3sin2t=3
2cos
2t−3
4
2.
2u ' ' 39.2 u=0, u0=0, u ' 0=0.1
3. a. undamped free motion: Simple Harmonic Motion
b. undamped motion with
0=1.8≈2=
: Beats
c. undamped motion with
0=2=
: Resonance
d. free damped motion; roots of the characteristic equation are complex: Under Damped
e. damped motion with forcing term: Steady State plus Transient
f. free damped motion; roots of the characteristic equation are repeated: Critically Damped
4. (a)
0=3=
(b)
ut= t
6sin 3t
5. (a)
0=2.8≈3=
(b)
1
1.16 cos2.8 t−cos3t= 2
1.16 sin 0.1 tsin2.9 t
(c) Length of beats =
2
20.1=10
Period of oscillation =
2
2.9
6.
2
2
7.
=
34
2≈2.91
The corresponding maximum value of the amplitude is
C
34
2
≈0.338
VI.
1.
x1'=x2, x2'=−5t x1−4x27−sin2t, x10=−2x20=1
2. (a)
y ' ' 7y ' 10 y=0, y0=−7, y ' 0=59
(b)
xt=4e−2t−3e−5t, yt=8e−2t−15e−5t
3. I: Solving
dy
dx =− x
y
yields
y2x2=C ,
hence the trajectories are circles and I matches A
II: Solving
dy
dx =x
y
yields
y2−x2=C ,
hence the trajectories are hyperbolas and II matches C
III: Solving
dy
dx =− x
2y
yields
y2x2
2=C ,
hence the trajectories are ellipses and III matches B
