Download Jim Lambers MAT 285 Spring Semester 2016

Jim Lambers
MAT 285
Spring Semester 2016-17
Practice Exam 2 Solution
1. Solve the initial value problem
y 00 + 4y 0 + 3y = 0,
y(0) = 2,
y 0 (0) = −1.
Solution The characteristic equation is
λ2 + 4λ + 3 = 0,
which factors into
(λ + 1)(λ + 3) = 0.
Therefore, the roots are λ1 = −1 and λ2 = −3. Since the roots are real and distinct, the
general solution is
y(t) = c1 e−t + c2 e−3t ,
which has the derivative
y 0 (t) = −c1 e−t − 3c2 e−3t .
From the initial conditions, we obtain the equations
y(0) = c1 + c2 = 2,
y 0 (0) = −c1 − 3c2 = −1.
By adding these two equations, we obtain −2c2 = 1, and therefore c2 = −1/2. Substituting
this value into either equation yieilds c1 = 5/2. We conclude that the solution is
5
1
y(t) = e−t − e−3t .
2
2
2. Solve the initial value problem
y 00 + 4y 0 + 5y = 0,
y(0) = 1,
y 0 (0) = 0.
Solution The characteristic equation is
λ2 + 4λ + 5 = 0,
which has roots
λ1,2 =
−4 ±
p
√
42 − 4(1)(5)
= −2 ± −1 = −2 ± i.
2
Since the roots are complex, the general solution is
y(t) = c1 e−2t cos t + c2 e−2t sin t,
which has the derivative
y 0 (t) = −2c1 e−2t cos t − c1 e−2t sin t − 2c2 e−2t sin t + c2 e−2t cos t.
From the initial conditions, we obtain the equations
y(0) = c1 = 1,
y 0 (0) = −2c1 + c2 = 0.
It follows that c1 = 1 and c2 = 2. We conclude that the solution is
y(t) = e−2t cos t + 2e−2t sin t.
3. Use reduction of order to find a second solution of the equation
xy 00 − y 0 + 4x3 y = 0,
y1 (x) = sin(x2 ).
Solution First, we divide both sides of the equation by the coefficient of x to obtain
y 00 −
1 0
y + 4x2 y = 0.
x
Then, this equation has the form
y 00 + p(x)y 0 + q(x)y = 0,
where p(x) = 1/x. We assume that the second solution y2 (x) has the form
y2 (x) = y1 (x)v(x).
Then v(x) is given by
Z
R
1
− p(x) dx
e
dx
[y (x)]2
Z 1
R
1
− − x1 dx
dx
=
e
[sin(x2 )]2
Z
1
=
eln x dx
[sin(x2 )]2
Z
x
=
dx
[sin(x2 )]2
v(x) =
Z
1
1
2 du,
2
sin
u
Z
1
=
csc2 u du
2
1
= − cot u
2
1
= − cot(x2 ).
2
u = x2
=
Neglecting the constant factor, we conclude that a second solution is
y2 (x) = sin(x2 ) cot(x2 ) = sin(x2 )
cos(x2 )
= cos(x2 ).
sin(x2 )
4. Solve the initial value problem
9y 00 − 12y 0 + 4y = 0,
y(0) = 2,
y 0 (0) = −1.
Solution The characteristic equation is
9λ2 − 12λ + 4 = 0,
which has roots
λ1,2 =
12 ±
p
(−12)2 − 4(9)(4)
2
= .
2(9)
3
Since this is a double root, the general solution is of the form
y(t) = c1 e2t/3 + c2 te2t/3 ,
which has the derivative
2
2
y 0 (t) = c1 e2t/3 + c2 e2t/3 + c2 te2t/3 .
3
3
From the initial conditions, we obtain the equations
y(0) = c1 = 2,
2
y 0 (0) = c1 + c2 = −1.
3
It follows that c1 = 2 and c2 = −1 − 2(2)/3 = −7/3. We conclude that the solution is
7
y(t) = 2e2t/3 − te2t/3 .
3
5. Find the general solution of the differential equation
y 00 + 2y 0 + y = 2e−t .
Solution The characteristic equation is
λ2 + 2λ + 1 = 0,
which factors into (λ + 1)2 = 0. It follows that the roots are both equal to −1, and therefore
the general solution of the homogeneous equation is
yh (t) = c1 e−t + c2 te−t .
Now, we need to find a particular solution of the inhomogeneous equation. Since the equation
has constant coefficients and the right-hand side g(t) = 2e−t is of an appropriate form, we
can use the method of undetermined coefficients.
Specifically, g(t) is of the form Pn (t)eat . Therefore, the particular solution is of the form
yp (t) = ts (A0 + A1 t + · · · + An tn )e−t .
In this case, n = 0, P0 (t) = 2, and a = −1. Since a is occurs twice as a root of the
characteristic equation, s = 2. Therefore, the particular solution has the form
yp (t) = t2 A0 e−t
where A0 is an undetermined coefficient.
We now substitute this form of yp (t) into the ODE. From
yp0 (t) = A0 (2t − t2 )e−t ,
yp00 (t) = A0 (t2 − 4t + 2)e−t ,
we obtain
A0 (t2 − 4t + 2)e−t + 2[A0 (2t − t2 )e−t ] + A0 t2 e−t = 2e−t ,
which simplifies to
2A0 = 2.
It follows that A0 = 1 and that
yp (t) = t2 e−t .
We conclude that the general solution is
y(t) = yh (t) + yp (t) = c1 e−t + c2 te−t + t2 e−t .
6. Find the general solution of the differential equation
t2 y 00 − 2ty 0 + 2y = 4t2 ,
y1 (t) = t,
y2 (t) = t2 .
Solution Since this inhomogeneous equation does not have constant coefficients, we cannot
use the method of undetermined coefficients. Instead, we can use variation of parameters.
First, we divide both sides of the equation by the coefficient of t2 , to obtain
2
2
y 00 − y 0 + 2 y = 4.
t
t
Then, compute the Wronskian of the homoegeneous solutions y1 and y2 , which is
W (y1 , y2 )(t) = y1 (t)y20 (t) − y2 (t)y10 (t) = t(t2 )0 − t2 (t0 ) = 2t2 − t2 = t2 .
Then, the particular solution is of the form
yp (t) = y1 (t)w1 (t) + y2 (t)w2 (t)
where
Z
Z
−y2 (t)g(t)
−4t2
w1 (t) =
dt =
dt = −4 dt = −4t,
W (y1 , y2 )(t)
t2
Z
Z
Z
y1 (t)g(t)
4t
1
w2 (t) =
dt =
dt
=
4
dt = 4 ln t.
W (y1 , y2 )(t)
t2
t
Z
We conclude that a particular solution is
yp (t) = −4t2 + 4t2 ln t,
and that the general solution is
y(t) = yh (t) + yp (t) = c1 t + c2 t2 + 4t2 ln t.
In the general solution, the −4t2 term has been dropped because it is a solution of the
homogeneous equation.
7. A object with a mass of 100 g stretches a spring 5 cm. If the object is set in motion from
its equilibrium position with a downward velocity of 10 cm/s, and if there is no damping,
determine the position u of the object at any time t. Find the frequency, period, and amplitude
of the motion.
Solution Since there is no damping and no external force, the motion of the object is modeled
by the equation
mu00 + ku = 0,
where u is measured in meters, m = 100 g, and the spring constant k satisfies the equation
kL − mg = 0,
where L = 0.05 m and g = 9.8m/s2 . Therefore,
k=
mg
(100 g)(9.8 m/s2 )
=
= 19, 600 g/s2 .
L
0.05 m
The initial conditions are
u(0) = 0,
u0 (0) = 0.1,
since the object is set in motion from its equilibrium position with a downward velocity of 10
cm/s, which is 0.1 m/s.
Dividing both sides of the ODE by the coefficient of u00 yields
u00 + 196u = 0.
The characteristic equation is
λ2 + 196 = 0,
which has complex roots λ1,2 = ±14i. It follows that the solution is of the form
u(t) = A cos 14t + B sin 14t,
which has a derivative of
u0 (t) = −14A sin 14t + 14B cos t.
From the initial conditions, we obtain the equations
u(0) = A = 0,
u0 (0) = 14B = 0.1,
which yields A = 0 and B = 1/140. Therefore, the solution is
u(t) =
1
sin 14t.
140
The frequency is
r
ω0 =
k
=
m
r
19600 √
= 196 = 14.
100
The period is
T =
2π
2π
π
=
= ,
ω0
14
7
and the amplitude is
R=
p
1
A2 + B 2 =
.
140