Taylor Series are very useful for approximating function values, much more effectively than standard linear approximations.

To me Taylor Series is very easy and straight forward. There is one important thing you need to remember about Taylor series and that is
y= a0 + a1x + a2x^(2) + a3x^(3) + a4x^(4)……….+anx^(n)
y’=a1 + 2A2x +3a3x^(2) +4a4x^(3)………..
y”=2a2 + 6a3x + 12a4x^(2)…………………
and with this all you do is plug it into a the equation.
For example
y”+2y’ +y=0 y(0)=0 y’(0)=2 x=1
the first step is to find y, y’, y” which are the given above from earlier. These do not change.
Then you plug it in the formula producing
(2a2 + 6a3x + 12a4x^(2))+ 2(a1 + 2A2x +3a3x^(2) +4a4x^(3)) + (a0 + a1x + a2x^(2) + a3x^(3) + a4x^(4)) …..= 0
(2a2 + 6a3x + 12a4x^(2))+ 2a1 + 4a2x +6a3x^(2) +8a4x^(3) + (a0 + a1x + a2x^(2) + a3x^(3) + a4x^(4)) ……= 0
Now you group by the power of x.
(2a2 + 2a1 + a0) + (6a3x + 4a2x + a1x) + (12a4x^(2) + 6a3x^(2) + a2x^(2) ) + a3x^(3) + a4x^(4))….. = 0
(2a2 + 2a1 + a0) + (6a3 + 4a2 + a1)X + (12a4 + 6a3 + a2 ) x^(2) + a3x^(3) + a4x^(4))….. = 0
Now we set to equal 0
2a2 + 2a1 + a0 =0
6a3 + 4a2 + a1 = 0
12a4 + 6a3 + a2 = 0

Remember y (0) =0 and y’(0)=2
Y(0) = 0 = a0
Y’(0)= 2 = a1
Now knowing that we will plug in a0 and a1
2a2 + 2(2) + 0 =0
2a2 + 4 =0 2a2 =-4 a2 =-2

6a3 + 4(-2) + (2) = 0
6a3 -6 = 0 a3=1
12a4 + 6a3 + a2 = 0
12a4 + 6(1) + (2) = 0
12a4 + 8 = 0 a4= -(1/3)
Now you substitute a0 , a1, a2, a3, and a4 into the first equation.(note we are only doing up to the 5th term if the problem wants up to the 6th term you would find up to a5 )
y= a0 + a1x + a2x^(2) + a3x^(3) + a4x^(4)……….+anx^(n)
Producing
y= 0+ 2x – 2x^(2) + 1x^(3) – (1/3)x^(4)……….+anx^(n)
Then you plug in x in this case x = 1
y(2)= 0+ 2(1) – 2(1)^(2) + 1(1)^(3) – (1/3)(1)^(4)
y(2)= 0+ 2 – 2+ 1 – (1/3)
y(2)= – (2/3)