Step 1: Formulate the Differential Equation

Let $P(x, y)$ be a point on the curve. The slope of the tangent is $\frac{dy}{dx}$.

The sum of coordinates is $x + y$.

According to the problem condition (correcting for the standard interpretation): Sum of coordinates $-$ Slope $= 5$.

$$x + y - \frac{dy}{dx} = 5$$

Rearranging this, we get:

$$\frac{dy}{dx} - y = x - 5$$

Step 2: Find the Integrating Factor

This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$.

Here, $P = -1$ and $Q = x - 5$.

The Integrating Factor (I.F.) is given by $e^{\int P \, dx}$.

$$I.F. = e^{\int -1 \, dx} = e^{-x}$$

Step 3: Apply the Solution Formula

The solution is given by:

$$y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + C$$

Substituting the values:

$$y e^{-x} = \int (x - 5) e^{-x} \, dx + C$$

Step 4: Integrate the Right-Hand Side

Using integration by parts, let $u = x - 5$ and $dv = e^{-x} dx$.

$$\int (x - 5) e^{-x} \, dx = (x - 5) \left( \frac{e^{-x}}{-1} \right) - \int (1) \left( \frac{e^{-x}}{-1} \right) dx$$ $$= -(x - 5) e^{-x} + \int e^{-x} dx$$ $$= -(x - 5) e^{-x} - e^{-x}$$ $$= -x e^{-x} + 5 e^{-x} - e^{-x}$$ $$= -x e^{-x} + 4 e^{-x}$$

Step 5: Find the General Solution

Substitute the integral back into the equation:

$$y e^{-x} = -x e^{-x} + 4 e^{-x} + C$$

Divide by $e^{-x}$ (or multiply by $e^x$):

$$y = -x + 4 + C e^x$$

Step 6: Find the Particular Solution

The curve passes through the point $(0, 2)$. Substitute $x = 0$ and $y = 2$ into the general solution.

$$2 = -(0) + 4 + C e^0$$ $$2 = 4 + C$$ $$C = -2$$

Step 7: Write the Final Equation

Substitute $C = -2$ back into the general solution:

$$y = -x + 4 - 2e^x$$

or

$$y = 4 - x - 2e^x$$