Series Solution of an ODE The ODE below is required to help compute the coefficients of function. 
There isnt any information about this topic in my textbook so i am just wondering how i would go about this question?

In this question you will construct the series solution of the following ODE $dy/dx = 2y$

Of course we are making a big assumption here – that the solution $y(x)$ can be written as a Taylor series. 

(a) Assume that $y(0) = A$ for some given number $A$. Use the ODE to compute the
successive derivatives of $y(x)$ at $x = 0$.

(b) Use the result of part (a) to deduce the coefficients $a_0, a_1, a_2, a_3$ and an in the infinite
series
$y(x)=a_0 +a_1x+a_2x^2 +a_3x^3 +···+a_nx^n$

(c) Use the result of part (b) to write down $y(x)$ as an infinite series. Your answer should contain only one constant $A$ and should include the general term in the infinite series.

(d) Solve the ODE using any other method. Does this solution agree with what you found in part (c)? Justify your answer.
you can differentiate the equation repeatedly to get the following: $$y' = 2y, y'' = 2y' = 2^2y, y^{(3)}= 2^3y, \cdots, y^{n} = 2^ny, \cdots \tag 1$$

suppose $y(0) = a.$ then substituting $x = 0,$ in $(1),$ we get $$y'(0) = 2a, y''(0)=2^2 a,\cdots,y^{(n)} = 2^n a, \cdots \tag2 $$

now using the  maclaurin series $$y =y(0)+xy'(0)+\frac{x^2}{2!}y''(0)+ \cdots+ \frac{x^n}{n!}y^{(n)}(0)+\cdots$$ which gives us 
$$\begin{align}y &amp;=a+2ax+2^2\frac{x^2}{2!}a+ \cdots+ 2^n\frac{x^n}{n!}a+\cdots
\\&amp;=a\left( 1+2x+\frac{(2x)^2}{2!}+ \cdots+ \frac{(2x)^n}{n!}+\cdots \right)\\
&amp;= ae^{2x}\end{align}$$
Series solution of a first order differential equation? Also you have written $dy/dy = 2y$:

Shurely shome mishtake mish Moneypenny?

EDIT Why downvote? I know it's fine to use series solution for first order; wlog that is; but it's not really needed. Also the typo is important.