Simplify the left-hand side (LHS) by multiplying the numerator and denominator by the conjugate of the denominator.

The conjugate of $1-j$ is $1+j$.

$$\frac{2+j}{1-j} = \frac{2+j}{1-j} \times \frac{1+j}{1+j}$$

Numerator: $(2+j)(1+j) = 2(1) + 2(j) + j(1) + j(j) = 2 + 2j + j + j^2$
Since $j^2 = -1$, this becomes $2 + 3j - 1 = 1 + 3j$.
Denominator: $(1-j)(1+j) = 1^2 - j^2 = 1 - (-1) = 1 + 1 = 2$.
So, the LHS simplifies to:

$$\frac{1+3j}{2} = \frac{1}{2} + \frac{3}{2}j$$

$$j(x+jy) = jx + j^2y$$

Since $j^2 = -1$, this becomes:

$$jx - y = -y + jx$$

Equating the real parts:

$$\frac{1}{2} = -y$$ $$y = -\frac{1}{2}$$

Equating the imaginary parts:

$$\frac{3}{2} = x$$

Thus, the values of $x$ and $y$ are:

$$x = \frac{3}{2}, \quad y = -\frac{1}{2}$$