MATHEMATICS 215/255, HOMEWORK 3 – SOLUTIONS
Section 3.2: #5, 11, 23, 26
5. The Wronskian of the given pair of functions is
et sin t
et cos t
t
t
W (e sin t, e cos t) = t
e (sin t + cos t) et (cos t − sin t)
= e2t (sin t cos t − sin2 t − sin t cos t − cos2 t) = − e2t. 11. We first divide through by x − 3 to get the equation in standard form, which gives
y ′′ +
ln |x|
x
y′ +
y = 0.
x−3
x−3
Since ln |x| is discontinuous at x = 0 and 1/x−3 is discontinuous at x = 3, the possible intervals are
(−∞, 0), (0, 3), and (3, ∞). As x0 = 1 ∈ (0, 3), the longest interval on which the given differential
equation has a unique, twice differentiable solution is (0, 3). 23. The characteristic equation of this differential equation is r 2 + 4r + 3 = (r + 1)(r + 3) = 0,
the roots of which are r = −1, −3. We may therefore form a fundamental set of solutions from the
functions y1 = e−t and y2 = e−3t . However, neither of these functions satisfy the specified initial
conditions at t0 = 1 as neither they nor their derivatives are ever zero. We must therefore find two
pairs of constants C1 , C2 and D1 , D2 such that u1 = C1 y1 + C2 y2 satisfies u1 (1) = 1, u′1 (1) = 0,
and u2 = D1 y1 + D2 y2 satisfies u2 (1) = 0, u′2 (1) = 1.
We first redefine y1 = e−(t−1) and y2 = e−3(t−1) for convenience, as then y1 (1) = y2 (1) = 1 (it is
easy to verify that these still form a fundamental set of solutions to the given differential equation).
We now must solve the two systems of two equations
u1 (1) =
u′1 (1)
and
C1
+
C2
= 1,
= −C1 − 3C2 = 0,
u2 (1) = D1 + D2 = 0,
u′2 (1) = −D1 − 3D2 = 1,
The necessary algebra leads to the solutions C1 = 3/2, C2 = −1/2, and D1 = 1/2, D2 =
−1/2. The fundamental set of solutions satisfying the initial conditions in Theorem 3.2.5 is thus
1
1
1
3
u1 (t) = e−(t−1) − e−3(t−1) and u2 (t) = e−(t−1) − e−3(t−1) . 2
2
2
2
26. We may verify that the given functions are solutions to the differential equation by direct
substitution. If y1 = x and y2 = xex , then y1′ = 1 and y2′ = (x + 1)ex , while y1′′ = 0 and
y2′′ = (x + 2)ex . We therefore have
x2 y1′′ − x(x + 2)y1′ + (x + 2)y1 = −x(x + 2) + x(x + 2) = 0,
and
x2 y2′′ − x(x + 2)y2′ + (x + 2)y2 = (x2 (x + 2) − x(x + 2)(x + 1) + (x + 2)x)ex = 0.
Homework 3 – SOLUTIONS
Page 2
We may also verify that y1 and y2 do constitute a fundamental set of solutions to the given
differential equation for x > 0 by observing that the Wronskian,
x
xex
W (y1 , y2 )(x) = = x2 ex ,
1 (x + 1)ex is nonzero for all x > 0. Section 3.3: #9, 19, 27, 32
9. The characteristic equation of this differential equation, r 2 + 2r − 8 = (r + 4)(r − 2) = 0, has
roots r = −4, 2. As the roots are real and distinct, the general solution is y = C 1 e−4t + C 2 e2t. 19. The characteristic equation for this differential equation, r 2 − 2r + 5 = 0, has roots r = 1 ± 2i,
and therefore the general solution has the form y = C1 et cos 2t + C2 et sin 2t. From the first initial
condition y(π/2) = C1 eπ/2 cos π + C2 eπ/2 sin π = −C1 eπ/2 = 0, we see that C1 = 0, and therefore
the specific solution must have the form y = C2 et sin 2t. It follows that y ′ = C2 et (sin 2t + 2 cos 2t),
and from the second initial condition y ′ (π/2) = C2 eπ/2 (sin π + 2 cos π) = −2C2 eπ/2 = 2, we see
that we must have C2 = −e−π/2 . The solution to the given initial value problem is therefore
y = −et−π/2 sin 2t. The graph of this function, which is a growing oscillation for increasing t,
looks as below. 50
40
y 30
20
10
0
0
−10
1
2
3
4
5
6
t
d λt
27. We may verify this easily by direct calculation. Since
e cos µt = eλt (cos µt − µ sin µt)
dt
d λt
λt
and
e sin µt = e (sin µt + µ cos µt), we have
dt
eλt cos µt
eλt sin µt
λt
λt
W (e cos µt, e sin µt) = λt
e (cos µt − µ sin µt) eλt (sin t + µ cos µt) = e2λt (sin µt cos µt + µ cos2 µt − sin µt cos µt + µ sin2 µt) = µe2λt.
We may also arrive at the same result more quickly by invoking Question 37 from Section 3.2, which
states that if f , g, and h are differentiable functions on an interval I, then W (f g, f h) = f 2 W (g, h)
on I. Letting f = eλt , g = cos µt, and h = sin µt, so that g′ = −µ sin µt and h′ = µ cos µt, we have
W (f g, f h) = e2λt · µ(cos2 µt + sin2 µt) = µe2λt . Homework 3 – SOLUTIONS
Page 3
32. If y = φ(t) = u(t) + iv(t), then by the linearity of the derivative (a combination of the results
stating that the derivative of a sum of two functions is the sum of their derivatives, while the
derivative of a constant multiple of a function is the constant multiple of the derivative of the
function), we have y ′ = u′ (t) + iv ′ (t) and y ′′ = u′′ (t) + iv ′′ (t). Substituting these expressions into
the original differential equation, we have
y ′′ + p(t)y ′ + q(t)y = (u′′ (t) + iv ′′ (t)) + p(t)(u′ (t) + iv ′ (t)) + q(t)(u(t) + iv(t))
= (u′′ (t) + p(t)u′ (t) + q(t)u(t)) + i(v ′′ (t) + p(t)v ′ (t) + q(t)v(t)) = 0.
As the functions p(t), q(t), u(t), and v(t) are all real valued (as are the first and second derivatives of
u(t) and v(t)), the final equation above is the sum of a real-valued function and a second real-valued
function multiplied by i. As a complex number can only be zero if both its real and imaginary
parts are zero, this means that
u′′ (t) + p(t)u′ (t) + q(t)u(t) = 0 and
v ′′ (t) + p(t)v ′ (t) + q(t)v(t) = 0.
In other words, u(t) and v(t) are solutions to the original differential equation, as desired.
Section 3.4: #4, 11, 16, 23
4. The characteristic equation 4r 2 + 12r + 9 = 0 has a double root at r = −3/2. The general
solution of the differential equation therefore has the form y = C 1 te−3t/2 + C 2 e−3t/2 . 11. The characteristic equation 9r 2 − 12r + 4 = 0 has a double root at r = 2/3. The general
solution to the differential equation is thus y = C1 te2t/3 + C2 e2t/3 . We may find the values of C1
and C2 from theinitial conditions. Since y(0) = C1 · 0 · 1 + C2 · 1 = C2 = 2, we must have C2 = 2.
2
2
2
t + 1 e2t/3 + C2 e2t/3 , we have y ′ (0) = C1 + C2 = −1, or C1 = −7/3. Hence
And as y ′ = C1
3
3
3
7 2t/3
+ 2e2t/3 . The graph of the solution
the solution to the given initial value problem is y = − te
3
therefore looks as below. We note that since the coefficient of the (dominating) te2t/3 term is
negative, we have y = −∞ as t → ∞. t
0.0
0
−5
y −10
−15
−20
0.5
1.0
1.5
2.0
2.5
Homework 3 – SOLUTIONS
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16. As established in Example 2, the characteristic equation r 2 − r + 0.25 = 0 has a double root at
r = 1/2, so the general solution has the form y = C1 tet/2 + C2 et/2 . From the
initial condition,
first 1
1
′
t + 1 et/2 + C2 et/2 ,
we once more have y(0) = C1 · 0 · 1 + C2 · 1 = C2 = 2. And since y = C1
2
2
1
the second initial condition now gives y ′ (0) = C1 + C2 = C1 + 1 = b, whence C1 = b − 1. The
2
solution to the given initial value problem is thus y = (b − 1)tet/2 + 2et/2 .
The sign of the coefficient of the dominating tet/2 term will determine whether y → ∞ or −∞
as t → ∞. In particular, if b − 1 > 0, or b > 1, the solution will grow positively, whereas if b < 1,
the solution will grow negatively. The critical value is thus b = 1. (Note that if b = 1, then the
solution is y = 2et/2 , which goes to ∞ as t → ∞.) 23. We first define y = v(t)y1 = v(t)t2 , where v(t) is a twice differentiable function on t > 0
which we will define later. Now, by the Product Rule, we have y ′ = v ′ (t)t2 + 2v(t)t and y ′′ =
v ′′ (t)t2 + 4v ′ (t)t + 2v(t). Substituting into the original differential equation gives
t2 y ′′ − 4ty ′ + 6y = t2 (v ′′ (t)t2 + 4v ′ (t)t + 2v(t)) − 4t(v ′ (t)t2 + 2v(t)t) + 6(v(t)t2 )
= t4 v ′′ (t) + (4t3 − 4t3 )v ′ (t) + (2t2 − 8t2 + 6t2 )v(t) = t4 v ′′ (t) = 0.
Since we are restricting ourselves to the interval t > 0, this means we must have v ′′ (t) = 0, or
v(t) = C1 t + C2 . It follows that y = C1 t3 + C2 t2 is a solution to the original differential equation,
and since the second term is a constant multiple of y1 , the desired second solution is y 2 = t3 . (It
is easy to verify both that t3 is a solution to the differential equation and that t2 and t3 form a
fundamental set of solutions for t > 0, as W (t2 , t3 ) = t4 6= 0 for all t > 0.)