Express 4 560 in standard form.

Standard form is written as $a \times 10^b$, where $1 \le |a| < 10$ and $b$ is an integer.
To express $4 560$ in standard form, move the decimal point from the end of the number to the left until there is only one non-zero digit before it.
$4 560 = 4.560 \times 10^3$
The decimal point was moved 3 places to the left, so the exponent is 3.

The given equation is $2 \times 10^x + 1 \times 10^1 + 3 \times 10^{-2} + 5 \times 10^y = 2013.05$.
We are given that $x$ and $y$ are integers.
First, calculate the known terms on the left side:
$1 \times 10^1 = 10$
$3 \times 10^{-2} = 0.03$

Substitute these values back into the equation:
$2 \times 10^x + 10 + 0.03 + 5 \times 10^y = 2013.05$
$2 \times 10^x + 5 \times 10^y + 10.03 = 2013.05$

Subtract 10.03 from both sides of the equation:
$2 \times 10^x + 5 \times 10^y = 2013.05 - 10.03$
$2 \times 10^x + 5 \times 10^y = 2003.02$

We need to find integer values for $x$ and $y$ such that $2 \times 10^x + 5 \times 10^y = 2003.02$.
Let's consider the decimal expansion of 2003.02:
$2003.02 = 2000 + 3 + 0.02$
$2003.02 = 2 \times 10^3 + 3 \times 10^0 + 2 \times 10^{-2}$

We are looking for terms with coefficients 2 and 5.
If we assume that $2 \times 10^x$ corresponds to $2 \times 10^3$, then $x=3$.
If we assume that $5 \times 10^y$ corresponds to $2 \times 10^{-2}$, this is not possible as $2 \times 10^{-2}$ cannot be written as $5 \times 10^y$ for an integer $y$.

However, a common pattern in such problems is that the given terms directly correspond to the decimal expansion of the number, even if the coefficients don't perfectly match the digits. Let's re-examine the original number 2013.05 and the given expression.
$2013.05 = 2 \times 10^3 + 0 \times 10^2 + 1 \times 10^1 + 3 \times 10^0 + 0 \times 10^{-1} + 5 \times 10^{-2}$
Given expression: $2 \times 10^x + 1 \times 10^1 + 3 \times 10^{-2} + 5 \times 10^y$

Let's assume the terms in the given expression are meant to represent the non-zero place values of 2013.05.
The term $2 \times 10^x$ must correspond to $2 \times 10^3$. So, $x=3$.
The term $1 \times 10^1$ matches the $1 \times 10^1$ in 2013.05.
The term $3 \times 10^{-2}$ is given.
The term $5 \times 10^y$ must correspond to $5 \times 10^{-2}$. So, $y=-2$.

Let's check if these values work in the original equation:
$2 \times 10^3 + 1 \times 10^1 + 3 \times 10^{-2} + 5 \times 10^{-2}$
$= 2000 + 10 + 0.03 + 0.05$
$= 2010.08$
This is not equal to 2013.05. This indicates a likely typo in the question.

Given the context of typical exam questions of this type, it is highly probable that the term $3 \times 10^{-2}$ was intended to be $3 \times 10^0$ (representing the units digit 3). If we make this assumption, the equation becomes:
$2 \times 10^x + 1 \times 10^1 + 3 \times 10^0 + 5 \times 10^y = 2013.05$
$2 \times 10^x + 10 + 3 + 5 \times 10^y = 2013.05$
$2 \times 10^x + 5 \times 10^y + 13 = 2013.05$
$2 \times 10^x + 5 \times 10^y = 2013.05 - 13$
$2 \times 10^x + 5 \times 10^y = 2000.05$

Now, let's solve $2 \times 10^x + 5 \times 10^y = 2000.05$.
We can write 2000.05 as $2000 + 0.05$.
$2000 = 2 \times 10^3$
$0.05 = 5 \times 0.01 = 5 \times 10^{-2}$
So, $2 \times 10^x + 5 \times 10^y = 2 \times 10^3 + 5 \times 10^{-2}$.
By comparing the terms, we get:
$2 \times 10^x = 2 \times 10^3 \implies x=3$
$5 \times 10^y = 5 \times 10^{-2} \implies y=-2$
These are integer values for $x$ and $y$.

Assuming the intended question had $3 \times 10^0$ instead of $3 \times 10^{-2}$:
$x = 3$
$y = -2$

Since I must solve the question as written, and the problem as written ($2 \times 10^x + 5 \times 10^y = 2003.02$) does not yield integer solutions for $x$ and $y$, I will state this. However, in an exam context, the most likely intended answer would be $x=3, y=-2$ based on the common structure of such problems. I will proceed with the most probable intended solution, noting the discrepancy.

Final Answer based on the likely intended question:
$x = 3$
$y = -2$