Math 3400
Test 2
Professor Carlson
SHOW ALL YOUR WORK
1. (20 pts) a) Find two independent exponential solutions y1 = er1 t
and y2 = er2 t of the equation
y ′′ − y = 0.
The characteristic equation is
r2 − 1 = 0,
with roots r1 = 1, r2 = −1. The two independent exponential solutions are
y 1 = et ,
y2 = e−t .
b) Show that the two solutions y1 and y2 are independent.
The Wronskian is
y1 y2′ − y1′ y2 = −et e−t − et e−t = −2.
Since the Wronskian is not zero, the solutions are independent.
c) Find the solution y of this equation satisfying
y ′ (0) = 2.
y(0) = 0,
Any solution has the form c1 et + c2 e−t . The initial conditions give
c1 + c2 = 0,
c1 − c2 = 2.
The solution we want is
y(t) = et − e−t .
2. (20 pts) a) Find two independent exponential solutions y1 = er1 t
and y2 = er2 t of the equation
y ′′ − 6y ′ + 10y = 0.
1
The characteristic equation is
r2 − 6r + 10 = 0,
with roots
r=
6±
√
36 − 40
= 3 ± i.
2
The two independent exponential solutions are
y1 = e(3+i)t ,
y2 = e(3−i)t .
b) Find two independent solutions z1 , z2 of this equation of the form
z1 = eαt cos(βt),
z2 = eαt sin(βt).
The Euler formula eiθ = cos(θ) + i sin(θ) leads to
z1 = e3t cos(t),
z2 = e3t sin(t).
c) Compute the Wronskian of z1 and z2 .
The Wronskian is
z1 z2′ − z1′ z2
= e3t cos(t)[3e3t sin(t) + e3t cos(t)] − [3e3t cos(t) − e3t sin(t)]e3t sin(t)
= e6t [cos(t)2 + sin(t)2 ] = e6t .
3. (20 pts) a) Find the general solution of the equation
y ′′ − y = e2t .
Using the annihilator method and (D − 2)e2t = 0 we find
(D − 2)(D + 1)(D − 1)y = 0,
so
y(t) = c1 et + c2 e−t + c3 e2t .
2
After plugging y in this form back into the original equation we find 3c3 = 1.
The general solution is
1
y(t) = c1 et + c2 e−t + e2t .
3
b) Find the solution y of this equation satisfying
y ′ (0) = 0.
y(0) = 0,
The equations are
1
= 0,
3
2
c1 − c2 + = 0,
3
c1 + c2 +
so
c1 = −1/2,
and
y(t) =
c2 = 1/6,
−1 t 1 −t 1 2t
e + e + e .
2
6
3
4. (15 pts) a) Show that
y1 (t) = sin(α) cos(t) + cos(α) sin(t)
is a solution of
y ′′ + y = 0
that satisfies
y1′ (0) = cos(α).
y1 (0) = sin(α),
Since α is a constant,
y1′ (t) = − sin(α) sin(t) + cos(α) cos(t),
y1′′ (t) = − sin(α) cos(t) − cos(α) sin(t),
so y1 solves the equation. In addition,
y1 (0) = sin(α) cos(0) + cos(α) sin(0) = sin(α),
3
y1′ (0) = − sin(α) sin(0) + cos(α) cos(0) = cos(α).
b) Show that
y2 (t) = sin(t + α)
is also a solution of y ′′ + y = 0 that satisfies
y2′ (0) = cos(α).
y2 (0) = sin(α),
y2′ (t) = cos(t + α),
y2′′ (t) = − sin(t + α),
so y2 solves the equation. In addition,
y2′ (0) = cos(α).
y2 (0) = sin(α),
c) What can you conclude about y1 and y2 . Explain.
The second order equation y ′′ + y = 0 has a unique solution satisfying
the initial conditions
y ′ (0) = cos(α).
y(0) = sin(α),
Since y1 and y2 are both solutions satisfying the same initial conditions,
they must be the same. That is
sin(t + α) = sin(α) cos(t) + cos(α) sin(t).
5. (25 pts) The functions y1 (t) = cos(t) and y2 (t) = sin(t) are independent solutions of y ′′ + y = 0. The variation of parameters formula says
that a particular solution of the equation
y ′′ + y = g(t),
is given by
yp (t) = −y1 (t)
Z
t
0
y2 (s)g(s)
ds + y2 (t)
W (y1 , y2 )(s)
a) Find yp (0) and yp′ (0).
4
Z
t
0
y1 (s)g(s)
ds.
W (y1 , y2 )(s)
First,
Z
yp (0) = −y1 (0)
0
y2 (s)g(s)
ds + y2 (0)
W (y1 , y2 )(s)
0
Z
0
0
y1 (s)g(s)
ds.
W (y1 , y2 )(s)
Since the limits of integration are both 0, the integrals are zero and yp (0) =
0.
Next, by the Fundamental Theorem of Calculus,
d
dt
Z
t
f (s) ds = f (t).
0
Applying this result,
yp′ (t)
=
−y1′ (t)
+y2′ (t)
Z
t
0
Z
t
0
y2 (t)g(t)
y2 (s)g(s)
ds − y1 (t)
W (y1 , y2 )(s)
W (y1 , y2 )(t)
y1 (s)g(s)
y1 (t)g(t)
ds + y2 (t)
.
W (y1 , y2 )(s)
W (y1 , y2 )(t)
Evaluating at zero as above,
yp′ (0) = 0 − y1 (0)
y2 (0)g(0)
y1 (0)g(0)
+ 0 + y2 (0)
= 0.
W (y1 , y2 )(0)
W (y1 , y2 )(0)
That is yp (0) = 0 and yp′ (0) = 0.
Find the general solution of
y ′′ + y = t
using
b) the method of undetermined coefficients (annihilator method),
A solution y will satisfy
D2 (D2 + 1)y = 0,
so
y(t) = c1 + c2 t + c3 cos(t) + c4 sin(t).
5
Plugging this expression into the equation gives
c1 + c2 t = t,
c1 = 0,
c2 = 1.
The general solution is
y(t) = t + c3 cos(t) + c4 sin(t).
c) the method of variation of parameters.
The Wronskian of y1 (t) = cos(t) and y2 (t) = sin(t) is W = y1 y2′ −
y1′ y2 = 1. Then using the formula above,
yp (t) = − cos(t)
Z
t
s sin(s) ds + sin(t)
0
Z
t
s cos(s) ds.
0
Integration by parts yields
yp (t) = − cos(t)[−s cos(s)
t
+
0
Z
t
cos(s) ds] + sin(t)[s sin(s)
0
t
−
0
Z
t
0
= t cos2 (t) − cos(t) sin(t) + t sin2 (t) + sin(t)[cos(t) − 1]
= t − sin(t).
The general solution is
y(t) = t + c3 cos(t) + c4 sin(t).
6
sin(s) ds.