Section 3.2
# 1 Find the Wronskian of the functions
y1 = e2t ,
y2 = e−3t/2 .
In this case
W (y1 , y2 ) = y1 y2′ − y1′ y2 = e2t (−3/2)e−3t/2 − 2e2t e−3t/2 =
−7 t/2
e .
2
# 2 Find the Wronskian of the functions
y1 = cos(t),
y2 = sin(t).
In this case
W (y1 , y2 ) = y1 y2′ − y1′ y2
= cos(t) ∗ cos(t) − (− sin(t)) sin(t) = cos2 (t) + sin2 (t) = 1.
# 13 A routine check.
# 16 Can Y (t) = sin(t2 ) be a solution of
y ′′ + p(t)y ′ + q(t)y = 0
with continuous coefficients?
According to the Existence and Uniqueness Theorem 3.2.1 solutions of
the equation are uniquely determined by their initial data. In this case we
consider the initial data at t = 0. Notice that Y (0) = 0 and Y ′ (0) = 0. If
Y (t) were a solution to such an equation, its initial data would agree with
that of the solution Z(t) = 0. Obviously, Y (t) 6= Z(t), so Y (t) can’t solve
such an equation.
# 22 Find the fundamental set of solutions satisfying
y1 (0) = 1,
y1′ (0) = 0,
y2 (0) = 0,
y2′ (0) = 1,
if
y ′′ + y ′ − 2y = 0.
1
The characteristic equation is
r2 + r − 2 = 0,
r = −2, 1.
The general solution is
y(t) = c1 et + c2 e−2t .
To find y1 solve
c1 + c2 = 1,
c1 − 2c2 = 0.
We get c2 = 1/3 and c1 = 2/3, so
y1 (t) =
2 t 1 −2t
e + e .
3
3
To find y2 solve
c1 + c2 = 0,
c1 − 2c2 = 1.
We get c2 = −1/3 and c1 = 1/3, so
y1 (t) =
1 t 1 −2t
e − e .
3
3
# 23 Find the fundamental set of solutions satisfying
y1 (1) = 1,
y1′ (1) = 0,
y2 (1) = 0,
y2′ (1) = 1,
if
y ′′ + 4y ′ + 3y = 0.
The characteristic equation is
r2 + 4r + 3 = 0,
r = −3, −1.
The general solution is
y(t) = c1 e−t + c2 e−3t .
2
To find y1 solve
c1 e−1 + c2 e−3 = 1,
−c1 e−1 − 3c2 e−3 = 0.
We get c2 = −e3 /2 and c1 = 3e1 /2, so
y1 (t) =
3 1−t 1 3−3t
e
− e
.
2
2
To find y2 solve
c1 e−1 + c2 e−3 = 0,
−c1 e−1 − 3c2 e−3 = 1.
We get c2 = −e3 /2 and c1 = e1 /2, so
y2 (t) =
1 1−t 1 3−3t
e
− e
.
2
2
# 24 Do the solutions y1 = cos(2t), y2 = sin(2t) form a fundamental
set of solutions of
y ′′ + 4y = 0?
The functions y1 , y2 do satisfy the equation. Checking the Wronskian
we get
W (y1 , y2 ) = y1 y2′ − y1′ y2 = 2 cos2 (2t) + 2 sin2 (2t) = 2,
so this is a fundamental set.
3