Solve (x+2)dy/dx − 2y = (x+2)⁻¹

The given equation is:

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

Divide the entire equation by $(x+2)$:

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

Here, $P(x) = -\frac{2}{x+2}$ and $Q(x) = (x+2)^{-2}$.

The integrating factor is given by $\mu(x) = e^{\int P(x) \, dx}$.
First, find $\int P(x) \, dx$:

$$\int -\frac{2}{x+2} \, dx = -2 \int \frac{1}{x+2} \, dx = -2 \ln|x+2| = \ln((x+2)^{-2})$$

Now, calculate $\mu(x)$:

$$\mu(x) = e^{\ln((x+2)^{-2})} = (x+2)^{-2}$$

Multiplying $\frac{dy}{dx} - \frac{2}{x+2}y = (x+2)^{-2}$ by $(x+2)^{-2}$:

$$(x+2)^{-2} \frac{dy}{dx} - 2(x+2)^{-3}y = (x+2)^{-4}$$

The left side of this equation is the derivative of the product $\mu(x)y$:

$$\frac{d}{dx} \left( (x+2)^{-2} y \right) = (x+2)^{-4}$$

$$\int \frac{d}{dx} \left( (x+2)^{-2} y \right) \, dx = \int (x+2)^{-4} \, dx$$ $$(x+2)^{-2} y = \frac{(x+2)^{-3}}{-3} + C$$ $$(x+2)^{-2} y = -\frac{1}{3(x+2)^3} + C$$

Multiply both sides by $(x+2)^2$:

$$y = (x+2)^2 \left( -\frac{1}{3(x+2)^3} + C \right)$$ $$y = -\frac{(x+2)^2}{3(x+2)^3} + C(x+2)^2$$ $$y = -\frac{1}{3(x+2)} + C(x+2)^2$$

The general solution is:

$$y = C(x+2)^2 - \frac{1}{3(x+2)}$$