1
Math 340
Test 2
Professor Carlson
SHOW ALL YOUR WORK
1. (20 pts) Consider the equation
y 00 − 16y = 0.
(1)
a) Find a pair of linearly independent solutions of the form ert . Show that
your solutions are linearly independent.
The characteristic equation is
r2 − 16 = (r + 4)(r − 4) = 0.
The exponential solutions are
z1 = e4t ,
z2 = e−4t .
To show independence we compute the Wronskian
W = y1y20 − y10 y2 = −4e4t e−4t − 4e4t e−4t = −8.
Since the Wronskian is not zero, the solutions are independent.
b) Describe the general solution of equation (1).
The general solution has the form
y(t) = c1 e4t + c2 e−4t .
c) Find the solution y(t) which satisfies the initial conditions
y(0) = 1,
y 0 (0) = 0.
We want to find c1 and c2 such that
c1 e0 + c2 e0 = c1 + c2 = 1,
4c1 e0 − 4c2 e0 = 4c1 − 4c2 = 0.
The second equation says c1 = c2 , so c1 = c2 =
1
2
1
1
y(t) = e4t + e−4t .
2
2
and
2
2. (20 pts) a) Find the exponential solutions y = ert of the equation
y 00 + 4y 0 − 3y = 0.
The characteristic equation is
r2 + 4r − 3 = 0.
The roots are
√
√
−4 ± 28
r=
= −2 ± 7.
2
The exponential solutions are
√
z1 = e(−2+
7)t
√
,
z2 = e(−2−
7)t
c) Find the solution y of this equation satisfying
y(0) = 0,
y 0 (0) = 1.
Solve the equations
c1 + c2 = 0,
(−2 +
√
These give
or
7)c1 + (−2 −
c1 = −c2 ,
√
2 7c1 = 1,
1
c1 = √ ,
2 7
1
c2 = − √ .
2 7
√
7)c2 = 1.
3. (20 pts) a)Find the general solution of the equation
y 00 − 5y 0 + 6y = t.
Write the equation as
[D2 − 5D + 6]y = t.
Apply D2 to get
D2 [D2 − 5D + 6]y = 0.
The roots of the characteristic equation are
r = 0, 0, 2, 3.
3
Look first for a solution y = c1 + c2 t. Putting this form into the equation
yields
−5c2 + 6(c1 + c2 t) = t.
That is
6c1 − 5c2 = 0,
c2 = 1/6,
c1 = 5/36.
The general solution is
y(t) =
1
5
+ t + c3 e3t + c4 e2t .
36 6
b) Find the solution y of this equation satisfying
y(0) = 0,
y 0 (0) = 0.
Evaluation gives
5
+ c3 + c4 = 0,
36
1
+ 3c3 + 2c4 = 0.
6
1
1
c4 = − , c3 = .
4
9
The solution is
y(t) =
5
1
1
1
+ t + e3t − e2t .
36 6
9
4
4. (20 pts) a) Find independent solutions y1 and y2 of
y 00 − 4y = 0.
y1 = e−2t ,
y2 = e2t .
The Wronskian is
W = y1 y20 − y10 y2 = 4.
b) The variation of parameters formula says that a particular solution of
the equation
y 00 − 4y = g(t),
(2)
4
is given by
Z
yp (t) = −y1 (t)
0
t
y2 (s)g(s)
ds + y2 (t)
W (y1 , y2 )(s)
Z
t
0
y1 (s)g(s)
ds.
W (y1 , y2 )(s)
Suppose that g(t) = e2t in (2). Find yp (t).
−2t
Z
t
yp (t) = −e
0
=−
e2s e2s
ds+e2t
4
Z
0
t
e−2s e2s
ds = −e−2t
4
Z
0
t
e4s
ds+e2t
4
Z
0
t
1
ds
4
1 −2t 4t
t
t
1
1
e [e − 1] + et = et − e2t + e−2t .
16
4
4
16
16
5. (10 pts) a) Show that an equation of the form
t2 y 00 + αty 0 + βy = 0
has at least one solution (and usually two) of the form
y = tr .
Putting tr into the equation gives
t2 r(r − 1)tr−2 + αtrtr−1 + βtr = 0.
Divide by tr and simplify to get
r2 + (α − 1)r + β = 0.
The quadratic formula gives at least one value of r.
(b) Find a solution y1 = tr for the equation
t2 y 00 − 3ty 0 + 4y = 0.
(3)
In this case we find
r2 − 4r + 4 = (r − 2)2 = 0.
So y = t2 is a solution.
(c) Show another solution for equation (3) has the form y2 = tr log(t).
Show that y1 and y2 are independent.
5
We get
y = t2 log(t),
y 0 = 2t log(t) + t,
y 00 = 2 log(t) + 2 + 1,
and
t2 [2 log(t) + 3] − 3t[2t log(t) + t] + 4t2 log(t) = 0.
To show independence, compute the Wronskian.
W = y1 y20 − y10 y2 = t2 [2t log(t) + t] − 2tt2 log(t) = t3 .
The solutions are independent on any interval not including t = 0.
6. (10 pts) Suppose that y1 (t) is a solution of the equation
y 00 + p(t)y 0 + q(t)y = 0.
(4)
Assume a second solution of equation (4) may be written as
y(t) = v(t)y1 (t).
By computing derivatives and using (4), find a first order linear equation
satisfied by v 0 (t).
If y = vy1 , then
y 0 = v 0 y1 + vy10 ,
y 00 = v 00 y1 + 2v 0 y10 + vy100
and the equation (4) is satisfied if
[v 00 y1 + 2v 0 y10 + vy100 ] + p(t)[v 0 y1 + vy10 ] + q(t)vy1 = 0.
Since y1 satisfies (4), this simplifies to
v 00 y1 + 2v 0 y10 + p(t)v 0 y1 = 0.
If w = v 0 we have the first order linear equation
y1 w0 + [2y10 + p(t)y1 ]w = 0.