= Solution
The <moment-generating function> is $M_Y(t)=\mathbb E[e^{tY}]$ for real $t$ where this expectation is finite; a finite neighborhood of zero is usually required when speaking of an existing <moment-generating function>. For a <standard normal distribution>, completing the square gives
$$
M_Y(t)=\frac1{\sqrt{2\pi}}\int_{\mathbb R}e^{ty-y^2/2}\,dy
=e^{t^2/2}\frac1{\sqrt{2\pi}}\int_{\mathbb R}e^{-(y-t)^2/2}\,dy.
$$
The translated <Gaussian density> integrates to one, so
$$
\boxed{M_Y(t)=e^{t^2/2}\quad(t\in\mathbb R).}
$$
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