$$ ( 1 - \frac { 2 } { y } + x ) \frac { d y } { d x } + y = \frac { 2 } { x } - 1 $$

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We start with the differential equation

$$
\left(1-\frac{2}{y}+x\right)\frac{dy}{dx}+y=\frac{2}{x}-1.
$$

Step 1. Rewriting as an equation in differentials

Subtract $$y$$ from both sides and write

$$
\left(1-\frac{2}{y}+x\right)\frac{dy}{dx} = \frac{2}{x}-1-y.
$$

Now multiply both sides by $$dx$$ to obtain

$$
\left(1-\frac{2}{y}+x\right)dy = \left(\frac{2}{x}-1-y\right)dx.
$$

Bring all terms to one side:

$$
\left(\frac{2}{x}-1-y\right)dx - \left(1-\frac{2}{y}+x\right)dy = 0.
$$

Step 2. Recognize the exact differential form

We identify

$$
M(x,y)=\frac{2}{x}-1-y,\quad N(x,y)=-\left(1-\frac{2}{y}+x\right)=-1+\frac{2}{y}-x.
$$

A differential equation of the form

$$
M(x,y)dx+N(x,y)dy=0
$$

is exact if

$$
\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}.
$$

Calculate these partial derivatives:

$$
\frac{\partial M}{\partial y}=-1,
$$
$$
\frac{\partial N}{\partial x}=-1.
$$

Since they are equal, the equation is exact.

Step 3. Find the potential function $$\Psi(x,y)$$

Since the equation is exact, there is a function $$\Psi(x,y)$$ such that

$$
\Psi_x(x,y)=M(x,y)=\frac{2}{x}-1-y, \quad \Psi_y(x,y)=N(x,y)=-1+\frac{2}{y}-x.
$$

Integrate $$M(x,y)$$ with respect to $$x$$ (treating $$y$$ as a constant):

$$
\Psi(x,y)=\int \left(\frac{2}{x}-1-y\right)dx = 2\ln|x| - x -xy + h(y),
$$

where $$h(y)$$ is an arbitrary function of $$y$$.

Step 4. Determine $$h(y)$$

Differentiate $$\Psi(x,y)$$ with respect to $$y$$:

$$
\Psi_y(x,y)=-x + h'(y).
$$

Set this equal to $$N(x,y)$$:

$$
-x+h'(y)=-1+\frac{2}{y}-x.
$$

Cancel $$-x$$ from both sides to find:

$$
h'(y)=-1+\frac{2}{y}.
$$

Integrate with respect to $$y$$:

$$
h(y)=\int \left(-1+\frac{2}{y}\right)dy = -y+2\ln|y|+C_1.
$$

We may absorb the constant $$C_1$$ into the final constant of integration.

Step 5. Write the general solution

Substitute $$h(y)$$ back into $$\Psi(x,y)$$:

$$
\Psi(x,y)=2\ln|x|-x-xy -y+2\ln|y|.
$$

Since $$\Psi(x,y)=C$$, the general solution is

$$
2\ln|x|-x-xy-y+2\ln|y|=C.
$$

This implicit equation represents the general solution to the differential equation.
The general solution to the differential equation is: $$ 2\ln|x| - x - xy - y + 2\ln|y| = C $$ where $$ C $$ is a constant.

\( ( 1 - \frac { 2 } { y } + x ) \frac { d y } { d x } + y = \frac { 2 } { x } - 1 \)

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