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Two probability distributions of the discrete random variable X and Y are given below.

$\text{X}$	$0$	$1$	$2$	$3$
$\text{P}(\text{X})$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{1}{5}$	$\frac{1}{5}$

$\text{Y}$	$0$	$1$	$2$	$3$
$\text{P}(\text{Y})$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$	$\frac{1}{10}$

Prove that $E(Y^2) = 2E(X)$.