Solving the Airy Equation using Laplace Transform I have been trying to read the book Airy Functions and Applications to Physics, Olivier Vallee & Manuel Soares, for research on the Airy Functions, but I am stuck on using the Laplace transform to solve the Airy equation to get the function in the correct form. 
The book states (pg 5):

We consider the following homogeneous second order differential equation called the Airy's equation $$y''-xy=0.$$ This differential equation may be solved by the method of Laplace, i.e. in seeking a solution as an integral $$y=\int_C e^{xz}v(z)dz,$$ this is equivalent to solve the first order differential equation $$ v'+z^2v=0.$$ We thus obtain the solution to the Airy's equation, except a normalisation constant, $$y=\int_C e^{xz-z^3/3}dz.$$

Below is my working:
Using the suggestion to find a solution to the Airy Equation $y''-xy=0$ in the form $y=\int_Ce^{xz}v(z)dz$, the first step would be to differentiate the expression for $y$ to find $y'$ & $y''$:
$$\frac{dy}{dx}=\int ze^{xz}v(z)dz,$$
$$\frac{d^2y}{dx^2}=\int z^2e^{xz}v(z)dz.$$
Then,
$$\int z^2e^{xz}v(z)dz-x\int e^{xz}v(z)dz=0,$$ so
$$\int e^{xz}v(z)(z^2-x)dz=0.$$
Then I tried to use integration by parts with $u=(z^2-x)v(z)=z^2v(z)-xv(z)$, so $\frac{du}{dz}=2zv(z)+z^2v'(z)-xv'(z)=2zv(z)+v'(z)(z^2-x)$, and $\frac{dv}{dz}=e^{xz}$, so $v=\frac{1}{z}e^{xz}$.
Following through with this gives:
$$v(z)(z^2-x)\frac{1}{z}e^{xz}-\int(2zv(z)+v'(z)(z^2-x))\frac{1}{z}e^{xz}dz=0.$$
This (sort of) simplified to: 
$$v(z)\frac{\left(z^2-x\right)e^{xz}}{z}-\int 2v(z)e^{xz}+v'(z)\frac{\left(z^2-x\right)e^{xz}}{z}dz=0.$$
At this point, I tried to differentiate this expression to reach the form $v'+z^2v=0$:
$$\frac{d}{dz}\left(zv(z)e^{xz}-\frac{xv(z)e^{xz}}{z}-\int 2v(z)e^{xz}dz-\int zv'(z)e^{xz}dz+\int\frac{xv'(z)e^{xz}}{z}dz\right)=0,$$
which gives
$$v(z)e^{xz}+zv'(z)e^{xz}+xzv(z)e^{xz}-\frac{xv(z)e^{xz}}{z^2}-\frac{xv'(x)e^{xz}}{z}-\frac{x^2v(z)e^{xz}}{z}-2v(z)e^{xz}-zv'(z)e^{xz}+\frac{xv'(z)e^{xz}}{z}=0.$$
The $e^{xz}$ terms all cancel through, leaving:
$$v(z)+zv'(z)+xzv(z)-\frac{xv(z)}{z^2}-\frac{xv'(z)}{z}-\frac{x^2v(z)}{z}-2v(z)-zv'(z)+\frac{xv'(z)}{z}=0.$$
From here, I removed the fractions:
$$z^2v(z)+z^3v'(z)+xz^3v(z)-xv(z)-xzv'(z)-x^2zv(z)-2z^2v(z)-z^3v'(z)+xzv'(z)=0,$$
and collected the $v(z)$ and $v'(z)$ terms:
$$v(z)\left(z^2+xz^3-x-x^2z-2z^2\right)+v'(z)\left(z^3-xz-z^3+xz\right)=0,$$
which leaves:
$$\left(xz^3-z^2-x^2z-x\right)v(z)=0.$$
This clearly does not match the form given from the book ($v'+z^2v=0$), but I cannot see where I have made the mistake.
Is my mistake something obvious I am missing, could it be due to the nature of the complex function (and if so, what), or is my error deeper? Any help would be much appreciated.
 A: You are overthinking it. Starting from the beginning
$$ \int_C z^2 e^{xz} v(z) \ dz - \int_C xe^{xz} v(z)\ dz = 0  $$
Integration by parts gives
$$ \int_C xe^{xz} v(z)\ dz = e^{xz}v(z) \Big \vert_C - \int_C e^{xz} v'(z) dz $$
Here $C$ is a vertical line on the complex plane, such that all singularities of $v(z)$ lies on the left of it.  If we hand-wave away that the first term goes to $0$, we can simplify
$$ \int_C e^{xz}(z^2 v(z) + v'(z)) \ dz = 0 $$
Notice the LHS is a function of $x$ but the RHS is identically $0$, so it must follow that
$$ v'(z) + z^2 v(z) = 0 $$
Which can be solved by separation of variables
$$ v(z) = e^{-z^3/3} $$
up to a multiplicative constant.
This answer doesn't justify everything, but hopefully it can explain some things.

It should also be noted that this text uses a less accessible form of the inverse Laplace transform which uses complex analysis. A more common definition is the forward transform
$$ v(z) = \int_0^{\infty} e^{-zx} y(x)\ dx $$
Applying the forward transform to the original equation, and using integration by parts, we can show that
\begin{align} 
\int_0^\infty e^{-zx} y''(x)\ dx &= -y'(0) + \int_0^\infty ze^{-zx} y'(x)\ dx \\ 
&= -y'(0) - zy(0) + \int_0^\infty z^2e^{-zx}y(x) dx \\ &= -zy(0) - y'(0) + z^2v(z) \\ 
\int_0^\infty e^{-zx} xy\ dx &= -\int_0^\infty \frac{d}{dz} e^{-zx} y(x)\ dx \\ 
&= -v'(z) 
\end{align}
where the second result is justified by differentiation under the integral. Then
$$ v'(z) + z^2v(z) = zy(0) + y'(0) $$
Letting $y(0) = y'(0) = 0$ leads to the given result.

Edit: Your mistake is in the very first integration by parts:
$$\frac{dv}{dz}=e^{xz} \implies v=\frac{1}{\color{red}{x}}e^{xz}$$
