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MA10192 Mathematics 1
Mark Opmeer
Solutions of Additional Exercises
Chapter 1
Functions and equations
1.1
Polynomials
1. x2 − 2x + 1 = 0.
This can be re-written as (x − 1)2 = 0, so x = 1 is the unique solution.
This can of course also be obtained using the quadratic formula.
2. x2 + 2x + 2 = 0.
This can be re-written as (x + 1)2 + 1 = 0. Since (x + 1)2 ≥ 0 for all x,
there are no solutions.
This can also be seen by the discriminant 22 − 4 × 1 × 2 = −4 being
negative.
For those of you who know complex numbers: there are of course complex
solutions and those are x = −1 ± i.
1.2
Exponentials and logarithms
1. ex e2x = 1.
Use the exponential rules to write this as e3x = 1. It follows that 3x = 0,
so that x = 0.
2. e2x − 2ex = 1.
With u = ex we obtain√the quadratic equation u2 − √
2u − 1 = 0. The
solutions are u = 1 ± 2. The equation ex = 1 − 2 does not have
√
solutions since the right-hand side √
is negative. The equation ex = 1 + 2
has as unique solution x = ln (1 + 2).
3. ex e3x = 2,
Using the rules for exponentials this is equivalent to e4x = 2. Taking
logarithms on both sides gives 4x = ln 2 which leads to x = ln42 .
1
4. ex = 2e−x .
Multiply both sides with ex to obtain e2x = 2. Take logarithms to obtain
2x = ln 2. So x = ln22 .
5. e2x − 2ex + 1 = 0,
This can be re-written as (ex − 1)2 = 0, so that ex = 1, so that x = 0 is
the unique solution.
6. ln x = 3.
Taking exponentials on both sides gives x = e3 .
7. ln |x − 2| + ln |x − 3| = 0.
2.5
-3
-2
-1
0
1
2
3
4
5
6
7
8
-2.5
-5
Figure 1.1: The graph of ln |x − 2| + ln |x − 3|
There are several ways of solving this.
One can use the logarithmic rule to obtain the equivalent ln |(x − 2)(x − 3)| =
0. Since the ln of something is zero if and only if that something is equal
to one, we obtain |(x − 2)(x − 3)| = 1. The absolute value of something
is equal to one if and only if it is one or minus one (don’t forget that
last possibility). This gives (x − 2)(x − 3) = 1 or (x − 2)(x − 3) = −1.
Writing this in standard form gives x2 − 5x + 5 = 0 and x2 − 5x + 7 = 0,
respectively. The first of these quadratic equations has as solutions
√
5± 5
x=
,
2
the second of these quadratic equations does not have solutions. The
solutions found are not equal to either 2 or 3 (the values of x for which
the original equation is meaningless), so both of these values are valid
solutions of the original equation.
One can also exponentiate the equation to obtain
eln |x−2|+ln |x−3| = 1
and use the exponential rules to obtain
eln |x−2| eln |x−3| = 1
2
and subsequently
|x − 2| |x − 3| = 1.
From here one can proceed as before.
A third way is to first multiply the equation by two
2 ln |x − 2| + 2 ln |x − 3| = 0
and use the logarithmic rules to obtain
ln |x − 2|2 + ln |x − 3|2 = 0
and using that the absolute value squared is just the thing itself squared:
ln (x − 2)2 + ln (x − 3)2 = 0.
Using the logarithmic rules again gives (x − 2)2 (x − 3)2 = 1, which by
taking square roots gives (x − 2)(x − 3) = 1 or (x − 2)(x − 3) = −1 and
then one can proceed as before.
8. ln (x + 1) + ln (x − 4) = 0.
3
2
1
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
-1
-2
-3
Figure 1.2: The graph of ln (x + 1) + ln (x − 4)
As in the previous exercise, one can use either the logarithmic or the
exponential rules. I will use the logarithmic rules here. Using these we
obtain the equivalent ln (x + 1)(x − 4) = 0, which gives (x + 1)(x − 4) = 1.
Writing this quadratic equation in standard form gives x2 − 3x − 5 = 0.
The solutions are
√
3 ± 29
x=
.
2
The original equation only makes sense if x > 4. The solution with a ‘+’
satisfies this inequality, but the solution with a minus does not. So only
√
3 + 29
x=
2
is a genuine solution of the original equation.
3
9. ln |x2 + 2x − 2| = 0,
Taking exponentials on both sides gives |x2 +2x−2| = 1. This means that
either x2 + 2x − 2 = 1 or x2 + 2x − 2 = −1. The first of these equations
factors as (x + 3)(x − 1) = 0, so that x = −3 or x = 1. The second
√
of these equations by the quadratic formula has solutions x = −1 ± 2.
Since there are absolute value signs in the original equation all of these
solutions are genuine solutions.
10. ln (x − 1) + ln (x + 2) = 0,
Note that this equation only makes sense if x > 1. Taking exponentials
leads to (x − 1)(x + 2) = 1. This can be re-written√as x2 + x − 3 = 0, which
by the quadratic formulas has solutions x = −1±2 13 . With the plus sign
this satisfies x > 1, but with the minus sign√it doesn’t. So the only valid
solution of the original equation is x = −1+2 13 .
1.3
Trigonometric functions
1.3.1
Definitions
1.3.2
Solving simple trigonometric equations
√
1. cos x =
3
2 .
One solution is π6 (the more complicated, but equal, expression arccos
will also do). It follows that all solutions are
x=
π
+ 2kπ,
6
x=−
√
3
2
π
+ 2nπ,
6
with k and n integers.
2. sin 2x =
√1 .
2
First substitute u = 2x to obtain the standard equation sin u = √12 . One
solution is π4 (the more complicated, but equal, expression arcsin √12 will
also do). It follows that all solutions are
u=
π
+ 2kπ,
4
u=
3π
+ 2nπ,
4
with k and n integers. Substituting u = 2x and solving for x gives
x=
π
+ kπ,
8
x=
3π
+ nπ,
8
with k and n integers.
3. tan 2x = 0.
Substitute u = 2x to obtain tan u = 0. This has solutions u = kπ with k
an integer. So 2x = kπ, so x = k π2 .
4
√
4. sin x = 23 ,
√
We have arcsin 23 = π3 . The other basic solution is π − π3 = 2π
3 . So the
+
2nπ
where
k
and
n
are
integers.
solutions are x = π3 + 2kπ and x = 2π
3
5. cos 2x = √12 .
Define u = 2x and rewrite this as cos u = √12 . We have arccos √12 = π4 .
The other basic solution is − π4 . So all solutions are given by u = π4 + 2kπ,
u = − π4 + 2nπ. Substituting u = 2x gives 2x = π4 + 2kπ, 2x = − π4 + 2nπ
so that we have x = π8 + kπ, x = − π8 + nπ where k and n are integers.
6. sin x = √12 ,
The basic solution is x = π4 , the second solution is then π − π4 = 3π
4 .
Because the sine function is 2π-periodic all solutions are given by x =
π
3π
4 + 2kπ, x = 4 + 2nπ, where k and n are integers.
7. cos x sin x = 0.
This implies that either cos x = 0 or sin x = 0. The solutions of the first
of these equations are x = π2 + kπ, the solutions of the second of these
equations are x = nπ. It follows that the solutions of the original equation
are x = π2 + m π2 , where m is an integer.
1.3.3
Inverse trigonometric functions
1. Solve sin x = 16 for x.
The basic solution is x = arcsin 16 (which cannot be simplified). The
second solution is then x = π − arcsin 16 . All solutions are then given by
x = arcsin 61 + 2kπ, x = π − arcsin 61 + 2nπ, where k and n are integers.
1.3.4
Polar coordinates
1. Write the point given in Cartesian coordinates by (1, −2) in polar coordinates.
p
√
The radius is r = 12 + (−2)2 = 5. Since the point is in the fourth
quadrant, the angle is θ = 2π − arctan 2.
2. Write the point given in polar coordinates as r = 2, θ = 3π
4 in Cartesian
coordinates.
√
√
We have x = r cos θ =√− √2, y = r sin θ = 2. So in Cartesian coordinates
the given point is (− 2, 2).
1.3.5
Harmonic form
1. 2 sin x + 4 cos x = 3.
Use the harmonic
√ form. We want to write (4, −2) in polar coordinates.
The radius is 20 and since the point is in the fourth quadrant, the angle
5
is 2π − arctan 12 . So we have 2 sin x + 4 cos x =
So the given equation is equivalent to
√
√
20 cos(x + 2π − arctan 12 ).
1
20 cos(x + 2π − arctan ) = 3.
2
The solutions satisfy
x+2π−arctan
1
1
3
3
= arccos √ +2kπ, x+2π−arctan = − arccos √ +2nπ,
2
2
20
20
and isolating x gives
1
3
1
3
x = −2π+arctan +arccos √ +2kπ, x = −2π+arctan −arccos √ +2nπ.
2
2
20
20
Here k and n are integers.
A second possibility is to use the sine harmonic form to solve this equation,
i.e. to write 2 sin x + 4 cos x = R sin(x + ϕ). What R and ϕ must be can
be derived similarly as was done in the lectures for the harmonic cosine
form. The calculations for solving the equation are also similar to those
when using the harmonic cosine form.
2. Solve sin x + 5 cos x = 3 for x.
To do this we want to write sin x + 5 cos x in harmonic form. √
That means
we want to write (5, −1) in polar coordinates. The radius is 26 and the
angle is θ = 2π − arctan 15 (since the point is in the fourth quadrant). So
√
sin x + 5 cos x = 26 cos(x + 2π − arctan 15 ).
√
√
So the equation to solve is 26 cos(x + 2π − arctan 51 ) = 3. Since 26 > 3
this has solutions. The solutions are
3
1
1
3
x = arccos √ −2π+arctan +2kπ, x = − arccos √ −2π+arctan +2nπ,
5
5
26
26
with k and n integers.
The alternative solutions presented in the solution of the previous exercise
are of course also possible here.
1.3.6
Solving more complicated trigonometric equations
2
1. sin x + cos2 2x = 1.
There are many ways to do this (depending on how you rewrite the equation using various trigonometric identities). The following is one possibility. Use that cos 2x = 1 − 2 sin2 x, so that, cos2 2x = 1 − 4 sin2 x + 4 sin4 x
to obtain the equivalent equation −3 sin2 x + 4 sin4 x = 0. This in turn is
equivalent to
−3
2
2
4 sin x
+ sin x = 0.
4
6
√
It follows that either sin x = 0 or sin x = ±
x = kπ, x =
3
2 .
This gives the solutions
2π
π
2π
π
+ 2mπ, x =
+ 2nπ, x = − + 2pπ, x = − + 2qπ,
3
3
3
3
where k, m, n, p, q are integers.
2. Solve cos2 x + sin2 2x = 1 for x.
Again, there are many different ways to solve this problem. Here is one.
Use the double angle formula for the sine (sin 2x = 2 cos x sin x) to obtain
cos2 x + 4 cos2 x sin2 x = 1. Use Pythagoras to write this as cos2 x +
4 cos2 x(1 − cos2 x) = 1, which simplifies to 4 cos4 x − 5 cos2 x + 1 = 0.
With u = cos2 x this is a quadratic equation, 4u2 − 5u + 1 = 0, with
solutions u = 14 and u = 1. So we want to solve cos2 x = 41 and cos2 x = 1.
This means that we want to solve cos x = 12 , cos x = − 12 , cos x = 1,
cos x = −1. The solutions are given by
x=
π
2π
2π
π
+2kπ, x = − +2nπ, x =
+2mπ, x = − +2pπ, x = qπ,
3
3
3
3
with k, n, m, p, q integers.
7
Chapter 2
Differentiation
2.1
Definition
1. Obtain the derivative of y = 2x3 from first principles (meaning: look at
the slope of chords etc.).
We compute
2x3 + 6x2 h + 6xh2 + 2h3 − 2x3
2(x + h)3 − 2x3
=
= 6x2 + 6xh + 2h2 .
h
h
Letting h → 0 gives that the derivative of 2x3 equals 6x2 .
2. Obtain the derivative of y = x4 from first principles (meaning: look at the
slope of chords etc.).
We have
(x + h)4 − x4
x4 + 4x3 h + 6x2 h2 + 4xh3 + h4 − x4
=
= 4x3 +6x2 h+4xh2 +h3 .
h
h
Setting h = 0 gives 4x3 for the derivative.
3. Obtain the derivative of y = x3 + x2 from first principles (meaning: look
at the slope of chords etc.).
We have
(x + h)3 + (x + h)2 − x3 − x2
x3 + 3x2 h + 3xh2 + h3 + x2 + 2xh + h2 − x3 − x2
=
h
h
2
2
= 3x + 3xh + h + 2x + h.
Putting h = 0 gives 3x2 + 2x, which is therefore the derivative of the given
function.
4. Find the tangent line of y = 3x5 in the point (1, 3) (you may use the rules
for differentiation).
The derivative of 3x5 is 15x4 , so the derivative in the given point is 15.
8
It follows that the tangent line is y = 15(x − 1) + 3 or equivalently y =
15x − 12.
5. Find the tangent line of y = x5 + x2 + 1 in the point (1, 3) (you may use
the rules for differentiation).
The derivative is 5x4 + 2x so that the slope of the tangent line, the value
of the derivative for x = 1, is 7. By the point-slope formula the tangent
line is y = 7(x − 1) + 3. This may alternatively be written as y = 7x − 4.
2.2
Notation
2.3
Derivatives of standard functions
2.4
Rules for differentiation
1. Calculate the derivative of x2 cos x.
Use the product rule to obtain for the derivative
−x2 sin x + 2x cos x.
2. Calculate the derivative of tan2 3x.
Use the chain rule with u = 3x, v = tan u, y = v 2 and
du
= 3,
dx
dv
1
=
,
du
cos2 u
dy
= 2v
dv
to obtain
6v
6 tan 3x
dy
=
=
.
2
dx
cos u
cos2 3x
This can alternatively be written as either of the following
6 sin 3x
,
cos3 3x
6 tan 3x + 6 tan3 3x.
3. Calculate the derivative of x12 .
Write this as x−2 and use the power-rule to obtain for the derivative
−2x−3 = −2
x3 or use the quotient rule to obtain the same.
4. Find the derivative of tan x + xe3x .
The derivative is
tan2 x + 1 + e3x + 3xe3x .
5. Calculate the derivative of x2 ex .
Using the product rule gives x2 ex + 2xex , which may be simplified as
(x2 + 2x)ex .
9
x
6. Calculate the derivative of x2tan
+x+2 .
By the quotient rule the derivative is
(x2 + x + 2)(tan2 x + 1) − (2x + 1) tan x
.
(x2 + x + 2)2
7. Compute the derivative of x sin x.
Using the product rule gives the answer sin x + x cos x.
8. Compute the derivative of cos(ex ).
Using the chain rule gives the answer −ex sin(ex ).
9. Compute the derivative of
ln x
x .
Using the quotient rule gives the answer
2.5
1
xx
−ln x
x2
=
1−ln x
x2 .
Derivatives of inverse functions
1. Calculate the derivative of arccos2 3x. This is the chain of three functions:
y(v) = v 2 , v(u) = arccos u, u(x) = 3x. The respective derivatives are:
dy
dv
√ −1 , du = 3. By the chain rule it follows that
dv = 2v, du =
1−u2 dx
dy
−6v
−6 arccos 3x
=√
= √
.
2
dx
1−u
1 − 9x2
2.6
Parametric differentiation
dy
at the point (1, 0) of the curve given by the equa1. Calculate the slope dx
t
2
tions x(t) = e + t , y(t) = tan t.
Use parametric differentiation to calculate
dy
1 + tan2 t
= t
.
dx
e + 2t
To find the t-value corresponding to the given point we have to solve
et + t2 = 1, tan t = 0. This second equation has as solutions all integer
multiples of π. The only one of those that gives a solution to the first
equation is zero. So t = 0. Substituting this gives that the slope in the
given point is 1.
dy
2. Calculate the slope dx
at the point (1, −1) of the curve given by the
equations x(t) = ln t + t4 , y(t) = sin πt + cos πt.
To find to which parameter value the given point corresponds we have to
solve ln t + t4 = 1, sin πt + cos πt = −1. For t = 1 the first equation holds.
Since ln t and t4 are increasing functions, so is their sum, so that this is
the only solution. We have sin π = 0 and cos π = −1; so t = 1 is indeed
10
the parameter value that corresponds to the given point. Use parametric
differentiation.
dy
ẏ
cos πt − sin πt
= =π
.
1
3
dx
ẋ
t + 4t
Substituting t = 1 gives that the slope in the given point is
−π
5 .
dy
3. Find the slope dx
at the point (x, y) = (0, 1) for the curve given parametrically by x(t) = t sin(t), y(t) = et cos t.
Using parametric differentiation we have
dy
=
dx
dy
dt
dx
dt
=
et cos t − et sin t
.
sin t + t cos t
The equation x = 0 gives t = kπ with k an integer. Substituting this in
the formula for y we see that only t = 0 gives the correct value y = 1. This
is because cos t = ±1 for such t, so that the formula for y in those points
becomes et = ±1, which only has t = 0 as a solution. So the unique
parameter value corresponding to the point (x, y) = (0, 1) is t = 0. It
follows that the slope in the given point is infinite (since in the expression
dy
the numerator is nonzero and the denominator is zero in t = 0)
for dx
2.7
Implicit differentiation
dy
1. Calculate the slope dx
at the point (0, 0) of the curve given by the equation
x
sin y + e cos 3y = 1.
Use implicit differentiation to obtain
dy
(cos y − 3ex sin 3y) + ex cos 3y = 0.
dx
Solving for
dy
dx
gives
ex cos 3y
dy
= x
.
dx
3e sin 3y − cos y
Substituting x = 0 and y = 0 gives for the slope at that point −1.
dy
2. Calculate the slope dx
at the point (0, 0) of the curve given by the equation
y
y sin x + e cos y = 1.
Use implicit differentiation to obtain
y 0 sin x + y cos x + y 0 ey cos y − y 0 ey sin y = 0.
Substituting x = 0 and y = 0 gives y 0 = 0. So the slope in the given point
is zero.
3. Compute the derivative
dy
dx
for the function given implicitly by
ey + ln(xy) + x2 = 0.
11
Implicit differentiation gives
dy y
dy
1
e + y+x
+ 2x = 0,
dx
dx xy
which can be re-written as
1 dy
1
ey +
+ + 2x = 0,
y dx x
and subsequently solved for
dy
dx
as
− 1 − 2x
dy
= xy 1 ,
dx
e +y
which can be simplified to
dy
−y − 2x2 y
=
.
dx
xyey + x
Re-writing ln (xy) = ln x + ln y before differentiating is probably a smarter
thing to do.
2.8
Maxima and minima
1. Determine the location of the local maxima and minima of the function
x3 + 3x2 − 9x + 5 for −4 ≤ x ≤ 4. Be sure to also determine whether the
points found are a local maximum or a local minimum.
Differentiate the function to obtain 3x2 + 6x − 9. This factors as 3(x +
3)(x − 1), so the derivative is zero in x = −3 and x = 1. The sign chart
of the derivative is
−
+
−4
−3
+
1
4
So x = −4 and x = 1 give local minima and x = −3 and x = 4 give local
maxima.
2. Determine when the function 2x3 − 15x2 + 36x + 2 is increasing and when
it is decreasing.
The derivative is 6x2 − 30x + 36 = 6(x − 2)(x − 3). The sign chart of the
derivative is
+
−
+
2
3
So the function is decreasing for 2 < x < 3 and increasing if x < 2 or
x > 3.
12
3. Determine the location of the local maxima and minima of the function
2x3 + 3x2 − 12x − 5 for −4 ≤ x ≤ 4. Be sure to also determine whether
the points found are a local maximum or a local minimum.
The derivative is 6x2 + 6x − 12 = 6(x2 + x − 2) = 6(x + 2)(x − 1). So
x = −2 and x = 1 are critical points. The sign chart is
−
+
−4
−2
+
1
4
(the signs are obtained by e.g. substituting x = −3, x = 0 and x = 2 in
the derivative). This patterns of signs means that x = −4 is a minimum,
x = −2 is a maximum, x = 1 is a minimum and x = 4 is a maximum.
4. Determine when the function x3 − 4x2 + 2x + 3 is increasing and when it
is decreasing.
The derivative
is 3x2 − 8x + 2. By the quadratic formula this has zeros
√
4± 10
x = 3 . The sign chart of the derivative is
+
√
4− 10
3
So the function is increasing for x <
√
√
for 4−3 10 < x < 4+3 10 .
−
√
4+ 10
3
√
4− 10
3
+
and x >
√
4+ 10
3
and decreasing
5. Determine the location of the local maxima and minima of the function
ex (x2 − 2x + 1) for −2 ≤ x ≤ 2. Be sure to also determine whether the
points found are a local maximum or a local minimum.
The derivative is ex (x2 − 2x + 1) + ex (2x − 2), which equals ex (x2 − 1). It
follows that the derivative is zero for x = ±1. The sign chart is
−
+
−2
−1
+
1
2
so that x = −2 and x = 1 are local minima and x = −1 and x = 2 are
local maxima.
2.9
Asymptotes
1. Determine the horizontal and vertical asymptotes of the function xx+3
2 −4 .
As x → ±∞ we have that the fraction converges to 0 (using e.g. L’Hopital).
So y = 0 is the horizontal asymptote. The denominator is zero in x = ±2
and the numerator is nonzero there. So x = −2 and x = 2 are the vertical
asymptotes.
2
+3
2. Determine the horizontal and vertical asymptotes of the function xx2 −9
.
As x goes to either plus or minus infinity, the fraction goes to the quotient
of the leading coefficients, which is 1. So y = 1 is the only horizontal
13
asymptote. For vertical asymptotes we look at where the denominator
x2 − 9 is zero. This happens for x = ±3. For these values of x the
numerator is nonzero, so x = −3 and x = 3 are vertical asymptotes.
x
3. Determine the horizontal and vertical asymptotes of the function exsin
−e−x .
As x → ±∞ the denominator converges to either plus or minus infinity.
The numerator stays bounded as x → ±∞. So the fraction converges to
zero. So y = 0 is the only horizontal asymptote of the function. The
denominator is zero only for x = 0. So x = 0 is a the only possible
candidate for a vertical asymptote. But since the numerator is also zero for
x = 0 further investigation is required. We apply l’Hopital. The derivative
of the numerator is cos x, which is 1 for x = 0 and the derivative of the
denominator is ex + e−x , which is 2 for x = 0. So
lim
x→0
1
sin x
= ,
ex − e−x
2
so x = 0 is not a vertical asymptote.
4. Determine the horizontal and vertical asymptotes of the function x22x
−1 .
Because the degree of the denominator is higher than that of the numerator, for x → ±∞ the function tends to zero, so y = 0 is the only horizontal asymptote. The denominator is zero in x = ±1 and the numerator is
nonzero there, so these are vertical asymptotes.
x
5. Determine the horizontal and vertical asymptotes of the function x arctan
.
x−1
Dividing the numerator and denominator by x we express the function as
arctan x
. The denominator then tends to 1 as x → ±∞, whereas the
1
1− x
numerator tends to − π2 as x → −∞ and to π2 as x → ∞. So y = ± π2
are the horizontal asymptotes. The denominator of the original function
is zero in x = 1 and the numerator is nonzero there, so this is a vertical
asymptote.
2.10
Numerical solution of equations
2.11
Taylor polynomials
1. Calculate the Taylor polynomial of degree 8 of x sin (x2 ) around x = 0.
3
The Taylor polynomial of degree 4 of sin u is u− u6 . So the Taylor polyno6
mial of degree 8 of sin x2 is x2 − x6 . It follows that the Taylor polynomial
7
of degree 8 of x sin x2 is x3 − x6 .
2. Calculate the Taylor polynomial of degree 10 of x cos (x3 ) around x = 0.
The Taylor series of cos u is
1−
u2
u4
+
+ ...
2
24
14
Substituting u = x3 gives that the Taylor series for cos (x3 ) is
1−
x6
x12
+
+ ...
2
24
Multiplying by x gives that the Taylor series for x cos (x3 ) is
x−
x7
x13
+
+ ...
2
24
We only have to go to degree 10, so the answer is x −
x7
2 .
3. Determine the Taylor polynomial of degree 3 of e2x (x2 + 2x + 1) around
x = 0.
3
2
The Taylor polynomial of degree 3 of eu is 1 + u + u2 + u6 , so that the
3
Taylor polynomial of degree 3 of e2x is 1 + 2x + 2x2 + 8x6 . Multiplying
2
this by x + 2x + 1 and ignoring terms that are of higher degree than 3
3
we obtain 1 + 4x + 7x2 + 44
6 x .
4. Determine the Taylor polynomial of degree 5 of ln x around x = 1.
In this case we use the definition of Taylor polynomial. The subsequent
derivative of the given function are x−1 , −x−2 , 2x−3 , −6x−4 , 24x−5 . So
the Taylor polynomial of degree 5 is: (x − 1) − 12 (x − 1)2 + 31 (x − 1)3 −
1
1
4
5
4 (x − 1) + 5 (x − 1) .
15
Chapter 3
Integration
3.1
Antiderivatives
R
1. Compute sin x dx.
− cos x + C.
2. Compute
ex + C.
3.2
R
ex dx.
Estimating area
1. Compute the area between the lines y = x, x = 0, x = 1 and y = 0
by approximating it from above and below by the area of rectangles and
taking the limit.
Use an equal distance partition: xk = nk for k = 0, . . . , n; the width of
each rectangle is then n1 . The function is increasing, so the minimum is
reached at the left end point and the maximum is reached at the right end
point. This gives for the upper and lower sums:
L(n) =
n−1
X
k=0
k
,
n2
U (n) =
k=
m(m + 1)
2
n
X
k
.
n2
k=1
Using the formula
m
X
k=0
from the lecture notes we obtain
L(n) =
(n − 1)n
,
2n2
U (n) =
n(n + 1)
.
2n2
As n → ∞ both expressions converge to 21 , so this is the desired area.
16
2. Compute the area between the lines y = x3 , x = 0, x = 1 and y = 0
by approximating it from above and below by the area of rectangles and
taking the limit.
The function is increasing, so the lower sums L(n) and upper sums U (n)
are
n−1 n−1 1X
k
1X
k+1
L(n) =
f
, U (n) =
f
n
n
n
n
k=0
k=0
(this is opposite to what we had in class where the function was decreasing). With the specific f , f (x) = x3 , this gives
n−1
1 X 3
L(n) = 4
k ,
n
k=0
n−1
n
1 X
1 X 3
3
U (n) = 4
(k + 1) = 4
j ,
n
n j=1
k=0
where j = k + 1. Using the formula for the sum of cubes given in class we
have
n2 (n + 1)2
(n − 1)2 n2
,
U
(n)
=
.
L(n) =
4n4
4n4
For both L(n) and U (n) the limit as n → ∞ is 14 (the quotient of the
leading coefficients). So the to be computed area is 14 .
3. Compute the area between the curves y = x2 , x = 0, x = 2 and y = 0
by approximating it from above and below by the area of rectangles and
taking the limit.
We divide the interval into equal pieces of length n2 . The k-th interval is
2k+2
[ 2k
n , n ] and k takes on integer values between 0 and n − 1.
The function is increasing, so the lower sums L(n) and upper sums U (n)
are
n−1 n−1 2X
2k
2k + 2
2X
f
, U (n) =
f
L(n) =
n
n
n
n
k=0
k=0
(this is opposite to what we had in class where the function was decreasing). With the specific f given, f (x) = x2 , this gives
L(n) =
n−1
2 X 2
4k ,
n3
U (n) =
k=0
n−1
n
2 X
2 X 2
2
(2k
+
2)
=
4j ,
n3
n2 j=1
k=0
where j = k + 1. Using the formula for the sum of squares given in class
we have
L(n) =
4(n − 1)n(2n − 1)
,
3n3
U (n) =
4n(n + 1)(2n + 1)
.
3n3
For both L(n) and U (n) the limits as n → ∞ is 83 (the quotient of the
leading coefficients). So the to be computed area is 83 .
17
3.3
Definition
3.4
The fundamental theorem of calculus
1. Compute
R π/2
2. Compute
R1
0
0
cos x dx.
Z π/2
π
π/2
cos x dx = [sin x]0 = sin − sin 0 = 1.
2
0
x2 dx
Z
1
x2 dx =
0
1 3
x
3
1
=
0
1
.
3
R1
3. Compute 0 ex dx.
R1
By the fundamental theorem of calculus 0 ex dx = [ex ]10 = e − 1.
3.5
Rules for integration
3.5.1
The substitution rule
R
1. Compute e2x dx.
Use the substitution u = 2x so that du = 2dx:
Z
1
e2x dx = e2x + C.
2
R1 x
2. Compute 0 x2 +1 dx.
Use the substitution u = x2 + 1 so that du = 2xdx:
2
Z 1
Z 2
1
x
1
1
dx
=
du
=
ln
|u|
= ln 2.
2
2
2
0 x +1
1 2u
1
R 3x
3. Compute e dx.
Use the substitution u = 3x, du = 3 dx to obtain
Z
Z
1
1
1
e3x dx =
eu du = eu + C = e3x + C.
3
3
3
R1
4. Compute 0 x2x+1 dx.
Use the substitution u = x2 + 1, du = 2x dx to obtain
Z
Z
x
1
1
1
1
dx =
dx = ln |u| + C = ln (x2 + 1) + C.
2
x +1
2
u
2
2
Then compute
Z
0
1
1
x
1
1
2
dx
=
ln
(x
+
1)
= ln 2.
x2 + 1
2
2
0
18
R
5. Compute sin 3x dx.
Substituting u = 3x we have du = 3dx so that
Z
Z
1
−1
−1
sin 3x dx =
sin u du =
cos u + C =
cos 3x + C.
3
3
3
R
6. Compute x cos(x2 ) dx.
Substituting u = x2 we have du = 2xdx so that
Z
Z
1
1
1
2
x cos x dx =
cos u du = sin u + C = sin x2 + C.
2
2
2
3.5.2
Integration by parts
R
1. Compute x sin x dx.
Integrate by parts with f = x, g 0 = sin x, f 0 = 1, g = − cos x:
Z
Z
x sin x dx = −x cos x + cos x dx = −x cos x + sin x + C.
R
2. Compute ex sin x dx.
Integrate by parts twice to obtain
Z
Z
Z
x
x
x
x
x
e sin x dx = e sin x − e cos x dx = e sin x − e cos x − ex sin x dx.
This gives
Z
2
so
Z
ex sin x dx = ex sin x − ex cos x + C,
ex sin x dx =
ex sin x − ex cos x
+ K.
2
R
3. Compute x2 cos x dx.
Integrate by parts with f (x) = x2 , g 0 (x) = cos x so that f 0 (x) = 2x,
g(x) = sin x which gives
Z
Z
x2 cos x dx = x2 sin x − 2x sin x dx.
Integrate by parts again with f (x) = 2x, g 0 (x) = − sin x so that f 0 (x) = 2,
g(x) = cos x which gives
Z
Z
− 2x sin x dx = 2x cos x − 2 cos x dx = 2x cos x − 2 sin x + C.
It follows that
Z
x2 cos x dx = x2 sin x + 2x cos x − 2 sin x + C.
19
R
4. Compute x3 ex dx.
This can be done by integration by parts.
Z
Z
3 x
3 x
x e dx = x e − 3 x2 ex dx.
The latter integral was computed in the lecture notes (also using integration by parts) and using that we obtain (x3 − 3x2 + 6x − 6)ex + C.
3.6
Anti-derivatives of inverse functions
R
1. Compute arctan x dx.
This is the integral of an inverse function, so we integrate by parts with
f (x) = arctan x, g 0 (x) = 1 so that f 0 (x) = x21+1 and g(x) = x which gives
Z
Z
arctan x dx = x arctan x −
3.7
1
x
dx = x arctan x − ln (x2 + 1) + C.
x2 + 1
2
Anti-derivatives of rational functions by partial fractions
R
x−2
1. Compute (x+1)
2 (x−1) dx.
Use partial fraction expansion
x−2
A
B
C
=
+
+
.
2
(x + 1) (x − 1)
x − 1 x + 1 (x + 1)2
Making denominators equal this gives
x − 2 = A(x + 1)2 + B(x + 1)(x − 1) + C(x − 1),
which in standard form is
x − 2 = (A + B)x2 + (2A + C)x + (A − B − C).
This gives the linear system of equations
A + B = 0,
2A + C = 1,
A − B − C = −2,
1
3
whose solution is A = −1
4 , B = 4 , C = 2 . For the integral we then get
Z
Z
−1 1
1 1
3
1
x−2
dx =
+
+
(x + 1)2 (x − 1)
4 x − 1 4 x + 1 2 (x + 1)2
=
−1
1
3 1
ln |x − 1| + ln |x + 1| −
+ C.
4
4
2x+1
20
2. Compute
We have
R
1
x2 +4
dx.
Z
1
1
dx =
2
x +4
4
Z
1
x 2
2
dx
+1
with the substitution u = x2 , du = 21 dx we have that this equals
Z
1
1
1
1
x
du = arctan u + C = arctan + C.
2
2
u +1
2
2
2
R
3. Compute x21−4 dx.
Use partial fraction expansion
1
A
B
=
+
.
x2 − 4
x−2 x+2
This gives the following equations for A and B:
A + B = 0,
2A − 2B = 1
with solution A = 14 , B = − 14 . So for the integral we have
Z
Z
1
1
1
1
1
1
dx
=
−
= ln |x − 2| − ln |x + 2| + C.
x2 − 4
4
x−2 x+2
4
4
R
4. Compute (x2x−4
2 +1)2 dx.
Rewrite
Z
Z
Z
2x − 4
2x
1
dx =
dx − 4
dx
(x2 + 1)2
(x2 + 1)2
(x2 + 1)2
This first integral is easy (u = x2 + 1) and the second integral was done
in class. So the answer is
−1
2x
− 2 arctan x − 2
+ C.
x2 + 1
x +1
R x+1
5. Compute x2 −3x+2
dx.
The denominator factors as x2 − 3x + 2 = (x − 2)(x − 1). So the form of
the partial fraction expansion is
A
B
x+1
=
+
.
x2 − 3x + 2
x−2 x−1
Multiplying with the original denominator gives
x + 1 = A(x − 1) + B(x − 2).
Substituting x = 2 gives A = 3 and substituting x = 1 gives B = −2. So
Z
Z
x+1
3
−2
dx =
+
dx = 3 ln |x − 2| − 2 ln |x − 1| + C.
x2 − 3x + 2
x−2 x−1
21
R
6. Compute xx+1
2 +4 dx.
Split the integral into two
Z
Z
Z
x
1
x+1
dx
=
dx
+
dx.
x2 + 4
x2 + 4
x2 + 4
For the first integral use the substitution u = x2 + 4, du = 2x dx to obtain
Z
Z
x
1
1
1
1
dx
=
du = ln |u| + C = ln (x2 + 4) + C.
x2 + 4
2
u
2
2
Re-write the second integral as
Z
Z
1
1
dx =
x2 + 4
4
1
x 2
2
dx
+1
and use the substitution u = x2 , du = 21 dx to obtain
Z
Z
1
1
1
x
1
1
1
dx
=
dx = arctan u + C = arctan + C.
2
2
x
4
2
u +1
2
2
2
+1
2
Combining the above we obtain
Z
x+1
1
1
x
dx = ln (x2 + 4) + arctan + C.
x2 + 4
2
2
2
R
7. Compute x21−1 dx.
This can be done by partial fractions. We have
1
A
B
=
+
.
x2 − 1
x−1 x+1
We obtain the following equation for A and B: A(x + 1) + B(x − 1) = 1,
so that
A + B = 0, A − B = 1.
Solving this gives A = 21 , B = − 12 . So
Z
Z
Z
1
1
1
1
1
1
1
dx =
dx−
dx = ln |x − 1|− ln |x + 1|+C
x2 − 1
2
x−1
2
x+1
2
2
8. Compute
Re-write
R
1
x2 +2
dx.
Z
and substitute u =
1
√
2
Z
1
1
dx =
2
x +2
2
x
√
2
so that du =
Z
√1
2
x
√
2
1
2
dx,
+1
dx, which gives
1
1
1
x
du = √ arctan u + C = √ arctan √ + C.
u2 + 1
2
2
2
22
3.8
Anti-derivatives of trigonometric rational functions
R
1. Compute cos x sin5 x dx.
Use the substitution u = sin x, du = cos xdx
Z
Z
1
1
5
cos x sin x dx = u5 du = u6 + C = sin6 x + C.
6
6
R
3
x
2. Compute cos
sin x dx.
Use the substitution u = sin x, du = cos x dx. Then we obtain
Z
Z
1
1
1 − u2
1
du =
− u du = ln |u| − u2 + C = ln | sin x| − sin2 x + C.
u
u
2
2
Alternatively use the substitution u = cos x, du = − sin x dx to obtain
Z
Z
Z
cos3 x
u3
u
1
1
dx =
du = u + 2
du = u2 + ln |u2 − 1| + C
sin x
u2 − 1
u −1
2
2
1
1
1
cos2 x + ln | sin2 x| + C = cos2 x + ln | sin x| + C,
2
2
2
where in the last equality we used the logarithmic rules (but you could
leave that part out and still get full credit).
R
3. Compute cos2 x dx.
2x
Use the double angle formula cos2 x = 1+cos
to write
2
Z
Z
1 + cos 2x
x sin 2x
cos2 x dx =
dx = +
+ C.
2
2
4
=
R
4. Compute cos3 x sin3 x dx.
Substitute u = sin x so that du = cos x dx and therefore
Z
Z
sin4 x sin6 x
u4 u6
3
3
−
+C =
−
+ C.
cos x sin x dx = (1 − u2 )u3 du =
4
6
4
6
3.9
Trigonometric substitution
R√
1. Calculate
3 − x2 dx. √
√
Use the substitution x = 3 sin u, dx = 3 cos udu:
Z p
Z p
Z √
√
√
2
2
3 − x dx =
3 − 3 sin u 3 cos u du =
3 cos2 u 3 cos u du
=
Z √
Z
√
3 cos u 3 cos u du = 3 cos2 u du
23
and then use the double angle formula
Z
1 + cos 2u
=3
du =
2
for the cosine
3
3 sin 2u
u+
+C
2
4
and the double angle formula for the sine to return to x:
√
3 sin u cos u
3
x
x 3 − x2
3
+ C = arcsin √ +
+ C.
= u+
2
2
2
2
3
R
2. Calculate (1 − x2 )3/2 dx.
Use the substitution x = sin u, dx = cos u du to obtain
Z
cos4 u du
Use the double angle formula for the cosine twice to write
cos4 u =
1 + 2 cos 2u +
(1 + cos 2u)2
1 + 2 cos 2u + cos2 2u
=
=
4
4
4
=
1+cos 4u
2
1
3 1
+ cos 2u + cos 4u.
8 2
8
So
Z
Z
3 1
1
3
1
1
(1−x2 )3/2 dx =
+ cos 2u+ cos 4u du = u+ sin 2u+ sin 4u+C.
8 2
8
8
4
32
We use the double angle formula for the sine to rewrite this in terms of x:
3
1
1
3
1
1
u + sin 2u +
sin 4u = u + sin u cos u +
sin 2u cos 2u
8
4
32
8
2
16
1
1
3
u + sin u cos u + sin u cos u(2 cos2 u − 1),
8
2
8
where in the last step we have also used the double angle formula for the
cosine. This can be simplified as
=
3
3
1
u + sin u cos u + sin u cos3 u.
8
8
4
We can
√ now obtain the solution in terms of x by noting that x = sin u
and 1 − x2 = cos u:
Z
3 p
1
3
(1 − x2 )3/2 dx = arcsin x + x 1 − x2 + x(1 − x2 )3/2 + C.
8
8
4
24
3.10
Hyperbolic substitution
R√
1. Compute
1 + x2 dx.
There are two possible substitutions. One possible substitution is x =
sinh u, dx = cosh u du which gives
Z p
Z
1 + x2 dx = cosh2 u du.
2u
The double angle formula for the hyperbolic cosine is cosh2 u = 1+cosh
2
(this is easily proven by the definition in terms of exponentials), which
gives
Z
Z
u sinh 2u
1 + cosh 2u
cosh2 u du =
du = +
+ C.
2
2
4
The double angle formula for the hyperbolic sine is sinh u = 2 sinh u cosh u
(again this is easily proven by the definition in terms of exponentials) so
that the above integral equals
u sinh u cosh u
+
+ C.
2
2
Using the formula for the inverse of the hyperbolic sine and cosh2 u −
sinh2 u = 1 gives
√
√
Z p
ln (x + 1 + x2 ) x 1 + x2
2
1 + x dx =
+
+ C.
2
2
The other possible substitution x = tan u, dx = cos12 u du gives
Z p
Z
1
2
1 + x dx =
du.
cos3 u
This integral can be solved by substituting w = sin u, dw = cos u du:
Z
Z
1
1
du =
dw
cos3 u
(1 − w2 )2
This last integral can be done by partial fractions
A
1
B
C
D
=
+
+
+
.
(1 − w2 )2
w − 1 (w − 1)2
w + 1 (w + 1)2
Multiplying with the original denominator gives
1 = A(w − 1)(w + 1)2 + B(w + 1)2 + C(w − 1)2 (w + 1) + D(w − 1)2 .
Substituting w = 1 gives B = 14 and substituting w = −1 gives D = 14 .
Substituting w = 0 gives 1 = −A + C + 21 and substituting w = 2 gives
25
1 = 9A + 3C +
C = 14 . So
10
4 .
Solving these equations for A and C gives A =
1
1
=
(1 − w2 )2
4
1
−1
1
1
+
+
+
w − 1 (w − 1)2
w + 1 (w + 1)2
−1
4 ,
.
This gives for the integral
Z
1
1
1
1
dw =
− ln |w − 1| −
+ ln |w + 1| −
+ C.
(1 − w2 )2
4
w−1
w+1
x
. Substituting this in
Since w = sin u and x = tan u we have w = √1+x
2
the antiderivative gives
Z p
1
x
1
x
1 + x2 dx =
− ln | √
− 1| − x
+ ln | √
+ 1| −
2
√
4
−
1
1+x
1 + x2
1+x2
3.11
Area between curves
1. Calculate the area bounded by the lines x = 2, x = 5 and the graphs of
y = x2 and y = 4x.
First find where the two curves intersect: x2 = 4x gives x = 0 and x = 4.
From x = 2 to x = 4 the function y = 4x is on top and from x = 4 to
x = 5 the function y = x2 is on top, so the area is given by
4 5
Z 4
Z 5
1 3
1 3
2
2
2
2
+ x − 2x
4x − x dx +
x − 4x dx = 2x − x
3
3
2
4
2
4
64
8 125
64
16 7
23
−8+ +
− 50 −
+ 32 =
+ =
.
3
3
3
3
3
3
3
2. Calculate the area in the quadrant x ≥ 0, y ≥ 0 bounded by the the
graphs of y = 2x3 and y = 3x.
We first calculate where these curves intersect. We q
have 2x3 = 3x if and
2
only if x(2x − 3) = 0 meaning that x = 0 or x = ± 32 . Since the area is
q
in the first quadrant we must have x = 0 and x = 32 . On this interval
y = 3x is on top, so the area is
√ 32
Z √ 32
9 9
3
1
9
= − = .
3x − 2x3 dx =
x2 − x4
2
2
4 8
8
0
0
= 32 −
3. Calculate the area in the quadrant x ≥ 0, y ≥ 0 that is bounded by the
curves y = x2 and y = 2x.
The intersection points satisfy x2 = 2x, so these are x = 0 and x = ±2.
The intersection points in the given quadrant are therefore x = 0 and
x = 2. So the integral that gives the area is
2
Z 2
x3
8
4
=4− = .
2x − x2 dx = x2 −
3 0
3
3
0
26
!
1
√ x
1+x2
+1
+C.
3.12
Numerical integration
3.13
Improper integrals
1. Calculate
1
Z
2. Calculate
1
√1
x
0
dx.
∞
R∞
1
√
x x
1
R1
1
0 x
a
√ 1
√
1
√ dx = lim 2 x a = lim 2 − 2 a = 2.
+
+
x
a→0
a→0
Z
b
dx.
1
√ dx = lim
b→∞
x x
3. Calculate
We have
1
Z
1
√ dx = lim
x
a→0+
0
Z
R1
1
b
−2
−2
1
√ dx = lim √
= lim √ + 2 = 2.
b→∞
x x
x 1 b→∞ b
dx.
Z
1
0
1
1
dx = [ln |x|]0 = ln 1 − ln 0 = − − ∞ = ∞,
x
or more explictly
Z 1
Z 1
1
1
1
dx = lim+
dx = lim+ [ln |x|]A = lim+ − ln A = ∞
x
x
A→0
A→0
A→0
A
0
R1
4. Compute 0 ln x dx.
We have (using e.g. integration by parts to obtain the anti-derivative)
1
Z
1
ln x dx = [x ln x − x]0 = −1,
0
where we have used that limx→0 x ln x = 0. This latter fact can be proven
by re-writing and using L’Hopital:
lim x ln x = lim
x→0
5. Compute
We have
ln x
1
x
x→0
R∞
1
1
x3
1
x
−1
x→0 2
x
= lim
x2 1
= − lim x = 0.
x→0 −1 x
x→0
= lim
dx.
Z
1
∞
∞
1
1
−1
dx =
= .
3
2
x
2x 1
2
27
Chapter 4
Ordinary differential
equations
4.1
Separable equations
dy
= yx.
1. Solve dx
This is a separable equation so separate variables
Z
Z
dy
= x dx.
y
Finding anti-derivatives gives
ln |y| =
1 2
x + C.
2
Solving for y gives (similarly as in class)
1
2
y = Ke 2 x .
dy
2. Solve dx
= xy with initial condition y(0) = −1.
Separation of variables gives
Z
Z
y dy = x dx,
so that
1 2
1
y = x2 + C.
2
2
√
2
Solving
for y gives y = ± x + 2C. The initial condition gives −1 =
√
± 2C, which implies√that we should have the minus sign and 2C = 1. So
the solution is y = − x2 + 1.
28
4.2
Linear equations
dy
1. Solve dx
− y = e2x .
Using integrating factors. Multiply by M to obtain
M
dy
− M y = M e2x .
dx
(4.1)
We want the equation to be equal to
(M y)0 = M e2x ,
(4.2)
which by the product rule is
M
dM
dy
+
y = M e2x .
dx
dx
(4.3)
Comparing (4.1) and (4.3) we then see that we must choose the integrating
factor M to satisfy
dM
= −M.
dx
To solve this we could use that it is a separable equation, but here we
easily see a solution to be M = e−x . So our equation for y becomes
(e−x y)0 = ex
when we write it in the form of (4.2). We simply integrate this last
equation to obtain
e−x y = ex + C
(remember to put the constant in at this point) and solve for y to obtain
y = e2x + Cex .
Using variation of parameters. We first solve the equation
dz
− z = 0.
dx
This can be done by separation of variables, but in this case it is easy
enough to see that z = ex is a solution. We then substitute y = Cex
into our equation to obtain an equation for C. Using the product rule we
obtain
C 0 ex + Cex − Cex = e2x .
Two terms cancel each other (as always happens when using this method)
so that we obtain
C 0 ex = e2x .
We divide by ex to obtain
C 0 = ex
29
and integrate to obtain
C = ex + K.
Since y = Cex we then have y = e2x + Kex .
Using the method of undetermined coefficients. We first solve the equation
dz
dx − z = 0. This can be done by separation of variables, but in this case
it is easy enough to see that z = Cex is the general solution. We then
try a solution of the same form as the right-hand side, in this case we try
y = Ae2x . Substituting this into the equation gives
2Ae2x − Ae2x = e2x .
It follows that A = 1, so that y = e2x is a solution and y = e2x + Cex is
the general solution.
dy
2. Solve dx
+ y tan x = sin 2x with initial condition y(0) = 1.
Using integrating factors. Multiply by M to obtain
M
dy
+ yM tan x = M sin 2x.
dx
(4.4)
We want the equation to be equal to
(M y)0 = M sin 2x,
(4.5)
which by the product rule is
M
dy
dM
+
y = M sin 2x.
dx
dx
(4.6)
Comparing (4.4) and (4.6) we then see that we must choose the integrating
factor M to satisfy
dM
= M tan x.
dx
To solve this we use that this is a separable equation. We separate variables to obtain
Z
Z
dM
= tan x dx.
M
We now carry out the integrations to obtain (we can omit the constant
since we only need one integrating factor):
ln |M | = − ln | cos x|,
(4.7)
sin x
where we use that tan x = cos
x and used substitution with u = cos x so
du
that dx = − sin x to obtain
Z
Z
−1
tan x dx =
du = − ln |u| + C = − ln | cos x| + C,
u
30
and we subsequently take C = 0 to obtain the simplest integration factor.
Using the logarithmic rules we write (4.7) as
1
ln |M | = ln
,
| cos x|
so that we see that M =
y becomes
1
cos x
is an integrating factor. So our equation for
1 dy
tan x
sin 2x
−y
=
,
cos x dx
cos x
cos x
when we write it in the form of (4.3), or more appropriately it becomes
(
1
sin 2x
y)0 =
.
cos x
cos x
when we write it in the form of (4.2). We use the double angle formula
for the sine to write the right-hand side as 2 sin x, so that we obtain the
equation
1
(
y)0 = 2 sin x,
cos x
which we simply integrate to obtain
1
y = −2 cos x + C.
cos x
(remember to put the constant in at this point) and solve for y to obtain
y = −2 cos2 x + C cos x.
We now determine the arbitrary constant from the initial value. We have
y(0) = −2 + C = 1,
so that C = 3. The unique solution then is y = −2 cos2 x + 3 cos x.
Using variation of parameters. We first solve the equation
dz
+ z tan x = 0.
dx
This can be done by separation of variables as follows.
Z
Z
dz
= − tan x dx.
z
We obtain from this
ln |z| = ln | cos x|,
sin x
cos x
(4.8)
and used substitution with u = cos x so
where we use that tan x =
that du
=
−
sin
x
to
obtain
dx
Z
Z
1
− tan x dx =
du = ln |u| + K = ln | cos x| + K,
u
31
and we subsequently take K = 0 to obtain the simplest solution. From
(4.8) it follows that z = cos x. We then substitute y = C cos x into our
equation to obtain an equation for C. Using the product rule we obtain
C 0 cos x − C sin x + C cos x tan x = sin 2x.
Two terms cancel each other (as always happens when using this method)
so that we obtain
C 0 cos x = sin 2x.
We divide by cos x and use the double angle formula for the sine to obtain
C 0 = 2 sin x
and integrate to obtain
C = −2 cos x + K.
Since y = C cos x we then have y = −2 cos2 x+K cos x. We now determine
the arbitrary constant from the initial value. We have
y(0) = −2 + K = 1,
so that K = 3. The unique solution then is y = −2 cos2 x + 3 cos x.
The method of undetermined coefficients is not really applicable in this
case since tan x appears as the coefficient of y in the differential equation
(this coefficient really needs to be constant for the method of undetermined
coefficients to work properly).
dy
+ 5y = x, y(0) = 2.
3. Solve dx
The integrating factor satisfies M 0 = 5M and therefore is e5x . This gives
0
e5x y = xe5x .
Integration by parts with f (x) = x, g 0 (x) = e5x so that f 0 (x) = 1 and
5x
g(x) = e5 gives
Z
xe5x dx =
xe5x
−
5
It follows that
e5x y =
Z
e5x
xe5x
e5x
dx =
−
+ C.
5
5
25
xe5x
e5x
−
+ C,
5
25
so that
x
1
−
+ Ce−5x .
5 25
Variation of parameters or the method of undetermined coefficients (with
guess y = Ax + B) can also be used to obtain the general solution.
y=
32
We have
y(0) = −
1
+ C,
25
since this should equals 2 we must have C =
of the initial value problem is
y=
4.3
51
25 .
1
51
x
−
+ e−5x .
5 25 25
Numerical methods
33
So the unique solution
Chapter 5
Functions of several
variables
5.1
Partial derivatives and extrema
1. Compute the partial derivatives
sin(xy) + y 2 + x.
Using the chain rule we obtain
∂f
∂x
∂f
= y cos(xy) + 1,
∂x
and
∂f
∂y
for the function f (x, y) =
∂f
= x cos(xy) + 2y.
∂y
2. Find the critical points (i.e. the points where ∂f
∂x = 0,
2
2
2
the function f (x, y, z) = x + y + 3z + x − 2y + 5.
We first compute the partial derivatives:
∂f
= 2x + 1,
∂x
∂f
= 2y − 2,
∂y
Putting these equal to zero gives x =
critical point.
34
−1
2 ,
∂f
∂y
and
∂f
∂z
= 0) for
∂f
= 6z.
∂z
y = 1, z = 0. This is the only