Calculus IV - HW 3
Due 7/13
Section 3.1
1. Give the general solution to the following differential equations:
(a) y 00 − 25y = 0
Solution: The characteristic equation is r2 − 25 = (r − 5)(r + 5). It follows
that the general solution is
y = c1 e5t + c2 e−5t .
(b) y 00 + 6y 0 + 8y = 0
Solution: The characteristic equation is r2 +6r +8 = (r +2)(r +4). It follows
that the general solution is
y = c1 e−2t + c2 e−4t .
(c) y 00 + 12y 0 = 0.
Solution: The characteristic equation is r2 + 12r = r(r2 + 12). It follows that
the general solution is
y = c1 e0t + c2 e−12t = c1 + c2 e−12t .
2. For each equation from problem 1; assume the constants in your general solution, φ(t),
are both positive, and find lim φ(t) and lim φ(t).
t→∞
t→−∞
1
Solution:
(a) lim φ(t) = lim c1 e5t + c2 e−5t = ∞
t→∞
t→∞
lim φ(t) = lim c1 e5t + c2 e−5t = ∞
t→−∞
t→−∞
(b) lim φ(t) = lim y = c1 e−2t + c2 e−4t = 0
t→∞
t→∞
lim φ(t) = lim y = c1 e−2t + c2 e−4t = ∞
t→−∞
t→−∞
(c) lim φ(t) = lim y = c1 + c2 e−12t = c1
t→∞
t→∞
lim φ(t) = lim y = c1 + c2 e−12t = ∞
t→−∞
t→−∞
Section 3.2
3. (7,11,12 from B&D) For each of the following differential equations with initial conditions, determine the longest interval in which a solution is guaranteed to exist.
(a) ty 00 + 3y = t;
y(1) = 1, y 0 (1) = 2
Solution: In order to apply the existence and uniqueness theorem for second
order linear differential equations we need to put the equation in the form
y 00 + p(t)y 0 + q(t)y = g(t).
Some manipulation of our equation yields
3
y 00 + y = 1.
t
Our task now is to find the largest open interval containing the t value at
which our initial condition is specified on which both functions q(t) = 3t and
g(t) = 1 are continuous. g is continuous everywhere and q is continuous for
t 6= 0. This gives us the option of two intervals: (−∞, 0) and (0, ∞). Choosing
the one containing 1 (where the initial condition is specified) leaves us with a
final answer of
(0, ∞).
(b) (x − 3)y 00 + xy 0 + (ln |x|)y = 0;
y(1) = 0, y 0 (1) = 1
Page 2
Solution: Some manipulation of our equation yields
y 00 +
x 0 ln |x|
y +
y = 0.
x−3
x−3
Now we need to find the largest open interval containing the x value at which
x
and
our initial condition is specified on which both functions p(t) = x−3
ln |x|
g(t) = x−3 are continuous. p is continuous for x 6= 3 and q is continuous
for x 6= 0, 3. This leaves us with three possible intervals: (−∞, 0), (0, 3), and
(3, ∞). Choosing the one containing 1 (where the initial condition is specified)
leaves us with a final answer of
(0, 3).
(c) (x − 2)y 00 + y 0 + (x − 2) tan(x)y = 0;
y(3) = 1, y 0 (3) = 2
Solution: Some manipulation of our equation yields
y 00 +
1
y 0 + tan(x)y = 0.
x−2
Now we need to find the largest open interval containing the x value at which
1
our initial condition is specified on which both functions p(t) = x−2
and
g(t) = tan(x) are continuous. p is continuous for x 6= 2 and q is continuous
for x 6= π2 + kπ with k an integer. This leaves us with infinitely many possible
intervals. We only need to find the one containing 3. The final answer is
(2,
3π
).
2
4. Optional (41 from B&D) The equation P (x)y 00 +Q(x)y 0 +R(x)y = 0 is said to be exact
if it can be written in the form [p(x)y 0 ]0 +[f (x)y]0 = 0, where f (x) is to be determined in
terms of P (x), Q(x), and R(x). The latter equation can be integrated once immediately,
resulting in a first order linear equation. By equating the coefficients of the preceding
equations and then eliminating f (x), show that a necessary condition for exactness is
P 00 (x) − Q0 (x) + R(x) = 0 (meaning exactness implies P 00 (x) − Q0 (x) + R(x) = 0). It
can be shown that this is also a sufficient condition.
Solution:
Page 3
Proof. To prove this we first assume that P (x)y 00 + Q(x)y 0 + R(x)y = 0 is exact.
In particular, this means that
P (x)y 00 + Q(x)y 0 + R(x)y = [p(x)y 0 ]0 + [f (x)y]0 .
Expanding the right hand side of the above equation we have
[p(x)y 0 ]0 + [f (x)y]0 = p0 (x)y 0 + p(x)y 00 + f 0 (x)y + f (x)y 0 .
Rearranging we have
p(x)y 00 + [p0 (x) + f (x)]y 0 + f 0 (x)y.
It follows that
P (x)y 00 + Q(x)y 0 + R(x)y = p(x)y 00 + [p0 (x) + f (x)]y 0 + f 0 (x)y.
Then by equating coefficients we have
P (x) = p(x),
Q(x) = p0 (x) + f (x),
and
R(x) = f 0 (x).
In particular,
Q0 (x) = p00 (x) + f 0 (x) = P 00 (x) + R(x)
and
P 00 (x) + R(x) − Q0 (x) = P 00 (x) − Q0 (x) + R(x) = 0.
5. (Problems 4,6 from B&D) Find the Wronskian of each of the following pairs of functions.
(a) x,
xex
(b) cos2 (θ), 1 + cos(2θ)
Solution:
(a)
x
x
xe
x
x
2 x
W = x = x(1 + x)e − xe = x e
1 (1 + x)e
Page 4
(b)
2
2
cos
(θ)
1
+
cos(2θ)
cos
(θ)
1
+
cos(2θ)
=
W = −2 cos(θ) sin(θ) −2 sin(2θ)
−2 cos(θ) sin(θ) −4 cos(θ) sin(θ)
= −4 cos3 (θ) sin(θ) − 1 + cos(2θ) − 2 cos(θ) sin(θ)
= −4 cos3 (θ) sin(θ) − 2 cos2 (θ) − 2 cos(θ) sin(θ)
= −4 cos3 (θ) sin(θ) + 4 cos3 (θ) sin(θ)
=0
Note that we used the identities sin(2θ) = 2 sin(θ) cos(θ) and
.
cos2 (θ) = 12 + cos(2θ)
2
6. (Problems 8,9,10 from B&D) For each of the following differential equations with initial
conditions, determine the longest interval on which the given initial value problem is
certain to have a unique twice differentiable solutions.
(a) (t − 1)y 00 − 3ty 0 + 4y = sin(t)
(b) y 00 + cos(t)y 0 + 3 ln |t|y = 0
y(−2) = 2,
y 0 (−2) = 1
y(2) = 3 y 0 (2) = 1
Solution:
(a) We need to divide through by t − 1 to put the equation into the standard
form. Then we are left with
y 00 −
3t 0
4
sin(t)
y +
y=
t−1
t−1
t−1
3t
4
. Then thinking of p(t) = − t−1
, q(t) = t−1
, and g(t) = sin(t)
, we see that
t−1
p(t), q(t), and g(t) are all continuous everywhere except at t = 1. Thus our
two intervals to choose between are I1 : −∞ < t < 1 and I2 : 1 < t < ∞.
However we need this interval to contain t0 which in this case is t0 = −2.
Therefore we want to pick I1 .
(b) Here p(t) = cos(t), q(t) = 3 ln |t|, and g(t) = 0. We see that p(t) and g(t) are
continuous everywhere. However q(t) has a discontinuity at t = 0. Thus we
must choose between the intervals I1 = −∞ < t < 0 and I2 = 0 < t < ∞.
Now our t0 = 2 so we choose I2 .
Page 5
7. (Problems (a) and (b) are 24 and 27 from B&D) For each of the following, verify
that the functions y1 and y2 are solutions of the given differential equation and state
whether or not they constitute a fundamental set of solutions.
(a) y 00 + 4y = 0;
y1 (t) = cos(2t) y2 (t) = sin(2t)
00
(b) 1 − x cot(x) y − xy 0 + y = 0
y1 (x) = x, y2 (x) = sin(x)
(c) t2 y 00 − 2ty 0 + 2y = 0
y1 (t) = t2 y2 (t) = −t2
Solution:
(a) We note that y100 = −4 cos(2t) and y200 = −4 sin(2t) so
y100 + 4y1 = −4 cos(2t) + 4 cos(2t) = 0
y200 + 4y2 = −4 sin(2t) + 4 sin(2t) = 0
so y1 and y2 are both solutions.
We have shown that two solutions constitute a fundamental set of solutions
if their Wronskian is not identically zero, meaning it is not zero everywhere.
Thus we compute
cos(2t)
sin(2t) = 2 cos2 (2t) + 2 sin2 (2t) = 2
W =
−2 sin(2t) 2 cos(2t)
Since W 6= 0, we have that y1 and y2 constitute a fundamental set.
(b) Note that y10 = 1 and y100 = 0 while y20 = cos(x) and y200 = − sin(x). Plugging
these in we get
1 − x cot(x) y100 − xy10 + y1 = 1 − x cot(x)(0) − x(1) + x
=0−x+x=0
and
1 − x cot(x) y200 − xy20 + y2 = 1 − x cot(x)(− sin(x)) − x cos(x) + sin(x)
= − sin(x) + x cos(x) − x cos(x) + sin(x) = 0
All that remains is to check their Wronkskian.
x sin(x) = x cos(x) − sin(x)
W =
1 cos(x)
which in particular is nonzero at x = 2π. Thus y1 and y2 form a fundamental
set of solutions.
Page 6
(c) We will demonstrate that y1 is a solution then note that since y2 is a constant
multiple of y1 , it must also be a solution. Note y1 = t2 , y10 = 2t, and y100 = 2.
Plugging this in we have
t2 y100 − 2ty10 + 2y1 = t2 (2) − 2t(2t) + 2t2 = 0
We then compute the Wronskian.
2
t −t2 = −2t3 + 2t3 = 0
W = 2t −2t
so y1 and y2 do not form a fundamental set.
Section 3.3
8. Use Euler’s Formula to write the following in the form a + ib.
(a) e2−3i
(b) eiπ
Solution:
(a) We use Euler’s Formula to calculate
e2−3i = e2 e−3i = e2 cos(−3) + i sin(−3) = e2 cos(3) − i sin(3)
= e2 cos(3) − ie2 sin(3).
(b) Similarly
eiπ = cos(π) + i sin(π)) = −1 + i(0) = −1.
9. (Problems 7,9,12 from B&D) Find the general solution of the following differential
equations.
(a) y 00 − 2y 0 + 2y = 0
(b) y 00 + 2y 0 − 8y = 0
(c) 4y 00 + 9y = 0
Page 7
Solution:
(a) We get the characteristic equation r2 −
√ 2r + 2 = 0. Using the quadratic
2± 4−4(2)
formula, we find that the roots are
= 1 ± i. However, we have
2
shown that if our roots are complex, namely r = λ ± iµ, then our general
equation is just y = c1 eλt sin µt + c2 eλt cos(µt). Here λ = 1 and µ = 1 so we
have the general solution
y = c1 et sin(t) + c2 et cos(t).
(b) Our characteristic equation is r2 + 2r − 8 = 0 which can be factored to
(r − 2)(r + 4) so it has two real roots, r = 2 and r = −4. Thus its general
solution is just
y = c1 e2t + c2 e−4t .
(c) The characteristic equation is 4r2 +9 = 0 which implies that the solutions are
r = ± 23 i. Following the example of part(a), we get that the general solution
is
3
3
t + c2 sin
t .
y = c1 cos
2
2
(Note that λ = 0 in this example.)
10. (Problem 17 and 22 from B&D) Find the solutions of the following initial value problems. Sketch the graph of each solution and describe its behavior as t → ∞.
(a) y 00 + 4y = 0,
y(0) = 0,
(b) y 00 + 2y 0 + 2y = 0
y 0 (0) = 1
y( π4 ) = 2,
y 0 ( π4 ) = −2
Solution:
(a) Note that we showed in Problem (3a) that y = c1 sin(2t) + c2 cos(2t) is a
general solution. Thus it only remains to determine the appropriate c1 and
c2 . We get
y(0) = c1 sin(0) + c2 cos(0) = c2 = 0
y 0 (0) = 2c1 cos(0) − 2c2 sin(0) = 2c1 = 1
so c1 = 12 and the solution of the initial value problem is y = 21 sin(2t). Thus
y(t) oscillates with fixed amplitude and frequency as t → ∞.
Page 8
(b) We find the characteristic
equation r2 + 2r + 2 = 0. We use the quadratic
√
formula to get r = −2± 24−4·2 = −1 ± i. Thus our general solution is
y = c1 e−t sin(t) + c2 e−t cos(t).
We calculate
y 0 (t) = −c1 e−t sin(t) − c2 e−t cos(t) + c1 e−t cos(t) − c2 e−t sin(t)
= (−c1 − c2 )e−t sin(t) + (c1 − c2 )e−t cos(t).
Then
y
π 4
− π4
π − π4
π = c1 e sin
+ c2 e cos
4
4
√
π
2
=
(c1 + c2 )e− 4 = 2
2
√ π
=⇒ c1 + c2 = 2 2e 4 .
Furthermore
π π π π
π
y0
= (−c1 − c2 )e− 4 sin
+ (c1 − c2 )e− 4 cos
4
4
4
√
√
2 −π
2 −π
=
e 4 (−c1 − c2 ) + (c1 − c2 ) =
e 4 (−2c2 )
2
2
√
π
= −c2 2e− 4 = −2
√ π
=⇒ c2 = 2e 4 .
√ π
√ π
However we showed earlier that c1 + c2 = 2 2e 4 =⇒ c1 = 2 2e 4 − c2 so
√ π √ π √ π
c1 = 2 2e 4 − 2e 4 = 2e 4 .
Therefore the solution to the initial value problem is
√ π
√ π
y = 2e 4 e−t sin(t) + 2e 4 −t cos(t)
√ π
√ π
= 2e 4 −t sin(t) + 2e 4 e−t cos(t).
Thus y(t) has a decaying oscillation as t → ∞.
Page 9
(This graph has been drastically vertically stretched in a non-uniform manner
to better illustrate the general trend of the graph.)
11. Optional (Problem 28 of Section 3.3 from B&D) In this problem we outline a different
derivation of Euler’s formula.
(a) Show that y1 (t) = cos(t) and y2 (t) = sin(t) are a fundamental set of solutions of
y 00 + y = 0; that is, show they are solutions and that their Wronskian is not zero.
(b) Show (formally) that y = eit is also a solution. Why does this imply that
eit = c1 cos(t) + c2 sin(t)
(1)
for some constants c1 and c2 ?
(c) Set t = 0 in equation (1) to show that c1 = 1.
(d) Differentiate equation (1) and set t = 0 to show that c2 = i. (It may be helpful to
remember that dtd ert = rert for any r ∈ C.) Use the values of c1 and c2 to arrive
at Euler’s Formula.
Solution:
(a) Since y100 (t) = − cos(t) and y200 (t) = − sin(t), it follows immediately that these
are solutions. We check and see
cos(t) sin(t) = cos2 (t) + sin2 (t) = 1
W =
− sin(t) cos(t)
so since W 6= 0, it follows that y1 and y2 are a fundamental set of solutions.
Page 10
(b) If y = eit , then y 0 = ieit and y 00 = i2 eit = −eit . Thus y 00 + y = −eit + eit = 0
meaning y = eit is a solution. However, by the definition of y1 and y2 being
a fundamental set, it is possible to write any other solution as c1 y1 + c2 y2 for
some c1 , c2 . Thus there exist c1 and c2 such that eit = c1 cos(t) + c2 sin(t).
(c) Setting t = 0 we get
ei0 = c1 cos(0) + c2 sin(t)
=⇒ 1 = c1
.
(d) Taking the derivative of equation (1), we get
ieit = −c1 sin(t) + c2 cos(t)
which if we plug in t = 0 gives i = c2 . Thus we find
eit = cos(t) + i sin(t)
which is Euler’s Formula.
Section 3.4
12. Find the general solution of the following differential equation.
2.25y 00 + 3y 0 + y = 0
Solution: The characteristic equation is
2
2.25r + 3r + 1 =
3
r+1
2
2
=0
which has the repeated root r = − 23 . It follows that the general solution is
y = c1 e−2t/3 + c2 te−2t/3 .
13. Find the solution of the following differential equation with initial conditions.
4y 00 − 20y 0 + 25y = 0;
y(1) = e5/2 , y 0 (1) =
Page 11
7e5/2
2
Solution: The characteristic equation is
4r2 − 20r + 25 = (2r − 5)2 = 0
which has the repeated root r = 52 . Hence the general solution is
y = c1 e5t/2 + c2 te5t/2 .
Plugging in the initial conditions we have
y(1) = (c1 + c2 )e5/2 = e5/2
implying that
c1 + c2 = 1,
and
0
y (1) =
(2)
5
7
7
c1 + c2 e5/2 = e5/2
2
2
2
implying that
5c1 + 7c2 = 7.
(3)
Solving the system of two equations comprised of (1) and (2) we have that c1 = 0
and c2 = 1. This leaves us with at final solution of
y = te5t/2 .
14. Suppose that we have the differential equation
ay 00 + by 0 + cy = 0.
(a) Write the characteristic equation and use the quadratic formula to find its roots
(in terms of a, b, and c).
(b) Recall that the discriminant is b2 − 4ac. Note that there are three possibilities:
i. b2 − 4ac > 0 implying there are two distinct real roots
ii. b2 − 4ac = 0 implying there is one repeated real root.
iii. b2 − 4ac < 0 implying there are two distinct complex roots.
For each of these cases, give a formula for the general solution that only involves
real-valued functions (meaning they don’t include imaginary numbers.)
Page 12
Solution:
(a) The characteristic equation is ar2 + br + c = 0 and it has roots
√
√
−b − b2 − 4ac
−b + b2 − 4ac
;
r2 =
.
r1 =
2a
2a
(b)
i. When b2 − 4ac > 0, we have two distinct real roots and we learned in
Section 3.1 that
√
y = c1 e
−b+
b2 −4ac
t
2a
√
+ c2 e
−b−
b2 −4ac
t
2a
is a general solution.
ii. When b2 − 4ac = 0, we have one repeated root and we saw in Section
3.4 that a general solution is given by
b
b
y = c1 e− 2a t + c2 te− 2a t .
iii. When b2 − 4ac < 0, we have two distinct complex roots. We showed in
Section 3.3 that a general solution is given by
√
√
b
b
4ac − b2
4ac − b2
− 2a
t
t
− 2a
t + c2 e
t
sin
cos
y = c1 e
2a
2a
15. Use your work in Problem 14 to write a general solution for each of the following
homogeneous differential equations.
(a) 9y 00 + 9y 0 − 4y = 0
(b) y 00 − 5y 0 + 6y = 0
(c) y 00 + 8y 0 + 16y = 0
Solution:
(a) Here a = 4, b = 0, and√ c = 9. Thus
b2 − 4ac = −4(4)(9) = −144 < 0. In
√
2
144
particular −b
= 0 and 4ac−b
= 2·4
= 12
= 32 . Thus our general solution is
2a
2a
8
3
3
y = c1 sin
t + c2 cos
t .
2
2
Page 13
(b) Here a = 1, b = −5, and c = 6. Thus b2 − 4ac = 25 − 4(1)(6) = 1 > 0. Then
have we have the general solution
√
y = c1 e
−b+
√
5+
= c1 e
b2 −4ac
t
2a
25−4(1)(6)
t
2
√
+ c2 e
−b−
√
5−
+ c2 e
b2 −4ac
t
2a
25−4(1)(6)
t
2
= c1 e3t + c2 e2t
(c) Here a = 1, b = 4, and 4. Thus b2 − 4ac = 42 − 4(1)4 = 0. Then have we
have the general solution
b
b
y = c1 e− 2a t + c2 te− 2a t = c1 e−2t + c2 te−2t
Section 3.5
16. Find the general solution of the following differential equation.
y 00 + 8y 0 + 25y = 40 cos(5t)
Solution: We begin by finding the solution to the associated homogeneous equation,
y 00 + 8y 0 + 25y = 0.
This has a characteristic equation of
r2 + 8r + 25 = 0
with roots
r = −4 ± 3i.
It follows that the general solution to the associated homogeneous equation is
e−4t (c1 cos(3t) + c2 sin(3t)).
Now we just need to find a particular solution to the nonhomogeneous case. We
do this with the method of undetermined coefficients. Let
y = A cos(5t) + B sin(5t)
and notice that
y 0 = −5A sin(5t) + 5B cos(5t),
Page 14
y 00 = −25A cos(5t) − 25B sin(5t).
Plugging into the equation we get
−25A cos(5t) − 25B sin(5t) + 8(−5A sin(5t) + 5B cos(5t)) + 25(A cos(5t) + B sin(5t))
= −40A sin(5t) + 40B cos(5t) = 40 cos(5t).
This implies that A = 0 and B = 1. Hence we have a particular solution of
y = sin(5t) and a general solution of
y = e−4t (c1 cos(3t) + c2 sin(3t)) + sin(5t).
17. Find the general solution of the following differential equations.
(a) 9y 00 + 12y 0 + 4y = 18et/3
Solution: We begin by finding the general solution to the associated homogeneous case. The characteristic equation is
9r2 + 12r + 4 = (3r + 2)2 = 0
which has the repeated root r = − 32 . It follows that the general solution to
the associated homogeneous case is
y = c1 e−2t/3 + c2 te−2t/3 .
We now find a general solution using the method of undetermined coefficients.
Let
y = Aet/3
and notice then that
A t/3
e ,
3
A
y 00 = et/3 .
9
Plugging into the equation we have
y0 =
Aet/3 + 4Aet/3 + 4Aet/3 = 9Aet/3 = 18et/3 .
It follows that A = 2 and we have a general solution of
y = c1 e−2t/3 + c2 te−2t/3 + 2et/3 .
Page 15
(b) 9y 00 + 12y 0 + 4y = 25 cos(t/3)
Solution: Since we already have the general solution to the associated homogeneous case, we only need to find a particular solution to the nonhomogeneous
case. Letting
y = A cos(t/3) + B sin(t/3)
we have
B
A
sin(t/3) + cos(t/3),
3
3
A
B
y 00 = − cos(t/3) − sin(t/3).
9
9
Plugging into the equation we have
y0 = −
−A cos(t/3)−B sin(t/3)−4A sin(t/3)+4B cos(t/3)+4A cos(t/3)+4B sin(t/3)
= (4B + 3A) cos(t/3) + (3B − 4A) sin(t/3) = 25 cos(t/3)
Equating coefficients we have the system of equations
4B + 3A = 25
3B − 4A = 0.
Solving this we have B = 4 and A = 3. It follows that we have a particular
solution of y = 3 cos(t/3) + 4 sin(t/3), and a general solution of
y = c1 e−2t/3 + c2 te−2t/3 + 3 cos(t/3) + 4 sin(t/3).
(c) 9y 00 + 12y 0 + 4y = 18et/3 + 25 cos(t/3)
Hint – Consider the principle of superposition.
Solution: Treating the left hand side of the equation as a differential operator, it follows that the general solution in this case is
y = c1 e−2t/3 + c2 te−2t/3 + 2et/3 + 3 cos(t/3) + 4 sin(t/3).
18. Solve the following differential equation using the two methods described, and then
solve for the values of the constants.
1
y 00 − 2y 0 = 6t2 − 6t − 12;
y(1) = , y 0 (1) = 1
2
0
(a) i. Let v = y and solve the resulting first order linear equation.
Page 16
Solution: Making the substitution we have the equation
v 0 − 2v = 6t2 − 6t − 12
which we can solve as a first order linear equation. Using the integrating
factor µ = e−2t we have
Z
µv = e−2t (6t2 − 6t − 12) dt
Z
1 −2t 2
1
= − e (6t − 6t − 12) +
e−2t (12t − 6) dt
2
2
Z
1
1 −2t
1
1 −2t 2
−2t
− e (12t − 6) +
12e
dt
= − e (6t − 6t − 12) +
2
2
2
2
3
3
−2t
2
−2t
= e (−3t + 3t + 6) − e
−3t +
− e−2t + c1
2
2
= e−2t (−3t2 + 6) + c1 .
Dividing through by µ = e−2t we have
v = −3t2 + 6 + c1 e2t .
ii. Notice that y =
R
v dt and use this to find the general solution.
Solution: We have that
Z
Z
y = v dt = −3t2 + 6 + c1 e2t dt
= −t3 + 6t + c1 e2t + c2 .
(b)
i. Solve the homogeneous case using the roots of the characteristic equation.
Solution: The characteristic equation of the associated homogeneous
equation is
r2 − 2r = r(r − 2) = 0.
It follows that the general solution to the homogeneous equation is
y = c1 e2t + c2 .
Page 17
ii. Referencing your solution from part (a), make a guess for a general form of
a particular solution and use the method of undetermined coefficients to find
the general solution.
Solution: From the above it is clear that the a particular solution to this
differential equation is y = −t3 + 6t. This is a bit odd because, based on
the methods we have developed, we would expect the particular solution
to be a quadratic. The reason that we need a cubic is that there is no
y term on the left hand side of the equation – hence if we plugged in a
general quadratic polynomial we the resulting polynomial would be linear.
So, playing the game that the question is asking us to play, let
y = At3 + Bt2 + Ct
and notice then that
y 0 = 3At2 + 2Bt + C,
y 00 = 6At + 2B.
Plugging into the equation we have
6At + 2B − 2(3At2 + 2Bt + C) = −6At2 + (6A − 4B)t − 2C = 6t2 − 6t − 12
It follows that A = −1, B = 0, and C = 6. Hence we have the particular
solution
y = −t3 + 6t
as expected. Again we find the general solution of
y = −t3 + 6t + c1 e2t + c2 .
(c) Solve for the constants in your general solution using the given initial values.
Solution: Plugging in the initial values we have
y(1) = 5 + c1 e2 + c2 =
1
2
and
y 0 (1) = 3 + 2c1 e2 = 1.
It follows that c1 = −e−2t and c2 = − 27 . The solution is then
7
y = −t3 + 6t − e2t−2 − .
2
Page 18
19. (28 from B&D) Optional Determine the general solution of
00
2
y +λ y =
N
X
am sin(mπt),
m=1
where λ > 0 and λ 6= mπ for m = 1, . . . , N .
Solution: We begin by solving for the general solution of the associated homogeneous equation
y 00 + λ2 y = 0.
The characteristic equation is
r2 + λ2 = 0
which has roots r = ±iλ. It follows that the general solution to the homogeneous
equation is
y = c1 sin(λt) + c2 cos(λt).
Treating the left hand side of the differential equation as a linear operator, we will
find a particular solution to
y 00 + λ2 y = am sin(mπt)
and then sum over m. It turns out, in this case, that we only need to plug
Am sin(mπt) in to solve for a particular solution. We then have
y = Am sin(mπt),
y 00 = −m2 π 2 Am sin(mπt).
Plugging into our equation we have
−m2 π 2 Am sin(mπt) + λ2 Am sin(mπt) = am sin(mπt)
It follows that
Am = am [λ2 − m2 π 2 ]−1 .
Summing over m and adding in the general solution to the associated homogeneous
equation we have the general solution
y = c1 sin(λt) + c2 cos(λt) +
N
X
am [λ2 − m2 π 2 ]−1 sin(mπt).
m=1
20. Use the method of undetermined coefficients to find a particular solution to each of
the following nonhomogeneous differential equations.
Page 19
(a) 9y 00 + 9y 0 − 4y = 3e−t + sin(2t)
(b) y 00 − 5y 0 + 6y = t2 − 4t + 2
(c) y 00 + 4y 0 + 4y = 4et cos(t)
Solution:
(a) We will find a particular solution Y (t) to 4y 00 + 9y = 3e−t + sin(2t) by finding
particular solutions Y1 (t) and Y2 (t) to 4y 00 + 9y = 3e−t and 4y 00 + 9y = sin(2t)
respectively.
For Y1 (t), our guess is Aet . Then
Y1 (t) = Ae−t
Y10 (t) = −Ae−t
Y100 (t) = Ae−t
Thus plugging these into the differential equation gives
4Ae−t + 9Ae−t = 3e−t
=⇒ 13A = 3
3
=⇒ A =
13
so we have Y1 =
3 −t
e .
13
For Y2 our guess is A sin(2t) + B cos(2t). Thus
Y1 (t) = A sin(2t) + B cos(2t)
Y10 (t) = 2A cos(t) − 2B sin(t)
Y100 (t) = −4A sin(t) + −4B cos(t)
Plugging these in we have
4 − 4A sin(t) + −4B cos(t) + 9 A sin(t) + B cos(t) = sin(2t)
=⇒ −7A sin(t) = sin(2t) and − 7B cos(2t) = 0 cos(2t)
=⇒ −7A = 1 and − 7B = 0
1
=⇒ A = − and B = 0.
7
1
Therefore we have Y2 = − 7 sin(2t).
Thus Y (t) = Y1 (t) + Y2 (t) =
3 −t
e
13
− 17 sin(2t).
Page 20
(b) Our guess is Y (t) = At2 + Bt + C so we have
Y (t) = At2 + Bt + C
Y 0 (t) = 2At + B
Y 00 (t) = 2A
Plugging this in we get
2A − 5 2At + B + 6 At2 + Bt + C = t2 − 4t + 2
=⇒ 6At2 + (6B − 10A)t + (2A − 5B + 6C) = t2 − 4t + 2
=⇒ 6A = 1,
6B − 10A = −4,
1
=⇒ A =
6
2A − 5B + 6C = 2
Plugging this in we get
6B − 10A = −4
1
=⇒ 6B − 10( ) = −4
6
14
=⇒ B = −
36
Finally we get
1
14
2A − 5B + 6C = 2( ) − 5( ) + 6C = 2
6
36
5
108
1 2
14
5
Thus the solution is Y (t) = 6 t − 36 t − 108 .
=⇒ C = −
(c) Here our guess is Y (t) = Aet cos(t) + Bet sin(t) so we have
Y (t) = Aet cos(t) + Bet sin(t)
Y 0 (t) = (A + B)et cos(t) + (B − A)et sin(t)
Y 00 (t) = (A + B) + (B − A) et cos(t) + (B − A) − (A + B) et sin(t)
= 2Bet cos(t) − 2Aet sin(t)
Thus plugging in we get
t
t
t
t
2Be cos(t) − 2Ae sin(t) + 4 (A + B)e cos(t) + (B − A)e sin(t)
Page 21
+4 Aet cos(t) + Bet sin(t) = 4et cos(t)
Collecting terms, we get
2B + 4A + 4B + 4A et cos(t) = 8A + 6B et cos(t) = 4et cos(t)
− 2A + 4B − 4A + 4B et sin(t) = − 6A + 8B et sin(t) = 0et sin(t)
Thus −6A + 8B = 0 which implies B = 34 A. Substituting this in we get
8
4 = 8A + 18
A which implies 50
A = 4 so A = 16
= 25
and it follows that
4
4
50
24
6
3
B = 4 A = 100 = 25 .
8 t
e
25
Therefore the solution is Y (t) =
cos(t) +
6 t
e
25
sin(t).
21. Use your work in Problem 14 and Problem 20 to find a general solution to the following
nonhomogeneous differential equations.
(a) 9y 00 + 9y 0 − 4y = 3e−t + sin(2t)
(b) y 00 − 5y 0 + 6y = t2 − 4t + 2
(c) y 00 + 4y 0 + 4y = 4et cos(t)
Solution: Recall that we get the general solution to a nonhomogeneous equation
by adding together the general solution to the homogeneous solution and a particular solution to the nonhomogenenous equation. However, for each of these, we
found the general solution to the homogeneous equation in Problem 2 and a particular solution the nonhomogeneous equation in Problem 3. Thus all that remains
is to add them together.
(a) y(t) = c1 sin
3
t
2
+ c2 cos
3
t
2
(b) y(t) = c1 e3t + c2 e2t + 61 t2 −
(c) y(t) = c1 e−2t + c2 te−2t +
+
14
t
36
8 t
e
25
3 −t
e
13
−
− 17 sin(2t)
5
108
cos(t) +
6 t
e
25
sin(t)
Section 3.6
22. Parts i-iv are optional. Parts v-vii are mandatory. This question explores the
general method of variation of parameters. Suppose we have a second order linear
differential equation in standard form,
y 00 + p(t)y 0 + q(t)y = g(t),
Page 22
and the general solution to the homogeneous case,
y = c1 y 1 + c2 y 2 .
We attempt to find the general solution of the non-homogeneous case by the following
means:
(a) replacing the constants, c1 and c2 , with functions of t, u1 (t) and u2 (t)
(b) stipulating that u01 y1 + u02 y2 = 0
(c) plugging the resulting expression into the non-homogeneous differential equation
(d) and solving for u1 and u2 .
Preform the steps below to derive a general solution method.
i. Let y = u1 y1 + u2 y2 and compute expressions for y 0 and y 00 bearing in mind the
condition given in (b) above. Hint – u001 and u002 should not appear in any of these
expressions.
ii. Using your results from i, plug y into the non-homogeneous differential equation
given at the onset of this problem.
iii. Rearrange the left hand side of the equation you found in part ii so that the
equation is in the form
u1 [· · · ] + u2 [· · · ] + u01 [· · · ] + u02 [· · · ] = g(t)
where each instance of [· · · ] represents an expression in y, y 0 , y 00 , p and q; all four
of these will be different expressions.
iv. Explain why the expressions from iii that are multiplied by u1 and u2 are zero and
notice that this leaves us with the equation
u01 y10 + u02 y20 = g(t).
v. Using the equations from iv and (b) we have the following system of linear (not
differential) equations for u01 and u02 .
u01 y1
u01 y10
+ u02 y2
+ u02 y20
= 0
= g
(1)
(2)
Solve the system using the following steps:
A.
B.
C.
D.
Multiply equation (1) by −y20 and equation (2)
Add the resulting equations together and solve
Multiply equation (1) by −y10 and equation (2)
Add the resulting equations together and solve
Page 23
by y2
for u01
by y1
for u02 .
vi. Recall that that the Wronskian of y1 and y2 is
y 1 y 2 = y1 y20 − y10 y2
W (y1 , y2 ) = 0
y1 y20 and show that your result from v. can be rewritten as
u01 = −
y2 g
,
W (y1 , y2 )
u02 =
y1 g
.
W (y1 , y2 )
vii. Find an expression for the general solution of the non-homogeneous equation (the
one we’ve been trying to solve this whole time) in terms of the indefinite integrals of
the expressions for u01 and u02 from part vi. Argue that this is the general solution.
Hint – remember the constants of integration!
Solution:
i. If y = u1 y1 + u2 y2 , the product rules gives us that
y 0 = u01 y1 + u1 y10 + u02 y2 + u2 y20 = (u1 y10 + u2 y20 ) + (u01 y1 + u02 y2 )
=⇒ y 0 = u1 y10 + u2 y20
with the last equivalence coming from the stipulation that u01 y1 + u02 y2 = 0.
Then
y 00 = u1 y100 + u01 y10 + u2 y200 + u02 y20
ii. We then get
(u1 y100 + u01 y10 + u2 y200 + u02 y20 ) + p(t)(u1 y10 + u2 y20 ) + q(t)(u1 y1 + u2 y2 ) = g(t)
iii. Separating out terms, we get
u1 (y100 + p(t)y10 + q(t)y1 ) + u2 (y200 + p(t)y20 + q(t)y2 ) + u01 y10 + u2 y20 = g(t).
iv. Both y1 and y2 are solutions of y 00 + p(t)y 0 + q(t)y = g(t) so
y100 + p(t)y10 + q(t)y1 = 0 = y200 + p(t)y20 + q(t)y2
and the equation becomes
u01 y10 + u02 y20 = g(t).
Page 24
v.
A. After multiplying, we get the system of equations
−u01 y1 y20 − u02 y2 y20 = 0
u01 y10 y2 + u02 y2 y20 = y2 g
B. Adding these together, we get
u01 (y10 y2 − y1 y20 ) = y2 g
Thus
u01 =
y2 g
− y1 y20
y10 y2
C. After multiplying, we get the system of equations
−u01 y1 y10 − u02 y2 y10 = 0
u01 y10 y1 + u02 y20 y1 = y1 g
D. Adding these together, we get
u02 (y1 y20 − y10 y2 ) = y1 g
Solving for u02 we get
u02 =
y1 g
.
− y10 y2
y1 y20
vi. This is immediate from the definition of the Wronskian.
vii. Recall that we set out to find y = u1 y1 + u2 y2 . To get u1 and u2 , we simply
integrate u01 and u02 . Thus we have
Z
Z
y1 (s)g(s)
y2 (s)g(s)
y = −y1
ds + y2
ds.
W (y1 , y2 )(s)
W (y1 , y2 )(s)
Remember that each of the indefinite integrals will have constant of integration, c1 and c2 respectively so the equation could equally well be written
Z
Z
y2 (s)g(s)
y1 (s)g(s)
y = −y1
ds + y2
ds + c1 y1 + c2 y2
W (y1 , y2 )(s)
W (y1 , y2 )(s)
R 2 (s)g(s)
R y1 (s)g(s)
ds+y
ds
which is the general solution since Y1 (t) = −y1 Wy(y
2
,y
)(s)
W (y1 ,y2 )(s)
1 2
is a particular solution to the nonhomogeneous equation and Y2 (t) = c1 y1 +
c2 y2 is a general solution to the homogeneous equation.
Page 25
23. (5 from B&D) Find the general solution of the following differential equation.
y 00 + y = tan(t),
t ∈ (0, π/2)
Solution: First we must find a fundamental set of solutions to the homogeneous
equation y 00 + y = 0. We see the characteristic equation is r2 + 1 = 0 so r = ±i
implying the general solution is y = c1 sin(t) + c2 cos(t). Since this is a general
solution, any choice of c1 and c2 gives a solution. Choosing c1 = 0 and c2 = 1 gives
a solution y1 = cos(t). Choosing c1 = 1 and c2 = 0 gives a solution y2 = sin(t).
Since we have two solutions, it only remains to see if they form a fundamental set.
Taking the Wronskian, we get
cos(t) sin(t) = cos2 (t) + sin2 (t) = 1 6= 0
W = − sin(t) cos(t)
so y1 and y2 do form a fundamental set. Note that since tan(t) is not continuous
everywhere, we must use the definite integral form of the theorem (see Theorem
3.6.1 on p. 188 of B&D). We already found that W (y1 , y2 ) = 1. Note that t0 is
strictly inside the interval (0, π/2). We get
Zt
y = − cos(t)
Zt
sin(s) tan(s)ds + sin(t) +
t0
cos(s) tan(s)ds + c1 cos(t) + c2 sin(t).
t0
We now need to find these integrals.
Zt
Zt
sin(s) tan(s)ds =
t0
sin2 (s)
ds =
cos(s)
Zt
1 − cos2 (s)
ds =
cos(s)
t0
t0
Zt
sec(s) − cos(s)ds
t0
t
= ln | sec s + tan s| − sin(s) = ln | sec t + tan t| − sin(t) − ln | sec t0 + tan t0 | + sin(t0 ).
t0
Thus noting that − ln | sec t0 + tan t0 | + sin(t0 ) is just some constant C1 , we have
that
Zt
sin(s) tan(s)ds = ln | sec t + tan t| − sin(t) + C1 .
t0
Next, we calculate
Zt
Zt
cos(s) tan(s)ds =
t0
t
sin(s)ds = − cos(s) = − cos(t) − cos(t0 )
t0
t0
Page 26
= − cos(t) + C2
Putting this all back together, we get
y = − cos(t) ln | sec t+tan t|−sin(t)+C1 +sin(t) −cos(t)+C2 +c1 cos(t)+c2 sin(t)
= − cos(t) ln | sec t + tan t| + sin(t) cos(t) − C1 cos(t) − sin(t) cos(t) + C2 sin(t)
+c1 cos(t) + c2 sin(t)
= − cos(t) ln | sec t + tan t| + (C2 + c2 ) sin(t) + (c1 − C1 ) cos(t).
= − cos(t) ln | sec t + tan t| + D1 cos(t) + D2 sin(t).
24. (10 from B&D) Find the general solution of the following differential equation.
y 00 − 2y 0 + y =
et
1 + t2
Solution: As before, we begin by finding the general solution to the homogeneous
equation
y 00 − 2y 0 + y = 0
which has characteristic equation r2 −2r +1 = 0 which has the repeated root r = 1.
Thus the general solution is y = c1 et + c2 tet . Thus we could take our two solutions
to be y1 = et and y2 = tet . We check the Wronskian and see
t
e
tet = (t + 1)e2t − te2t = e2t 6= 0
W = t
e (t + 1)et so y1 and y2 do in fact form a fundamental set.
Since each part is continuous on the whole real line, we may simply use the indefinite
integral form to get
Z
Z t et
et
tet · 1+t
e · 1+t2
2
t
t
y = −e
dt + te
ds + c1 et + c2 tet
2t
e
e2t
Z
Z
t
1
t
t
= −e
dt
+
te
dt + c1 et + c2 tet
1 + t2
1 + t2
t 1
2
t
= −e
ln |1 + t | + C1 + te arctan t + C2 + c1 et + c2 tet
2
1 t
= − e ln |1 + t2 | + tet arctan(t) + (c1 − C − 1)et + (c2 + C2 )tet
2
1
= − et ln |1 + t2 | + tet arctan(t) + D1 et + D2 tet .
2
Page 27
25. Find the general solution to the following differential equation.
y 00 + y = sec4 (t),
Hint
R
t ∈ (0, π/2)
sec3 (t) dt = 12 sec(t) tan(t) + 12 ln | sec(t) + tan(t)| + c.
Solution: Looking at Problem 2, we get that y1 = cos(t) and y2 = sin(t) form a
fundamental set of solutions with W (y1 , y2 ) = 1. Since sec(t) has discontinuities,
we use the definite integral form of variation of parameters to get that
Zt
y = − cos(t)
sin(s) sec4 (s)ds + sin(t)
t0
Zt
cos(s) sec4 (s)ds + c1 sin(t) + c2 cos(t)
t0
We calculate the integrals below:
Zt
Zt
4
sin(s) sec (s)ds =
t0
tan(s) sec3 (s)ds.
t0
We now use u-substitution with u = sec(s) and du = sec(s) tan(s)ds to write the
integral as
Z
1
1
u2 du = u3 = sec3 (s)
3
3
Thus we have
Zt
− cos(t)
t0
t 1 3 sin(s) sec (s)ds = − cos(t) · sec (s)
3
t0
4
1
1
= − cos(t) sec3 (t) + cos(t) sec3 (t0 )
3
3
1
= − sec2 (t) + C1 cos(t)
3
For the other integral we have
Zt
4
Zt
cos(s) sec (s)ds =
t0
sec3 (s)ds
t0
t
1
1
= sec(s) tan(s) + ln | sec(s) + tan(s)|
2
2
t0
Page 28
1
1
sec(t) tan(t) + ln | sec(t) + tan(t)|
2
2
1
1
− sec(t0 ) tan(t0 ) + ln | sec(t0 ) + tan(t0 )|
2
2
1
1
= sec(t) tan(t) + ln | sec(t) + tan(t)| + C2
2
2
=
Thus
Zt
sin(t)
cos(s) sec4 (s)ds
t0
=
1
1
sin(t) sec(t) tan(t) + sin(t) ln | sec(t) + tan(t)| + C2 sin(t)
2
2
1
1
= tan2 (t) + sin(t) ln | sec(t) + tan(t)| + C2 sin(t)
2
2
Putting this all together we get
1
1
1
y = − sec2 (t) + tan2 (t) + sin(t) ln | sec(t) + tan(t)| + C1 cos(t) + C2 sin(t)
3
2
2
Page 29