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FP2 Revision Notes
Inequalities
Key Words: sketch, positive
 Use a sketch to best evaluate points of intersection
 Only multiply by POSITIVE values

First order differential equations
Key Words: family of solution curves, separating the variables, integrating factor, transformations
𝑑𝑦

1




Series
Key Words: method of differences, partial fractions, sigma notation rules

If

For a 1st order D
...
in the form

𝑑𝑥

= 𝑓(𝑥)𝑔(𝑦), then ∫

𝑔(𝑦)

𝑑𝑦 = ∫ 𝑓(𝑥) 𝑑𝑥 + 𝑐
𝑑𝑦
𝑑𝑥

+ 𝑃𝑦 = 𝑄 where P and Q are functions of x, multiply through by the

integrating factor to obtain general solution

𝑛



When evaluating ∑1 𝑓(𝑟) consider r=1, r=2, r=3 … then sum and terms will cancel!





𝑑𝑦

When using substitutions get y and

in terms of other variables and it should drop out!

If the general term 𝑢 𝑟 = 𝑓(𝑟) − 𝑓(𝑟 + 1) then ∑1𝑛 𝑢 𝑟 = ∑1𝑛 𝑓(𝑟) − 𝑓(𝑟 + 1)

Further Complex Numbers
Key Words: modulus-argument form, principal argument, complex exponential form, de Moivre’s theorem,
binomial expansion, locus of points: circle; perpendicular bisector,
 If 𝑧 = 𝑥 + 𝑖𝑦 then the complex number can be written as 𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
 Principal argument: −𝜋 < 𝜃 ≤ 𝜋

𝑒 𝑖𝜃 = 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃 (can be proved using Maclaurin expansion of 𝑠𝑖𝑛𝑥, 𝑐𝑜𝑠𝑥 and 𝑒 𝑖𝑥 )
 Thus a complex number can be written in complex exponential form 𝑧 = 𝑟𝑒 𝑖𝜃

𝑐𝑜𝑠(𝑥) = 𝑐𝑜𝑠(−𝑥), − 𝑠𝑖𝑛(𝑥) = 𝑠𝑖𝑛(−𝑥)
 For 𝑧1 = 𝑟1 (𝑐𝑜𝑠𝜃1 + 𝑖𝑠𝑖𝑛𝜃1 ) and 𝑧2 = 𝑟2 (𝑐𝑜𝑠𝜃2 + 𝑖𝑠𝑖𝑛𝜃2 )
o
𝑧1 𝑧2 = 𝑟1 𝑟2 (𝑐𝑜𝑠(𝜃1 + 𝜃2 ) + 𝑖𝑠𝑖𝑛(𝜃1 + 𝜃2 ))
o
𝑧1 /𝑧2 = 𝑟1 /𝑟2 (𝑐𝑜𝑠(𝜃1 − 𝜃2 ) + 𝑖𝑠𝑖𝑛(𝜃1 − 𝜃2 ))
o
Can be proved using trig identities
𝑧

Second order differential equations
Key Words: auxiliary quadratic, general solution, complementary function, particular integral





For roots to the aux equation, the general solution to the 2nd order D
...
is…
o
𝑦 = 𝐴𝑒 𝛼𝑥 + 𝐵𝑒 𝛽𝑥 (distinct roots 𝛼 and 𝛽)
o
𝑦 = (𝐴 + 𝐵𝑥)𝑒 𝛼𝑥 (repeated root 𝛼)
o
𝑦 = 𝐴𝑐𝑜𝑠𝜔𝑥 + 𝐵𝑠𝑖𝑛𝜔𝑥 (imaginary roots ±𝑖𝜔)
o
𝑦 = 𝑒 𝑝𝑥 (𝐴𝑐𝑜𝑠𝑞𝑥 + 𝐵𝑠𝑖𝑛𝑞𝑥) (complex roots 𝑝 ± 𝑖𝑞)



For 𝑎

𝑑2𝑦
𝑑𝑥 2

o






𝑧 − = 2𝑖 𝑠𝑖𝑛𝜃

o

𝑧𝑛−

1
𝑧

𝑛

o
Can be proved using 𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃)
𝑧 = 𝑟(𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃) = 𝑟(𝑐𝑜𝑠(𝜃 + 2𝑘𝜋) + 𝑖𝑠𝑖𝑛(𝜃 + 2𝑘𝜋))
To remove a modulus (using Pythagoras’ theorem):
|𝑧| = 𝑘
o
o
∴ |𝑥 + 𝑖𝑦| = 𝑘
o
∴ 𝑥 2 + 𝑦2 = 𝑘2
To remove an argument:
o
𝑎𝑟𝑔(𝑧) = 𝜃
o
𝑎𝑟𝑔(𝑥 + 𝑖𝑦) = 𝜃
𝑦
o
= 𝑡𝑎𝑛 𝜃 (adjust accordingly depending on quadrant)
𝑥




For a complex number w, 𝑤 = 𝑢 + 𝑖𝑣
For a transformation T from the z-plane to the w-plane:
o
𝑤 = 𝑧 + 𝑎 + 𝑖𝑏 is a translation ( 𝑏𝑎)
o
o

𝑤 = 𝑘𝑧 is an enlargement scale factor k centre (0,0)
𝑤 = 𝐾𝑧 + 𝑎 + 𝑖𝑏 is an enlargement scale factor k centre (0,0) followed by translation ( 𝑏𝑎)

+ 𝑐𝑦 = 𝑓(𝑥)
𝑑2𝑦

𝑑𝑦

Then solve for particular integral
If 𝑓(𝑥) is in the form… then try…

𝑘→ 𝑎

𝑘𝑥 → 𝑎𝑥 + 𝑏

𝑘𝑥 2 → 𝑎𝑥 2 + 𝑏𝑥 + 𝑐

𝑘𝑒 𝑝𝑥 → 𝐴𝑒 𝑝𝑥

𝑚𝑐𝑜𝑠𝜔𝑥 → 𝑎𝑐𝑜𝑠𝜔𝑥 + 𝑏𝑠𝑖𝑛𝜔𝑥

𝑚𝑠𝑖𝑛𝜔𝑥 → 𝑎𝑐𝑜𝑠𝜔𝑥 + 𝑏𝑠𝑖𝑛𝜔𝑥

𝑚𝑐𝑜𝑠𝜔𝑥 + 𝑛𝑠𝑖𝑛𝜔𝑥 → 𝑎𝑐𝑜𝑠𝜔𝑥 + 𝑏𝑠𝑖𝑛𝜔𝑥
General solution is 𝑦 = 𝐶
...
E

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