Further Pure Maths A2 Level
Series and Induction
These are some examples of finite and infinite series where it is possible to find an exact expression for the
sum. In general, finding an expression for the sum of the first n terms of a series can be surprisingly
difficult, if not impossible. For example, it is not possible to express a formula for the sum
𝑛
∑
𝑘=1
1
1
1
1
1
= 2 + 2 + 2 + ⋯+ 2
2
𝑘
1
2
3
𝑛
in terms of standard functions.
In cases where you manage to guess the formula for the sum of a series, you can then try to prove it by
induction.
The inductive step relies on making the connection between the sum of the first k terms and the sum of
the first k + 1 terms; this is done simply by adding the next term of the series.
If 𝑆𝑘 = 𝑢1 + 𝑢2 + ⋯ + 𝑢𝑘 then
𝑆𝑘+1 = 𝑆𝑘 + 𝑢𝑘+1
Using Standard Series
You can use the following formulae without proof (unless the question explicitly asks you to prove them).
Remember that a constant, c,
summed n times is nc and not
just c.
𝑛
∑1 = 1 × 𝑛
𝑟=1
𝑛
1
∑ 𝑟 = 𝑛(𝑛 + 1)
2
𝑟=1
𝑛
1
∑ 𝑟 2 = 𝑛(𝑛 + 1)(2𝑛 + 1)
6
𝑟=1
𝑛
𝑛
𝑟=1
𝑟=1
1
∑ 𝑟 3 = 𝑛2 (𝑛 + 1)2 = (∑ 𝑟)
4
Remember you can manipulate series in several ways:
∑(𝑢𝑟 + 𝑣𝑟 ) = ∑ 𝑢𝑟 + ∑ 𝑣𝑟
∑ 𝑐𝑢𝑟 = 𝑐 ∑ 𝑢𝑟
𝑛
∑ 𝑐 = 𝑐𝑛
𝑟=1
2
Where c is a constant
The Method of Differences
Alternative (shorter) Method
As you keep adding more and more terms of a series, the sum could keep increasing without a limit, or it
could approach a finite value. In the latter case you say that the series converges (is convergent) and you
could try to find its sum to infinity. To do this, you can find an expression for the sum of the first n terms
and consider what this expression tends to when n gets very large.
Powers and Roots of Complex Numbers
De Moivre’s Theorem
For a complex number, z, with modulus r and argument θ:
𝑛
𝑧 𝑛 = (𝑟(cos 𝜃 + 𝑖 sin 𝜃)) = 𝑟 𝑛 (cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃)
For every integer power n.
For positive integer powers, you can prove this result by induction.
You can use de Moivre’s theorem to evaluate powers of complex numbers.
Complex Exponents
𝑒 𝑖𝜃 ≡ cos 𝜃 + 𝑖 sin 𝜃
The complex conjugate of 𝑧 = 𝑟𝑒 𝑖𝜃 is 𝑧 ∗ = 𝑟𝑒 −𝑖𝜃
Roots of Complex Numbers
In Pure Core Student Book 1, you learnt how to find the two square roots of a complex number by writing
𝑧 = 𝑥 + 𝑖𝑦 and comparing real and imaginary parts. You also know that a polynomial equation of degree
n has n complex roots. Just as a complex number has two square roots, it will have three cube roots, four
fourth roots, and so on.
You can’t always use the algebraic method to find all those roots. De Moivre’s theorem gives an alternative
method.
To solve 𝑧 𝑛 = 𝑤:
•
•
•
•
•
write w in modulus–argument form
use de Moivre’s theorem to write 𝑧 𝑛 = 𝑟 𝑛 (cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃)
compare moduli, remembering that they are always real
compare arguments, remembering that adding 2π onto the argument does not change the number
write n different roots in modulus–argument form.
Roots of Unity
Further Factorising
In Pure Core Student Book 1, Chapter 5, you learnt that complex roots of a real polynomial come in
conjugate pairs, and how you can use this fact to factorise a polynomial. You used the important result
that, for any complex number w,
(𝑧 − 𝑤)(𝑧 − 𝑤 ∗ ) = 𝑧 2 − 2𝑧𝑅𝑒(𝑤) + |𝑤|
You can now combine this with your knowledge of roots of complex numbers to factorise expressions of
the form 𝑧 𝑛 + 𝑐
Geometry of complex numbers


Multiplication by 𝑟 cis 𝜃 corresponds to a rotation about the origin though angle θ and an
enlargement with scale factor r.
Division by 𝑟 cis 𝜃 corresponds to a rotation about the origin though angle -θ and an enlargement
1
with scale factor 𝑟 .
Complex Numbers and Trigonometry
Deriving Multiple Angle Formulae
Application to Polynomial Equations
Powers of Trigonometric Functions
𝑒 𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃
𝑒 −𝑖𝜃 = cos −𝜃 + 𝑖 sin −𝜃
= cos 𝜃 − 𝑖 sin 𝜃
By adding and subtracting these two equations you can establish two very useful identities:
2 cos 𝜃 = 𝑒 𝑖𝜃 + 𝑒 −𝑖𝜃
2𝑖 sin 𝜃 = 𝑒 𝑖𝜃 − 𝑒 −𝑖𝜃
You can further generalise these expressions: If 𝑧 = 𝑒 𝑖𝜃 , then
𝑧𝑛 +
1
= 2 cos 𝑛𝜃
𝑧𝑛
𝑧𝑛 −
1
= 2𝑖 sin 𝑛𝜃
𝑧𝑛
Trigonometric Series
Is it possible to simplify a sum such as sin 𝑥 + sin 2𝑥 + sin 3𝑥 + sin 4𝑥 ?
You can simplify certain sums of this type using the exponential form of complex numbers and the formula
for the sum of geometric series. This is because sin 𝑘𝑥 is the imaginary part of 𝑒 𝑖𝑘𝑥 , and the numbers
𝑒 𝑖𝑥 , 𝑒 2𝑖𝑥 , 𝑒 3𝑖𝑥 , 𝑒 4𝑖𝑥 form a geometric series.
If the modulus of the common ratio is smaller than 1, a geometric series also has a sum to infinity.
Hyperbolic Functions
Defining Hyperbolic Functions
Trigonometric functions are sometimes called circular functions. This is
because of the definition that states: a point on the unit circle (with equation
𝑥 2 + 𝑦 2 = 1) defining a radius at an angle θ to the positive x-axis, has
coordinates (cos 𝜃 , sin 𝜃).
Related to the circle is a curve with equation 𝑥 2 − 𝑦 2 = 1, called a hyperbola.
Points on this hyperbola have coordinates (cosh 𝜃 , sinh 𝜃) although θ can no
longer be interpreted as an angle.
sinh 𝑥 =
𝑒 𝑥 − 𝑒 −𝑥
2
cosh 𝑥 =
𝑒 𝑥 + 𝑒 −𝑥
2
tanh 𝑥 =
The graphs of each look like this;
Inverse Hyperbolic Functions
The graphs look like this:
sinh 𝑥 𝑒 𝑥 − 𝑒 −𝑥
=
cosh 𝑥 𝑒 𝑥 + 𝑒 −𝑥
sinh−1 𝑥 = ln (𝑥 + √𝑥 2 + 1)
cosh−1 𝑥 = ln (𝑥 + √𝑥 2 − 1)
tanh−1 𝑥 =
1
1+𝑥
ln (
)
2
1−𝑥
These will be given in your formula book
Proof
These results can all be proved in the same way. The proof for cosh−1 𝑥 is given here.
Note
When you are trying to solve equations involving hyperbolic cosines,
using the inverse function does not give all the solutions: it just gives the
positive one. This is because the cosh function is not one-to-one. As can
be seen from the graph, there is a second, negative solution.
Hyperbolic Identities
cosh2 𝑥 − sinh2 𝑥
Proof
Addition
Formulae
cosh(𝐴 + 𝐵) ≡ cosh 𝐴 cosh 𝐵 + sinh 𝐴 sinh 𝐵
sinh(𝐴 + 𝐵) ≡ sinh 𝐴 cosh 𝐵 + cosh 𝐴 sinh 𝐵
cosh(𝐴 − 𝐵) ≡ cosh 𝐴 cosh 𝐵 − sinh 𝐴 sinh 𝐵
sinh(𝐴 − 𝐵) ≡ sinh 𝐴 cosh 𝐵 − cosh 𝐴 sinh 𝐵
2 cosh2 𝐴 − 1
cosh 2𝐴 ≡ { 1 + 2 sinh2 𝐴
cosh2 𝐴 + sinh2 𝐴
sinh 2𝐴 ≡ 2 sinh 𝐴 cosh 𝐴
Solving Harder Hyperbolic Equations
When you are solving equations involving hyperbolic functions you have several options:
•
•
•
Rearrange to get a hyperbolic function that is equal to a constant and use inverse hyperbolic
functions.
Use the definition of hyperbolic functions to get an exponential function that is equal to a constant
and use logarithms.
Use an identity for hyperbolic functions to simplify the situation to one of the two preceding
options.
When you are dealing with the sum of difference of two hyperbolic functions, it is often useful to use
exponential form
Differentiation
𝑑
(sinh 𝑥) = cosh 𝑥
𝑑𝑥
𝑑
(cosh 𝑥) = sinh 𝑥
𝑑𝑥
𝑑
1
(tanh 𝑥) =
𝑑𝑥
cosh2 𝑥
Integration
∫ sinh 𝑥 𝑑𝑥 = cosh 𝑥 + 𝑐
∫ cosh 𝑥 𝑑𝑥 = sinh 𝑥 + 𝑐
∫ tanh 𝑥 𝑑𝑥 = ln cosh 𝑥 + 𝑐
Lines and Planes in Space
Equation of a Plane
The vector equation of a plane containing a point with position vector a and parallel to the directions of
vectors 𝒅𝟏 and 𝒅𝟐 is 𝒓 = 𝒂 + 𝜆𝒅𝟏 + 𝜇𝒅𝟐
The scalar product equation of the plane is given by:
𝒓∙𝒏 =𝒂∙𝒏
Where n is the normal to the plane and a is the position vector of appoint in the plane
The scalar product 𝒂 ∙ 𝒏 is a constant, denoted d below
The cartesian equation of a plane can be written in the form:
𝑛1 𝑥 + 𝑛2 𝑦 + 𝑛3 𝑧 = 𝑑
You can also convert from a Cartesian to a vector equation by finding any two vectors that are
perpendicular to the normal.
Intersection Between a Line and a Plane
Angles Between Lines and Planes
The angle between a line l and a plane Π is the smallest possible angle that l makes with any of the lines in
Π.
The angle between the line with direction vector d and the plane with normal n is 90𝑜 − 𝜙, where 𝜙 is the
acute angle between d and n
The angle between two planes is equal to the angle between their normals
Distances Between Points, Lines and Planes
Distance between a Point and a Plane
The shortest distance between the point with position vector b and the plane with equation 𝒓 ∙ 𝒏 = 𝑝 is
given by:
𝐷=
|𝒃 ∙ 𝒏 − 𝑝|
|𝒏|
This will be given in your formula book
Distance between a Point and a Line
The shortest distance between the point with coordinates (𝑥1 , 𝑦1 ) and the line with equation 𝑎𝑥 + 𝑏𝑦 = 𝑐
is given by
𝐷=
|𝑎𝑥1 + 𝑏𝑦1 − 𝑐|
√𝑎2 + 𝑏 2
This will be given in your formula book
Distance between Two Skew Lines
The shortest distance between two skew lines with equations 𝒓 = 𝒂 + 𝜆𝒅𝟏 and 𝒓 = 𝒃 + 𝜇𝒅𝟐 is given by
𝐷=
|(𝒃 − 𝒂) ∙ 𝒏|
|𝒏|
Where 𝒏 = 𝒅𝟏 × 𝒅𝟐
This will be given in your formula book
Distance between Two Parallel Lines
The distance between parallel lines with equations 𝒓 = 𝒂 + 𝜆𝒅 and 𝒓 = 𝒃 + 𝜇𝒅 is given by:
𝐷 = |𝒂 − 𝒃| sin 𝜃 where cos 𝜃 =
(𝒂 − 𝒃) ∙ 𝒅
|𝒂 − 𝒃||𝒅|
This formula will not be given in the formula book so it is a good idea to draw a diagram to make sure you
get it right
Simultaneous Equations and Planes
Linear Simultaneous Equations
For a system of simultaneous equations in matrix form, 𝑴𝒓 = 𝒂:


If det 𝑴 ≠ 0 the equations have a unique solution
If det 𝑴 = 0 there is no unique solution. Use elimination to distinguish between two cases:
o Consistent equations: there are infinitely many solutions
o Inconsistent equations: there are no solutions
Intersections of Planes
Two planes are parallel if their normal vectors are multiples of each other.
With three distinct planes there are several possibilities; altogether, there are five different arrangements:
1. Consistent system: unique solution
The three planes meet at a single point
2. Consistent system: infinitely many solutions
The three planes intersect in a line (the planes form a sheaf)
3. Inconsistent system: no solutions
There is no point common to all three planes
a. All three planes are parallel
b. Two of the planes are parallel
c. The planes enclose a triangular prism, so that each pair of planes intersects in a line, with the
three distinct lines running parallel to each other
To determine the geometrical arrangement of three planes described by a set of simultaneous equations,
use 𝑴𝒓 = 𝒂:


If det 𝑴 ≠ 0 the three planes meet at a single point
If det 𝑴 = 0 then:
o Consistent equations: the planes meet in a line (form a sheaf)
o Inconsistent equations: then either some of the lanes are parallel or the three planes form a
prism
Further Calculus Techniques
Differentiation of Inverse Trigonometric Functions
𝑑
1
sin−1 𝑥 =
, |𝑥| < 1
𝑑𝑥
√1 − 𝑥 2
𝑑
1
cos−1 𝑥 = −
, |𝑥| < 1
𝑑𝑥
√1 − 𝑥 2
𝑑
1
tan−1 𝑥 =
𝑑𝑥
1 + 𝑥2
These will be given in your formula book
Differentiation of Inverse Hyperbolic Functions
𝑑
1
sinh−1 𝑥 =
𝑑𝑥
√𝑥 2 + 1
𝑑
1
cosh−1 𝑥 =
,𝑥 > 1
𝑑𝑥
√𝑥 2 − 1
𝑑
1
tanh−1 𝑥 =
, |𝑥| < 1
𝑑𝑥
1 − 𝑥2
These will be given in your formula book
Using inverse trigonometric and hyperbolic functions in integration
∫
1
√𝑎2 − 𝑥 2
∫
𝑑𝑥 = sin−1
𝑥
+ 𝑐 , |𝑥| < 1
𝑎
1
1
𝑥
−1
𝑑𝑥
=
𝑡𝑎𝑛
+𝑐
𝑎2 + 𝑥 2
𝑎
𝑎
∫
∫
1
𝑥
𝑑𝑥 = sinh−1 + 𝑐
𝑎
√𝑎2 + 𝑥 2
1
𝑥
𝑑𝑥 = cosh−1 + 𝑐 , 𝑥 > 1
𝑎
√𝑥 2 − 𝑎2
These will be given in your formula book
You also need to know how to derive these results using trigonometric or hyperbolic substitutions.
Using Partial Fractions in Integration
In general, when there is a quadratic factor in the denominator, there are three possibilities:

The quadratic factorises into two different linear factors, (𝑥 − 𝑝)(𝑥 − 𝑞). The corresponding partial
𝐴
𝐵
fractions are 𝑥−𝑝 + 𝑥−𝑞
𝐴
𝐵

The quadratic is a perfect square, (𝑥 − 𝑝)2 . The corresponding partial fractions are 𝑥−𝑝 + (𝑥−𝑝)2

The quadratic does not factorise (the quadratic factor is irreducible). For example, (𝑥 2 + 1) or
(𝑥 2 + 2𝑥 + 5) Then there is only one corresponding partial fraction, with a numerator of the form
𝐵𝑥 + 𝐶
If 𝑓(𝑥) is a polynomial of order less than or equal to 2, then:
𝑓(𝑥)
𝐴
𝐵𝑥 + 𝐶
=
+ 2
2
2
(𝑥 − 𝑝)(𝑥 + 𝑞 ) 𝑥 − 𝑝 𝑥 + 𝑞 2
Applications of Calculus
The Maclaurin Series
Therefore, the Maclaurin series of a function 𝑓(𝑥) is given by:
𝑓(𝑥) = 𝑓(0) + 𝑓 ′ (0)𝑥 +
𝑓 ′′ (0) 2 𝑓 ′′′ (0) 3
𝑓 (𝑟) (0) 𝑟
𝑥 +
𝑥 + ⋯+
𝑥
2!
3!
𝑟!
Not every function has a Maclaurin series. For example, for 𝑓(𝑥) = ln(𝑥), 𝑓(0) doesn’t exist (nor do any of
the derivatives of ln 𝑥 at 𝑥 = 0). However, 𝑓(𝑥) = ln(1 + 𝑥) does have a Maclaurin series as now 𝑓(0),
and all the derivatives at x = 0, do exist.
Using Standard Maclaurin Series
The Maclaurin series of a few standard functions are given in your formula book. These can often be used,
without needing to be derived, to find the series for more complicated functions.
𝑒𝑥 = 1 +
sin 𝑥 =
𝑥 𝑥2 𝑥3
𝑥𝑟
+ + + ⋯+
, 𝑓𝑜𝑟 𝑥 ∈ ℝ
1! 2! 3!
𝑟!
𝑥 𝑥3 𝑥5 𝑥7
𝑥 2𝑟+1
− + − … + (−1)𝑟
, 𝑓𝑜𝑟 𝑥 ∈ ℝ
(2𝑟 + 1)!
1! 3! 5! 7!
cos 𝑥 = 1 −
𝑥2 𝑥4 𝑥6
𝑥 2𝑟
+ − + ⋯ + (−1)𝑟
, 𝑓𝑜𝑟 𝑥 ∈ ℝ
(2𝑟)!
2! 4! 6!
1
1
1
1
ln(1 + 𝑥) = 𝑥 − 𝑥 2 + 𝑥 3 − 𝑥 4 + ⋯ + (−1)𝑟−1 𝑥 𝑟 , 𝑓𝑜𝑟 − 1 < 𝑥 ≤ 1
2
3
4
𝑟
𝑛(𝑛 − 1) 2 𝑛(𝑛 − 1)(𝑛 − 2) 3
𝑥 +
𝑥 +⋯
2!
3!
𝑛(𝑛 − 1)(𝑛 − 2) … (𝑛 − (𝑟 − 1)) 𝑟
+
𝑥 , 𝑓𝑜𝑟 |𝑥| < 1, 𝑛 ∈ ℝ
𝑟!
(1 + 𝑥)𝑛 = 1 + 𝑛𝑥 +
These will be given in your formula book
The question will make it clear whether you are required to find the Maclaurin series of a function from
first principles or whether you can use one of the standard results in the formula book to find the series
you need.
Don’t overlook the information on the values of x for which these series are valid; this is a very important
part of each result.
Improper Integrals
Integrals where the range of integration extends to infinity
To evaluate an improper integral, you need to replace the infinite limit by b, find the value of the integral
in terms of b and then consider what happens when 𝑏 → ∞:
∞
The value of the improper integral ∫𝑎 𝑓(𝑥) 𝑑𝑥 is
𝑏
lim ∫ 𝑓(𝑥) 𝑑𝑥
𝑏→∞ 𝑎
If this limit is infinite you say the improper integral diverges (does not have a value)
Integrals where the integrand is undefined at a point within the range of integration
2
5
Examples of such integrals are ∫0 𝑥 3 ln 𝑥 𝑑𝑥, which isn’t defined at 𝑥 = 0, and ∫0
1
√𝑥−3
𝑑𝑥, which isn’t
defined at 𝑥 = 3.
To evaluate the first of these integrals, you need to replace 0 by b as the lower limit, find the value of the
integral in terms of b and then consider the limit 𝑏 → ∞:
If 𝑓(𝑥) is not defined at 𝑥 = 𝑘, then:
𝑘
𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = lim ∫ 𝑓(𝑥) 𝑑𝑥
𝑏→𝑘 𝑎
𝑎
And
𝑐
𝑐
∫ 𝑓(𝑥) 𝑑𝑥 = lim ∫ 𝑓(𝑥) 𝑑𝑥
𝑏→𝑘 𝑏
𝑘
If the limit is not finite, then the improper integral diverges (does not have a value).
If the point where the integrand is not defined is not an end point, you need to split the integral into two.
If 𝑓(𝑥) is not defined at 𝑥 = 𝑘 ∈ (𝑎, 𝑐), then:
𝑐
𝑏
𝑐
∫ 𝑓(𝑥) 𝑑𝑥 = lim ∫ 𝑓(𝑥) 𝑑𝑥 + lim ∫ 𝑓(𝑥) 𝑑𝑥
𝑎
𝑏→𝑘 𝑎
𝑏→𝑘 𝑏
If either limit is not finite, then the improper integral diverges (does not have a value).
Volumes of Revolution
When the curve 𝑦 = 𝑓(𝑥) between 𝑥 = 𝑎 and 𝑥 = 𝑏 is rotated 360𝑜 about the x-axis, the volume of
revolution is given by
𝑏
𝑉 = 𝜋 ∫ 𝑦 2 𝑑𝑥
𝑎
When the curve 𝑦 = 𝑓(𝑥) between 𝑦 = 𝑎 and 𝑦 = 𝑏 is rotated 360𝑜 about the y-axis, the volume of
revolution is given by
𝑏
𝑉 = 𝜋 ∫ 𝑥 2 𝑑𝑥
𝑎
Proof
The volume of revolution of the region between the curves 𝑔(𝑥) and 𝑓(𝑥) is:
𝑏
𝑉 = 𝜋 ∫ (𝑔(𝑥)2 − 𝑓(𝑥)2 ) 𝑑𝑥
𝑎
Where 𝑔(𝑥) is above 𝑓(𝑥) and the curves intersect at 𝑥 = 𝑎 and 𝑥 = 𝑏
Make sure that you square each term within the brackets and do not make the mistake of squaring the
𝑏
2
whole expression inside the brackets: the formula is not 𝜋 ∫𝑎 (𝑔(𝑥) − 𝑓(𝑥) ) 𝑑𝑥
When the part of a curve with parametric equations 𝑥 = 𝑓(𝑡), 𝑦 = 𝑔(𝑡), between points with parameter
values t1 and t2, is rotated about one of the coordinate axes, the resulting volume of revolution is.
𝑡2
𝜋 ∫ 𝑦2
𝑡1
𝑑𝑥
𝑑𝑡
𝑑𝑡
For rotation about the x-axis
𝑡2
𝜋 ∫ 𝑥2
𝑡1
𝑑𝑦
𝑑𝑡
𝑑𝑡
For rotation about the y-axis
The Mean Value of a Function
The mean value of a function 𝑓(𝑥) between 𝑎 and 𝑏 is:
𝑏
1
∫ 𝑓(𝑥) 𝑑𝑥
𝑏−𝑎 𝑎
Polar Coordinates
There may be values of 𝜃 for which r is not defined. This happens when the expression for r has a negative
value; r is a distance and so it must be non-negative. An effective way to identify such areas is to sketch the
graph of r against 𝜃
If you use graphing software/a graphical calculator to plot polar curves, you will find that some of them will
allow negative values of r – remember that in this course, we require that r takes non-negative values.
Features of Polar Curves
𝑑𝑟
The minimum and maximum values for r occur where 𝑑𝜃 = 0
For a curve with polar equation 𝑟 = 𝑓(𝜃), the line 𝜃 = 𝛼 is a tangent at the pole if 𝑓(𝛼) = 0 but 𝑓(𝛼) > 0
on one side of the line (for r cannot take negative values).
Changing between Polar and Cartesian Coordinates
A point with polar coordinates (𝑟, 𝜃) had Cartesian coordinates (𝑟 cos 𝜃 , 𝑟 sin 𝜃)
For a point with Cartesian coordinates (𝑥, 𝑦) the polar coordinates satisfy:

𝑟 = √𝑥 2 + 𝑦 2

tan 𝜃 = 𝑥
𝑦
Area Enclosed by a Polar Curve
Finding the area bounded by a polar curve is similar to finding the area bounded by a Cartesian curve,
except rather than being the area between the curve, the x-axis and the vertical lines 𝑥 = 𝑎 and 𝑥 = 𝑏,
now it is the area between the curve, the pole, and the lines from the pole 𝜃 = 𝛼 and 𝜃 = 𝛽. Think of it in
terms of sectors rather than rectangles:
The area enclosed between a polar curve and the half-lines 𝜃 = 𝛼 and 𝜃 = 𝛽 is
𝛽
∫
𝛼
Proof
1 2
𝑟 𝑑𝜃
2
To find the area enclosed between two polar curves, find the intersection points of the curves and
calculate the part of the area bounded by each curve separately
Differential Equations


The order of a differential equation is the largest number of times the dependent variable is
differentiated
A linear differential equation is one in which the dependent variable only appears to the power of
𝑑𝑦
1. Any differential equation involving 𝑦 2 , sin 𝑦 , 𝑦 𝑑𝑥 is non-linear

A homogenous differential equation is one where every term involves the dependent variable
The general solution to an nth-order differential equation has n arbitrary constants.
For a linear differential equation, the general solution is given by:
𝑦 = 𝑦𝑐 + 𝑦𝑝
Where 𝑦𝑐 is the complementary function and 𝑦𝑝 is a particular integral.
Integrating Factor for First Order Differential Equations
Given a first order linear differential equation:
𝑑𝑦
+ 𝑃(𝑥)𝑦 = 𝑄(𝑥)
𝑑𝑥
multiply through by the integrating factor:
𝐼(𝑥) = 𝑒 ∫ 𝑃(𝑥) 𝑑𝑥
𝐼(𝑥)
𝑑𝑦
+ 𝐼(𝑥)𝑃(𝑥)𝑦 = 𝐼(𝑥)𝑄(𝑥)
𝑑𝑥
𝑑
(𝐼(𝑥)𝑦) = 𝐼(𝑥)𝑄(𝑥)
𝑑𝑦
𝑦=
1
∫ 𝐼(𝑥) 𝑄(𝑥)𝑑𝑥
𝐼(𝑥)
Solving 2 nd order Linear Differential Equations with Constant Coefficients
When solving 2nd order non-homogenous linear differential equations:






Find the auxiliary equation
Solve the auxiliary equation
Find the complementary function
If Non-homogenous:
o Select a Trial Function
o Calculate the Particular Integral
Find the general solution
(Find the particular solution if given limits)
Auxiliary Equation
The auxiliary equation of the differential equation
𝑎
𝑑2𝑦
𝑑𝑦
+𝑏
+ 𝑐𝑦 = 0
2
𝑑𝑥
𝑑𝑥
Is
𝑎𝑘 2 + 𝑏𝑘 + 𝑐 = 0
Proof:
Complementary Function
Proof:
Trial Function
𝒇(𝒙)
𝒑𝒙 + 𝒒
Polynomial
𝒑𝒆𝒃𝒙
𝒑 𝐜𝐨𝐬 𝒃𝒙
𝒑 𝐬𝐢𝐧 𝒃𝒙
Trial Function
𝑎𝑥 + 𝑏
General polynomial of the same order
𝑎𝑒 𝑏𝑥
𝑎 sin 𝑏𝑥 + 𝑏 cos 𝑏𝑥
If your trial function is already part of the complementary function, try multiplying the trial function by 𝑥
Applications of Differential Equations
Forming Differential Equations
Simple Harmonic Motion
The differential equation for simple harmonic motion is
𝑑2𝑥
= −𝜔2 𝑥
𝑑𝑡 2
The general solution to the simple harmonic motion differential equation is
𝑥 = 𝐴 sin 𝜔𝑡 + 𝐵 cos 𝜔𝑡
If initially the object is:


At the equilibrium position, then the solution will be 𝑥 = 𝑎 sin 𝜔𝑡
At the maximum displacement from the equilibrium position, then the solution will be 𝑥 = 𝑎 cos 𝜔𝑡
The general solution to the simple harmonic motion can also be written as 𝑅 sin(𝜔𝑡 + 𝜑)
The period, T, of a particle moving with SHM is
𝑇=
2𝜋
𝜔
The relationship between velocity and displacement for a particle moving with simple harmonic motion is
𝑣 2 = 𝜔2 (𝑎2 − 𝑥 2 )
Proof
Damping and Damped Oscilations
The drag force on an object moving with speed v is given by
𝐷 = −𝐾𝑣
Where K is a constant.
If you add this to the standard equation for simple harmonic motion, the differential equation becomes:
𝑑2𝑥
𝑑𝑥
+𝑘
+ 𝜔2 𝑥 = 0
2
𝑑𝑡
𝑑𝑡
𝐾
Where 𝑘 = 𝑚
Linear Systems