$X\sim { \mathcal{N}( \mu , \sigma^2) } $ a normal r.v, $Y = \frac{X-\mu}{\sigma} $
The solution says $\mathbb{E}(e^{\lambda X})= \mathbb{E}(e^{\sigma \lambda Y}) e^{\lambda \mu} = e^{\sigma^2 \lambda^2 + \lambda \mu} $
But what I find is :
$\mathbb{E}(e^{\lambda X}) = e^{ \frac{ \sigma^2 \lambda^2}{2} + \lambda \mu} $
I can't see that I made a calculation mistake:
$ \mathbb{E}( \frac{X-\mu}{\sigma} ) = \int y \frac{1}{\sqrt{2 \pi}} e^{-y^2/2} $ which gives $Y$ ~ ${ \mathcal{N}( 0 , 1) } $
$ \mathbb{E}(e^{ \lambda Y} ) = \int e^{\lambda y} \frac{1}{\sqrt{2 \pi}} e^{-y^2/2} = e^{\lambda^2/2} \int \frac{1}{\sqrt{2 \pi}} e^{-(y-\lambda)^2/2} = e^{\lambda^2/2}$
Now for $e^{\lambda X}$:
$ \mathbb{E}(e^{ \lambda X} ) = \mathbb{E}(e^{ \lambda \sigma Y + \mu \lambda} ) $ $= e^{\lambda \mu}\mathbb{E}(e^{ \lambda \sigma Y } ) $ $= e^{\lambda \mu} \int e^{\lambda \sigma y } \frac{1}{\sqrt{2 \pi}} e^{-y^2/2} $
$= e^{\lambda \mu + \mu ^2\lambda^2/2} \int \frac{1}{\sqrt{2 \pi}} e^{-(y-\sigma \lambda)^2/2} = e^{\lambda \mu + \mu ^2\lambda^2/2}$
EDIT : I made a typing mistake, I meant :
$= e^{\lambda \mu + \sigma ^2\lambda^2/2} \int \frac{1}{\sqrt{2 \pi}} e^{-(y-\sigma \lambda)^2/2} = e^{\lambda \mu + \sigma ^2\lambda^2/2}$
Your last two equalities are wrong. Note that $\lambda \sigma y-y^{2}/2=-(y-\lambda \sigma )^{2}/2+\lambda ^{2} \sigma ^{2}/2$.
EDIT: Your edited answer is correct and the given answer is wrong.