1
Math 340
Test 2
Professor Carlson
SHOW ALL YOUR WORK
1. (20 pts) Consider the equation
y 00 − 9y = 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 − 9 = (r + 3)(r − 3) = 0.
The exponential solutions are
z1 = e3t ,
z2 = e−3t .
To show independence we compute the Wronskian
W = y1y20 − y10 y2 = −3e3t e−3t − 3e3t e−3t = −6.
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 e3t + c2 e−3t .
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,
3c1 e0 − 3c2 e0 = 3c1 − 3c2 = 0.
The second equation says c1 = c2 , so c1 = c2 =
1
2
1
1
y(t) = e3t + e−3t .
2
2
and
2
2. (20 pts) a) Find the exponential solutions y = ert of the equation
y 00 + 2y 0 + 3y = 0.
The characteristic equation is
r2 + 2r + 3 = 0.
The roots are
r=
−2 ±
√
4 − 12
2
√
= −1 ± i 2.
The exponential solutions are
z1 = e(−1+i
√
2)t
,
z2 = e(−1−i
√
2)t
b) Find two independent solutions of the form
y1 = eαt cos(βt),
where α and β are real numbers.
The solutions we want are
√
y1 = e−t cos( 2t),
y2 = eαt sin(βt),
√
y2 (t) = e−t sin( 2t).
c) Find the solution y of this equation satisfying
y(0) = 0,
y 0 (0) = 1.
Notice that y2 (0) = 0 and
√
√
√
y20 (t) = −e−t sin( 2t) + e−t 2 cos( 2t).
The function
√
1
y(t) = √ e−t sin( 2t)
2
is the one we want.
3. (20 pts) a)Find the general solution of the equation
y 00 − 3y 0 + 2y = t.
3
Write the equation as
[D2 − 3D + 2]y = t.
Apply D2 to get
D2 [D2 − 3D + 2]y = 0.
The roots of the characteristic equation are
r = 0, 0, 2, 1.
Look first for a solution y = c1 + c2 t. Putting this form into the equation
yields
−3c2 + 2(c1 + c2 t) = t.
That is
2c1 − 3c2 = 0,
c2 = 1/2,
c1 = 3/4.
The general solution is
y(t) =
3 1
+ t + c3 et + c4 e2t .
4 2
b) Find the solution y of this equation satisfying
y(0) = 0,
y 0 (0) = 0.
Evaluation gives
3
+ c3 + c4 = 0,
4
1
+ c3 + 2c4 = 0.
2
1
c4 = , c3 = −1.
4
The solution is
y(t) =
3 1
1
+ t − et + e2t .
4 2
4
4. (20 pts) a) Find independent solutions y1 and y2 of
y 00 − y = 0.
y1 = e−t ,
y2 = et .
4
The Wronskian is
W = y1 y20 − y10 y2 = 2.
b) The variation of parameters formula says that a particular solution of
the equation
y 00 − y = g(t),
(2)
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).
yp (t) = −e
−t
Z
0
t
es e2s
ds + et
2
Z
0
t
e−s e2s
ds = −e−t
2
Z
t
0
e3s
ds + et
2
Z
0
t
es
ds
2
1
1
1
1
1
= − e−t [e3t − 1] + et [et − 1] = e−t − et + e2t .
6
2
6
2
3
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)
5
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.
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.