Math 315 - Homework #3 Solutions
Note: B&D refers to our course textbook written by Boyce and DiPrima.
Section 3.2
1. In each of Problems (a) and (b), determine the longest interval in which the given initial value problem
is certain to have a unique twice-differentiable solution. Do not attempt to find the solution.
(a) y 00 + (cot t)y = et ,
y(π/2) = 0, y 0 (π/2) = 1
Solution: Note that the given equation is written in standard form. The function q(t) = cot t is
continuous for all t 6= kπ where k is any integer. The function et is continuous on R. Therefore,
by Theorem 3.2.1, the largest interval containing the initial point t0 = π/2 such that the initial
value problem is guaranteed to have a unique twice-differentiable solution is 0 < t < π.
(b) y 00 + tan(t − π/2) y 0 + 3 ln(t − 2)y = 5,
y(3) = 0, y 0 (3) = 2
Solution: Note that the given equation is written in standard form. The function p(t) =
tan(t − π/2) is continuous for all t 6= kπ where k is any integer. The function q(t) = ln(t − 2) is
continuous for all t > 2. Therefore, by Theorem 3.2.1, the largest interval containing the initial
point t0 = 3 such that the initial value problem is guaranteed to have a unique twice-differentiable
solution is 2 < t < π.
2. (B&D Problem 22) Find the fundamental set of solutions specified by Theorem 3.2.5 for the following
differential equation and initial point.
y 00 + y 0 − 2y = 0,
t0 = 0
Solution: Let’s first determine the general solution of the differential equation. We solve the corresponding characteristic equation:
r2 + r − 2 = 0
=⇒
(r − 1)(r + 2) = 0
=⇒
r1 = 1, r2 = −2
Therefore, the general solution is
y(t) = c1 et + c2 e−2t .
Note that y1 (t) = et and y2 (t) = e−2t form a fundamental set of solutions. However, they do not
satisfy the prescribed initial conditions specified by Theorem 3.2.5 at t0 = 0. Therefore, they are not
the fundamental solutions that we want. We seek two fundamental solutions, y3 (t) and y4 (t), of the
differential equation such that the following initial conditions are satisfied:
For y3 :
y3 (0) = 1, y30 (0) = 0
For y4 :
y4 (0) = 0, y40 (0) = 1
Let’s determine y3 . Since y3 is a solution, we have
y3 (t) = c1 et + c2 e−2t
=⇒
1
y30 (t) = c1 et − 2c2 e−2t
Applying the initial conditions for y3 yields c1 = 2/3 and c2 = 1/3. Therefore,
2 t 1 −2t
.
e + e
3
3
y3 (t) =
Similarly, for y4 , we have
y4 (t) = c1 et + c2 e−2t
=⇒
y40 (t) = c1 et − 2c2 e−2t
Applying the initial conditions for y4 yields c1 = 1/3 and c2 = −1/3. Therefore,
y4 (t) =
1 t 1 −2t
e − e
.
3
3
It is easy to show that W (y3 , y4 )(t) = e−t , which is positive (and therefore nonzero) for all t. Hence,
y3 and y4 form a fundamental set of solutions. Note that W (y3 , y4 )(0) = 1 as expected.
3. (B&D Problem 26) Verify that the functions y1 and y2 are solutions of the given differential equation.
Do they constitute a fundamental set of solutions?
x2 y 00 − x(x + 2)y 0 + (x + 2)y = 0,
x > 0;
y1 (x) = x, y2 (x) = xex
Solution: Plugging y1 into the equation yields
x2 · (0) − x(x + 2)(1) + (x + 2)x = −x2 − 2x + x2 + 2x = 0
Therefore, y1 is a solution. Plugging y2 into the equation yields
x2 (xex + 2ex ) − x(x + 2)(xex + ex ) + (x + 2)xex = (x + 2)ex x2 − x(x + 1) + x = 0
|
{z
}
0
Therefore, y2 is a solution. The Wronskian of y1 and y2 is
x
xex = x(xex + ex ) − xex (1) = x2 ex .
W (y1 , y2 )(x) = 1 xex + ex Since W (y1 , y2 )(x) 6= 0 for all x > 0, y1 and y2 form a fundamental set of solutions.
Section 3.3
4. (B&D Problems 12, 14, and 16) In each of Problems (a), (b), and (c), find the general solution of the
given differential equation.
(a) 4y 00 + 9y = 0
Solution: We solve the associated characteristic equation:
4r2 + 9 = 0
=⇒
3
r=± i
2
Therefore, the general solution is
3
3
y(t) = c1 cos t + c2 sin t
2
2
2
(b) 9y 00 + 9y 0 − 4y = 0
Solution: We solve the associated characteristic equation:
9r2 + 9r − 4 = 0
=⇒
(3r − 1)(3r + 4) = 0
=⇒
r1 =
1
4
, r2 = −
3
3
Therefore, the general solution is
y(t) = c1 et/3 + c2 e−4t/3
(c) y 00 + 4y 0 + 6.25y = 0
Solution: We solve the associated characteristic equation:
p
√
−4 ± 42 − 4(1)(6.25)
−4 ± −9
3
2
r + 4r + 6.25 = 0 =⇒ r =
=
=2± i
2(1)
2
2
Therefore, the general solution is
3
3
y(t) = e−2t c1 cos t + c2 sin t
2
2
5. (B&D Problem 18) Find the solution of the given initial value problem. Sketch the graph of the
solution and describe its behavior for increasing t.
y 00 + 4y 0 + 5y = 0,
y(0) = 1, y 0 (0) = 0
Solution: We solve the associated characteristic equation:
r2 + 4r + 5 = 0
=⇒
r=
√
−4 ± −4
= −2 ± i
2
Therefore, the general solution is
y(t) = e−2t [c1 cos t + c2 sin t]
Applying the first initial condition yields
y(0) = c1 = 1
Therefore,
y(t) = e−2t [cos t + c2 sin t]
which implies that
y 0 (t) = −2e−2t [cos t + c2 sin t] + e−2t [− sin t + c2 cos t]
Applying the second initial condition yields
y 0 (0) = −2 + c2 = 0
=⇒
c2 = 2
Therefore,
y(t) = e−2t [cos t + 2 sin t]
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The solution is a decaying oscillation. This is expected since the amplitude is controlled by the exponential term which tends to zero as t → ∞.
Section 3.4
6. (B&D Problems 6 and 8) In each of Problems (a) and (b), find the general solution of the given
differential equation.
(a) y 00 − 6y 0 + 9y = 0
Solution: We solve the associated characteristic equation:
r2 − 6r + 9 = 0
=⇒
(r − 3)2 = 0
=⇒
r=3
(repeated real root)
Therefore, the general solution of the differential equation is
y(t) = c1 e3t + c2 te3t
(b) 16y 00 + 24y 0 + 9y = 0
Solution: We solve the associated characteristic equation:
16r2 + 24r + 9 = 0
=⇒
(4r + 3)2 = 0
=⇒
r=−
3
4
(repeated real root)
Therefore, the general solution of the differential equation is
y(t) = c1 e−3t/4 + c2 te−3t/4
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7. (B&D Problem 27) Use the method of reduction of order to find a second solution of the following
differential equation.
xy 00 − y 0 + 4x3 y = 0,
y1 (x) = sin x2
x > 0;
Solution: The differential equation in standard form is
y 00 −
1 0
y + 4x2 y = 0.
x
A second solution has the form y2 (x) = v(x)y1 (x) where
R
e− p(x) dx
dx.
y12 (x)
Z
v(x) =
Here, p(x) = −1/x. Hence
R
Z
e 1/x dx
dx
sin2 (x2 )
Z
x
dx
sin (x2 )
v(x) =
=
2
Z
=
x csc2 (x2 ) dx
1
= − cot(x2 )
2
Therefore,
1
1
y2 (x) = v(x)y1 (x) = − cot(x2 ) · sin(x2 ) = − cos(x2 )
2
2
Note that the coefficient −1/2 can be removed from y2 (x) since y2 (x) is eventually multiplied by an
arbitrary constant c2 in the general solution. Therefore, we can take our second solution to be
y2 (x) = cos(x2 ) .
Section 3.5
8. (B&D Problem 8) Find the general solution of the following differential equation.
y 00 + 2y 0 + y = 2e−t
Solution: We first find the complementary solution, yc (t). The associated characteristic equation is
r2 + 2r + 1 = 0
=⇒
(r + 1)2 = 0
=⇒
r = −1
(repeated root)
Therefore, yc (t) = c1 e−t + c2 te−t . Next, we find a particular solution yp (t) of the nonhomogeneous
equation by the method of undetermined coefficients. A natural first guess for a particular solution
would be Ae−t . However, since this guess is already present in yc (t), we know that it is a solution
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of the homogeneous equation. The form Ate−t will not work either since it is also present in yc (t).
Therefore, we will try yp (t) = At2 e−t . We plug our guess back into the differential equation to obtain
yp00 + 2yp0 + yp = 2Ae−t − 4Ate−t + At2 e−t + 2(2Ate−t − At2 e−t ) + At2 e−t
= (A − 2A + A)t2 e−t + (−4A + 4A)te−t + 2Ae−t
= 2Ae−t = |{z}
2e−t
g(t)
Therefore, we must have A = 1, and so yp (t) = t2 e−t . It follows that the general solution is
y(t) = c1 e−t + c2 te−t + t2 e−t .
9. Find the solution of the following initial value problem.
y 00 + 4y = 4 sin 2t,
y(0) = 2, y 0 (0) = −1
Solution: We first find the complementary solution, yc (t). The associated characteristic equation is
r2 + 4 = 0
=⇒
r = ±2i
Therefore, yc (t) = c1 cos 2t + c2 sin 2t. Next, we look for a particular solution, yp (t), of the nonhomogeneous equation by the method of undetermined coefficients. A natural first guess for a particular
solution is A cos 2t + B sin 2t. However, since this function is already present in yc (t), it is a solution
of the associated homogeneous equation. Hence, we will assume a particular solution of the form
yp (t) = At cos 2t + Bt sin 2t. Computing the first and second derivatives of yp yields
yp0 (t) = A cos 2t − 2At sin 2t + B sin 2t + 2Bt cos 2t
yp00 (t) = −4A sin 2t + 2B cos 2t − 4At cos 2t − 4Bt sin 2t
It follows that
yp00 + 4yp = −4A sin 2t + 2B cos 2t − 4At cos 2t − 4Bt sin 2t + 4(At cos 2t + Bt sin 2t)
= −4A sin 2t + 2B cos 2t = |4 sin
{z 2t}
g(t)
After equating the cofficients of the sine terms, we see that A = −1. Since g(t) has no cosine term, we
have B = 0. Therefore, yp (t) = −t cos 2t. Thus, the general solution is
y(t) = c1 cos 2t + c2 sin 2t − t cos 2t.
Differentiating with respect to t yields
y 0 (t) = −2c1 sin 2t + 2c2 cos 2t − cos 2t + 2t sin 2t.
Applying the initial conditions yields
y(0) = c1 = 2
y 0 (0) = 2c2 − 1 = −1
=⇒
c2 = 0
Therefore, the solution of the initial value problem is
y(t) = 2 cos 2t − t cos 2t .
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