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Sample Final Questions: Solutions
Math 21B, Winter 2013
1. Evaluate the following integrals:
Z
Z
1
tan−1 x
(a)
dy;
(b)
dx;
y − y3
x2
Z
(c)
1
p
√ dx.
1+ x
Solution.
• (a) We use partial fractions:
1
−1
= (
3
y−y
y y − 1)(1 + y))
A
B
C
+
+
.
=
y
y−1 y+1
Putting the partial fractions expansion over a common denominator,
we get
A
B
C
(A + B + C) y 2 + (B − C) y − A
+
+
=
.
y
y−1 y+1
y(y − 1)(y + 1)
• Hence, we require that
B − C = 0,
A + B + C = 0,
A = 1.
The solution is
A = 1,
1
B=− ,
2
1
C=− .
2
• Thus,
Z
Z 1
1
1
−
−
dy
y 2(y − 1) 2(y + 1)
1
1
= ln |y| − ln |y − 1| − ln |y + 1| + C
2
22 1 y =
ln
+C
2 y2 − 1 1
dy =
y − y3
• (b) To simplify the integral, we use integration by parts to differentiate
the inverse tangent. Setting
u = tan−1 x,
du =
1
dx
x2
1
v=−
x
dv =
1
dx,
1 + x2
in the formula for integration by parts, we get
Z
Z
1
tan−1 x
tan−1 x
+
dx.
dx = −
2
x
x
x (1 + x2 )
• The partial fractions expansion of the new integrand is
1
A Bx + C
=
+ 2
2
x (1 + x )
x
x +1
(A + B) x2 + Cx + A
=
x (x2 + 1)
Hence, we choose
A + B = 0,
C = 0,
A = 1,
meaning that A = 1, B = −1, C = 0.
• Thus,
Z
1
x
−
dx
x x2 + 1
1
= ln |x| − ln x2 + 1 + C.
2
1
dx =
x (1 + x2 )
Z • It follows that
Z
tan−1 x
tan−1 x
1
2
+
ln
|x|
−
dx
=
−
ln
x
+
1
+ C.
x2
x
2
• (c) We use the substitution u =
√
x, with
1
du = √ dx.
2 x
It follows that dx = 2u du, so
Z
Z
1
2u
p
√
du.
√ dx =
1+u
1+ x
• Next, we use the substitution
v=
with
√
1 + u,
1
dv = √
du.
2 1+u
It follows that u = v 2 − 1 and du = 2v dv, so
Z
Z
2u
2 (v 2 − 1)
√
· 2v dv
du =
v
1+u
Z
= 4
v 2 − 1 dv
=
4 3
v − 4v + C.
3
• Combining these results, and expressing the answer in terms of x, we
get
Z
4 3
1
p
v − 4v + C
√ dx =
3
1+ x
4
=
(1 + u)3/2 − 4 (1 + u)1/2 + C
3
√ 3/2
√ 1/2
4
=
1+ x
−4 1+ x
+ C.
3
• We would
get
p
√ the same result if we combined the substitutions and set
v = 1 + x at the start.
Remark. The integrals can be evaluated by other (equally correct) methods.
The answers, however, must always agree up to an additive constant.
2. Evaluate the following integrals:
Z
Z
2
(a) x ln x dx;
(b) x cos3 x2 sin x2 dx;
(c)
Z √
36 − 4t2 dt.
Solution.
• (a) We use integration by parts to differentiate the logarithm. Setting
u = ln x,
1
du = dx,
x
dv = x2 dx
1
v = x3
3
in the formula for integration by parts, we get
Z
Z
1 3
1 3 1
2
x ln x dx =
x ln x −
x · dx
3
3
x
Z
1
1 3
x ln x −
x2 dx
=
3
3
1
1 3
x ln x − x3 + C.
=
3
9
• (b) First, we use the substitution u = x2 with du = 2x dx, which gives
Z
Z
1
3
2
2
x cos x sin x dx =
cos3 u sin u du.
2
• Next, we use the substitution v = cos u with dv = − sin u du, which
gives
Z
Z
1
1
3
cos u sin u du = −
v 3 dv
2
2
1
= − v 4 + C.
8
• Expressing this answer in terms of x, we get
Z
1
x cos3 x2 sin x2 dx = − cos4 x2 + C.
8
• (c) Using the trigonometric substitution
t = 3 sin θ,
dt = 3 cos θ dθ,
we get
Z √
36 −
4t2
dt = 2
Z √
Z
9−
t2
dt = 18
cos2 θ dθ.
• Using the double angle formula, we have
Z
Z
9
2
18 cos θ dθ = 9 (1 + cos 2θ) dθ = 9θ + sin 2θ + C.
2
• We have θ = sin−1 (t/3) and, by the double angle formula for sine,
r
p
2t
t2
2t √
2
sin 2θ = 2 sin θ cos θ = 2 sin θ 1 − sin θ =
1− =
9 − t2 .
3
9
9
• It follows that
Z √
36 −
4t2
dt = 9 sin
−1
√
t
+ t 9 − t2 + C.
3
3. Let R be the region of the (x, y)-plane bounded by the curves
y = ln x,
x = e2 .
y = 1,
(a) Express the area of R as an integral with respect to x.
(b) Express the area of R as an integral with respect to y.
(c) Evaluate the integrals in (a) and (b). Check that you get the same answer.
(d) Find the volume of the solid obtained by revolving R about the the y-axis.
Solution.
• (a) The region R consists of the points (x, y) such that
e ≤ x ≤ e2 ,
1 ≤ y ≤ ln x.
Thus, splitting R into vertical rectangles with of height (ln x − 1) and
width dx, we may express the area A of R as
Z e2
A=
(ln x − 1) dx.
e
(b) Since y = ln x implies that x = ey , the region R consists of the
points (x, y) such that
1 ≤ y ≤ 2,
ey ≤ x ≤ e2 .
Thus, splitting R into horizontal rectangles with of width (2 − ey ) and
height dy, we may express the area A of R as
Z 2
A=
e2 − ey dy.
1
• (c) For the integral in (a), using the standard integral of ln x, we get
Z e2
(ln x − 1) dx
A =
e
=
=
=
=
2
[x ln x − x − x]ee
e2 ln e2 − 2e2 − (e ln e − 2e)
2e2 − 2e2 − (e − 2e)
e.
Alternatively, you can use integration by parts.
• For the integral in (b), we get
Z 2
A =
e2 − ey dy
1
2
2
= e y − ey 1
= 2e2 − e2 − e2 − e
= e.
Thus, we get the same answer both ways.
• (d) We can compute the volume of the solid of revolution in two different ways, depending on how we split up R. Either method would be
fine (though, as is often the case, the ‘washer’ method is simpler here).
• Method 1: Shells. The rotation of a vertical strip about the y-axis gives
a ‘shell’ of height h = ln x − 1, radius r = x, width dx, and volume
dV = 2πrh dx.
Hence, the volume V of the resulting solid of revolution is given by
Z e2
V = 2π
x (ln x − 1) dx.
e
• We evaluate this integral by use of integration by parts, with
u = ln x − 1,
1
du = dx,
x
dv = x dx
1
v = x2 ,
2
which gives
e2
Z e2
1 2
1 2 1
= 2π x (ln x − 1) − 2π
x · dx
2
2
x
e
e
Z e2
= π e4 ln e2 − 1 − e2 (ln e − 1) − π
x dx
V
e
1 2
x
2
π 4
=
e + e2 .
2
= πe4 − π
e2
e
• Method 2: Washers. The rotation of a horizontal strip about the y-axis
gives a ‘washer’ of inner radius R = ey , outer radius r = e2 , height dy,
and volume
dV = π r2 − R2 dy.
Hence, the volume V of the solid of revolution is given by
Z 2
e4 − e2y dy
V = π
1
2
1 2y
4
= π e y− e
2
1
π 4
2
e +e .
=
2
4. Express the length of the curve with parametric equation
x=
t
,
1 + t3
y=
t2
1 + t3
(0 ≤ t ≤ 1)
as an integral with respect to t. (You don’t need to evaluate the integral.)
Solution.
• The element of arclength is
s
ds =
dx
dt
2
+
dy
dt
2
dt.
• We have
dx
1 · (1 + t3 ) − t · 3t2
=
dt
(1 + t3 )2
1 − 2t3
=
,
(1 + t3 )2
2t · (1 + t3 ) − t2 · 3t2
dy
=
dt
(1 + t3 )2
2t − t4
.
=
(1 + t3 )2
• Hence
dx
dt
2
+
dy
dt
2
2
2
(1 − 2t3 ) + (2t − t4 )
(1 + t3 )4
1 + 4t2 − 4t3 − 4t5 + 4t6 + t8
=
(1 + t3 )4
=
• The length L of the curve is given by
Z 1s
1 + 4t2 − 4t3 − 4t5 + 4t6 + t8
L=
dt.
(1 + t3 )4
0
5. The curve y = x3 , with 0 ≤ x ≤ 1, is rotated about the x-axis.
(a) Express the surface area of the resulting surface of revolution as an integral with respect to x.
(b) Compute the surface area by evaluating this integral.
(c) Compute the volume of the corresponding solid of revolution.
Solution.
• (a) The surface area A is given by
Z
A = 2πr ds
where r = y is the radius of the surface of revolution, and ds is the
element of arclength on the curve.
• We have
s
ds =
=
1+
q
dy
dx
2
dx
1 + (3x2 )2 dx.
• The surface area is therefore given by the integral
Z 1 √
A = 2π
x3 1 + 9x4 dx.
0
• (b) Using the substitution u = 1 + 9x4 , with du = 36x3 dx, we get
Z
2π 10 √
A =
u du
36 1
10
π 2 3/2
=
u
18 3
1
π
3/2
=
10 − 1 .
27
• (c) The volume V of the solid of revolution about the x-axis is
Z 1
πy 2 dx
V =
Z0 1
πx6 dx
=
0
1
1 7
= π x
7
0
π
=
.
7
6. Let bxc denote the floor function (that is, the greatest integer less than or
equal to x e.g. b3/2c = 1, b14/5c = 2). Is this function Riemann integrable
on the interval 0 ≤ x ≤ 3? If so, evaluate
Z 3
bxc dx.
0
Solution.
• The function is Riemann integrable, since it is piecewise continuous
with a finite number of jump discontinuities (at x = 0, 1, 2, 3).
• Using the additive property of the integral and the fact that

 0 if 0 ≤ x < 1
1 if 1 ≤ x < 2
bxc =

2 if 2 ≤ x < 3
we get
Z
3
Z
1
bxc dx =
0
Z
2
bxc dx +
1
1
Z
0 · dx +
=
0
3
bxc dx
0
Z
Z
bxc dx
Z 23
2
1 · dx
1
2 · dx
2
= 0+1+2
= 3.
• Note that the values of bxc at the endpoints of these integrals do not
affect them, since changing the values of a function at finitely many
points does not change its integral.
7. Suppose that the velocity v(t) of a particle at time t is given by
√
v(t) = t t + 1.
(a) Compute the distance traveled by the particle for 0 ≤ t ≤ 3.
(b) What is the average velocity of the particle over that time interval?
Solution.
• (a) The distance traveled L is equal to the integral of the velocity with
respect to time,
Z 3
√
L=
t t + 1 dt.
0
• Using the substitution u =
√
t + 1, with
1
dt,
du = √
2 t+1
t = u2 − 1
and dt = 2u du, we get
2
Z
u2 − 1 u · 2u du
L =
1
Z
2
u4 − u2 du
1
2
1 5 1 3
= 2 u − u
5
3
1
1 1
32 8
− −
−
= 2
5
3
5 3
116
=
.
15
= 2
• (b) The average velocity v¯ is given by
Z
1 3
v¯ =
v(t) dt
3 0
1 116
=
·
3 15
116
=
.
45
8. Suppose that an empty reservoir has the shape of a circular cone of radius
a at its top and depth h, where the top of the reservoir is at a height 2h above
sea level and the bottom (the vertex of the cone) is at height h. Calculate
the work done against gravity required to fill the reservoir by pumping water
from sea level into the reservoir. (Assume that the water only has to be
pumped up to the current water level in the reservoir, not all the way up
to the top.) Express your answer in terms of a, h, and the weight density
w = ρg of water.
Solution.
• Let z be the height measure from above the bottom of the reservoir.
(It would be equally correct to use height y = z + h above sea level
instead.)
• The reservoir cross section at height z is a circle of radius r = az/h
(since the radius varies linearly from r = 0 at z = 0 to r = a at z = h).
• The volume dV of a ‘slice’ of the reservoir of thickness dz at height
z is A(z) dz where A = πr2 is the cross-sectional area. The mass dm
of water required required to fill this ‘slice’ is ρA(z) dz, where ρ is the
mass-density of water.
• Water at height z has to be pumped a vertical distance (z + h) from
sea level. The work dW required to pump a mass dm of water is
g(z + h) dm, or
dW = ρg(z + h)A(z) dz = w(z + h)A(z) dz.
• Thus, the total work W required is
Z h
W =
w(z + h)A(z) dz
0
Z h
az 2
dz
w(z + h)π
=
h
0
Z
πwa2 h 3
2
z
+
hz
dz
=
h2 0
h
πwa2 1 4 1 3
=
z + hz
h2
4
3
0
7
=
πwa2 h2 .
12
9. As a simplified model of the earth’s atmosphere, consider a stationary
atmosphere in which the pressure p(z) and mass density ρ(z) vary with height
0 ≤ z < ∞. Let g be the gravitational acceleration (assumed constant).
(a) Explain briefly why
Z ∞
ρ(s) ds.
p(z) = g
z
(You can assume that the improper integral converges at infinity.)
(b) Explain briefly why it follows that
dp
= −gρ(z).
dz
(c) Suppose that ρ(z) = cp(z) for some constant c, meaning that the density
is proportional to the pressure, and p(0) = p0 where p0 is the atmospheric
pressure at the ground z = 0. Find p(z).
(d) At what height H is the pressure p(H) = p0 /e reduced by a factor e from
its value at the ground?
Solution.
• (a) In hydrostatic equilibrium, the pressure p(z) at some height z in a
fluid is equal to the gravitational force per unit horizontal area exerted
by the fluid above. This force is equal to g times the mass per unit
area of the fluid above, or
Z ∞
p(z) = g
ρ(s) ds.
z
• (b) This equation follows from the fundamental theorem of calculus,
and the fact that the integral changes sign when we exchange the limits:
Z ∞
Z z
d
d
dp
= g
ρ(s) ds = −g
ρ(s) ds = −gρ(z).
dz
dz z
dz ∞
• (c) Using the formula ρ = cp to eliminate the density ρ from the equation in (b), and the condition at z = 0, we get
dp
= −cgp,
dz
p(0) = p0 .
• The solution of this initial value problem is the exponential function
p(z) = p0 e−cgz .
• (d) The pressure decreases by a factor of e over a height
H=
1
.
cg
Remark. The pressure p, specific volume V , and absolute temperature T
of a unit mass of an ideal gas are related by the equation
pV = RT,
where R is a gas constant. (For dry air, R = 287 JKg−1 K−1 , where the
absolute temperature T is measured in Kelvins K.) The density of the gas
is given by ρ = 1/V . Thus, ρ is proportional to p if the temperature T is
constant, and in that case the constant of proportionality is given by
c=
1
.
RT
In an isothermal atmosphere in hydrostatic equilibrium, the density and pressure decrease exponentially with height,
p(z) = p0 e−z/H ,
ρ(z) = ρ0 e−z/H .
The height H over which they decrease by a factor of e is called the scale
height of the atmosphere, and it is given by
H=
RT
.
g
Although the earth’s atmosphere is not exactly isothermal, its temperature
in the first 70 km above the surface is approximately within 15% of 250 K.
The corresponding scale height is H = 7.2 km.
Humans require an air pressure of at least 0.5 atmospheres to survive on
a long-term basis. This sets an upper limit on habitable altitudes of about
5.8 km or 19, 000 ft e.g. for some communities in the Andes or Tibet. The
peak of Mt. Everest, at an altitude of 8.85 km or 29, 028 ft, is not habitable.
10. Define a function F (x) by
√
Z
x
2
e−t dt.
F (x) =
0
Compute F 0 (2).
Solution.
• Write F as a composition,
F (x) = G
where
Z
G(y) =
y
√ x ,
2
e−t dt.
0
• By the fundamental theorem of calculus,
2
G0 (y) = e−y .
• By the chain rule,
√ √ 0
x ·
x
√ 1
= G0
x · √
2 x
−x
e
= √ .
2 x
F 0 (x) = G0
• It follows that
1
F 0 (2) = √ .
2 2e2
11. Suppose that the function f (x) is defined on the interval a ≤ x ≤ b with
continuous first and second derivatives. If f (a) = f (b) = 0, show that
b
Z
f (x)f 00 (x) dx ≤ 0.
a
Hint. Integrate by parts.
Solution.
• We use the integration by parts formula
Z b
Z b
b
v du,
u dv = [uv]a −
a
a
where a, b are understood to be x-values, with
u = f,
dv = f 00 dx
du = f 0 dx,
v = f 0.
This gives
Z b
00
f (x) · f (x) dx = [f (x)f
0
b
(x)]a
Z
−
a
b
f 0 (x) · f 0 (x) dx
a
• The boundary terms vanish because
b
[f (x)f 0 (x)]a = f (b)f 0 (b) − f (a)f 0 (a) = 0 · f 0 (b) − 0 · f 0 (a) = 0.
• It follows that
Z
b
00
Z
f (x)f (x) dx = −
a
b
2
[f 0 (x)] dx.
a
• Since a square is nonnegative and the integral of a nonnegative function
is nonnegative, the right-hand side of this equation is less than or equal
to zero, so
Z
b
f (x)f 00 (x) dx ≤ 0,
a
12. Show that
2
Z
1≤
0
1
dx ≤ 2.
1 + sin4 (πx)
Solution.
• Since | sin θ| ≤ 1 for any −∞ < θ < ∞, we have
1 ≤ 1 + sin4 (πx) ≤ 2.
Therefore
1
1
≤
≤ 1.
2
1 + sin4 (πx)
• It follows from the positivity properties of the integral that
Z 2
Z 2
Z 2
1
1
1 dx,
dx ≤
dx ≤
4
0
0 1 + sin (πx)
0 2
which implies that
Z
1≤
0
2
1
dx ≤ 2.
1 + sin4 (πx)
13. (a) Use summation notation to write down the Riemann sum for
Z 2
x5 dx
0
that is obtained by partitioning the interval 0 ≤ x ≤ 2 into n equal subintervals and evaluating the integrand at the left endpoint of each subinterval.
Does this Riemann sum provide a lower bound or an upper bound of the
integral, or neither?
(b) Write down an integral that corresponds to the following limit of Riemann
sums:
!
n
1
2X
.
lim
n→∞
n k=1 1 + 2k/n
Solution.
• (a) We partition the interval 0 ≤ x ≤ 2 into n subintervals of equal
length
2
∆x =
n
with endpoints
{x0 , x1 , x2 , . . . , xk , . . . , xn }
where
xk =
2k
n
for k = 0, 1, 2, . . . , n.
• The left endpoint of the k th -interval, xk−1 P
≤ x ≤ xk , is xk−1 . Thus,
taking ck = xk−1 in the Riemann sum In = nk=1 f (ck ) ∆x, we get
In =
n
X
(xk−1 )5 ∆x
k=1
5
n X
2
2(k − 1)
=
n
n
k=1
n
26 X
=
(k − 1)5
n6 k=1
• We can write this sum in other equivalent forms by changing the summation index; for example,
n−1
26 X 5
In = 6
j .
n j=0
The j = 0 (or k = 1) term could be omitted from the sum because it
is 0.
• Since x5 is an increasing function, its value at the left endpoint of a
subinterval underestimates the function on the subinterval, and these
Riemann sums provide a lower bound for the value of the integral.
• (b) Partitioning the interval 0 ≤ x ≤ 2 into n subintervals with equal
length ∆x = 2/n, and endpoints xk = 2k/n for k = 0, 1, 2, . . . , n, we
may write
n
n
X
1
1
2X
=
∆x.
n k=1 1 + 2k/n k=1 1 + xk
These are Riemann sums (using the right endpoints of the subintervals)
for the integral
!
Z 2
n
2X
1
1
dx = lim
.
n→∞
n k=1 1 + 2k/n
0 1+x
Remark 1. The Riemann sums in (a) that use the right end point are
Jn =
n
X
k=1
(xk )5 ∆x =
n
26 X 5
k .
n6 k=1
Since x5 is increasing, these Riemann sums provide an upper bound for the
value of the integral.
Remark 2. The Riemann sums for x5 can be evaluated exactly by use of
the summation formula
n
X
k=1
k5 =
1 2
n (n + 1)2 2n2 + 2n − 1 .
12
This gives
26 n2 (n + 1)2 (2n2 + 2n − 1)
Jn =
.
12
n6
Dividing the powers of n in the denominator into the numerator and taking
the limit as n → ∞, we get
2 26
1
2
1
26
lim Jn =
lim 1 +
2+ − 2 = .
n→∞
12 n→∞
n
n n
6
The value of this limit (which defines the Riemann integral) agrees with the
value of the integral computed by use of the fundamental theorem of calculus:
2
Z 2
26
1 6
5
= .
x dx = x
6
6
0
0
Note that we obtain In by replacing n by n − 1 in the formula for Jn , which
gives
26 (n − 1)2 n2 (2n2 − 2n − 1)
In =
.
12
n6
The limit of the lower Riemann sums
2 26
1
1
2
26
lim In =
lim 1 −
2− − 2 = .
n→∞
12 n→∞
n
n n
6
is therefore the same as the limit of the upper Riemann sums, consistent with
the fact that x5 is Riemann integrable (because it is continuous).
14. (a) Show that the limit
(Z
Z
π/2−ε
lim+
tan x dx +
ε→0
0
)
π
tan x dx
π/2+ε
exists and evaluate it.
(b) Define the improper integral
Z
π
tan x dx.
0
Does it converge?
Solution.
• For 0 < b < π/2 and π/2 < a < π, we consider the integrals
Z b
Z π
Ib =
tan x dx,
Ja =
tan x dx.
0
a
These Riemann integrals exist since tan x is continuous on the intervals
[0, b] and [a, π].
• Using the formula for the antiderivative of tan x,
Z
tan x dx = − ln |cos x| + C,
or writing tan x = sin x/ cos x and using logarithmic integration, we
get
Ib = [− ln |cos x|]b0 = − ln |cos b| ,
Ja = [− ln |cos x|]πa = ln |cos a| .
• (a) From these equations, with b = π/2 − ε and a = π/2 + ε, it follows
that
Z π/2−ε
Z π
tan x dx +
tan x dx
0
π/2+ε
π
π
+ ε − ln cos
−ε = ln cos
2
2
cos (π/2 + ε) = ln cos (π/2 − ε) • As ε → 0+ , we have
π
π
− ε → 0,
cos
+ ε → 0.
cos
2
2
Therefore, we may use l’Hospitˆal’s rule to find the limit of the fraction
by differentiating the numerator and denominator with respect to ε.
We get
lim+
ε→0
− sin (π/2 + ε)
ε→0
sin (π/2 − ε)
− sin (π/2)
=
sin (π/2)
= −1
cos (π/2 + ε)
=
cos (π/2 − ε)
lim+
• Since the function ln |x| is continuous at x = 1, it follows that
cos (π/2 + ε) cos
(π/2
+
ε)
= ln lim
lim+ ln ε→0+ cos (π/2 − ε) ε→0
cos (π/2 − ε)
= ln |−1|
= 0.
Thus, the stated limit exists, and its value is
)
(Z
Z π
π/2−ε
tan x dx = 0.
lim+
tan x dx +
ε→0
π/2+ε
0
(b) The integrand tan x is discontinuous at x = π/2, and the improper
integral is defined by
Z π
Z b
Z π
tan x dx = lim −
tan x dx + lim +
tan x dx.
0
b→π/2
0
a→π/2
a
• From above, we see that
Z b
tan x dx = − ln |cos b| .
0
This does not converge as b → π/2− , since cos b → cos(π/2) = 0, and
− ln |x| diverges to ∞ as x → 0. Similarly,
Z π
tan x dx = ln |cos a|
a
diverges to −∞ as a → π/2+ .
• Thus, the improper integral does not converge.
Remark. Note that the improper integrals
Z
π/2
π
Z
tan x dx
tan x dx,
π/2
0
are required to exist separately. The limit
(Z
Z
π/2−ε
lim+
tan x dx +
ε→0
0
π
)
tan x dx
π/2+ε
is finite because of a cancelation between the two terms when b = π/2 − ε,
a = π/2 + ε are chosen symmetrically about the discontinuity at x = π/2.
This is an example of a singular integral called a principal value integral, but
it is not a convergent improper Riemann integral.
15. Solve the following initial value problem for y(t):
dy
= ty 3 ,
dt
y(0) = 1.
Sketch a graph of the solution. What happens to the solution?
Solution.
• Separating variables, we get
Z
Z
dy
=
y3
t dt
Evaluation of the integrals gives
−
1
1
= t2 + C
2
2y
2
• The solution for y is
y(t) = √
1
C − t2
(where C is a different arbitrary constant from before).
• The initial condition y(0) = 1 implies that
1
1= √
C
or C = 1, so the solution is
y(t) = √
1
.
1 − t2
• The solution goes to infinity as t → 1− and t → −1+ , so it is only
defined for −1 < t < 1.