Is this Taylor series correct taken correctly? Confused reasoning I have $dx/dy=-ay, x(0)=1$ initial value problem. 
Then $x(y)=\frac{x(0)}{0!}y^0+ \frac{x'(0)}{1!}y+\frac{x''(0)}{2!}y^2=1+(-a)y+a^2y^2...$

Given
$$
x'(y) = \frac{dx}{dy} = -ay.
$$
The second derivative can be found with
$$
x''(y) = \frac{d^2x}{dy^2} = \frac{d}{dy} \frac{dx}{dy} = -a,
$$
and thus the third derivative will be equal to
$$
x^{(3)}(y) = \frac{d^3x}{dy^3} = \frac{d}{dy} \frac{d^2x}{dy^2} = 0,
$$
any higher derivatives will also be zero.
So the solution for $x(y)$ will just be a second order polynomial of the form
$$
x(y) = x(0) + x'(0) y + \frac{x''(0)}{2} y^2 = x(0) - 0 \cdot y - \frac{a}{2} y^2 = 1 - \frac{a}{2} y^2.
$$
The same solution can be found by integrating on both sides
$$
\int{dx} = \int{-aydy},
$$
$$
x = - \frac{a}{2} y^2 + c.
$$
