If $$y(x)$$ is the solution of the differential equation $$(x + 2)\frac{dy}{dx} = x^2 + 4x - 9$$, $$x \neq -2$$ and $$y(0) = 0$$, then $$y(-4)$$ is equal to

We are given the differential equation $$(x + 2) \frac{dy}{dx} = x^2 + 4x - 9$$ with $$x \neq -2$$ and the initial condition $$y(0) = 0$$. We need to find $$y(-4)$$.

First, rearrange the equation to solve for $$\frac{dy}{dx}$$:

$$\frac{dy}{dx} = \frac{x^2 + 4x - 9}{x + 2}$$

Simplify the right-hand side by performing polynomial division. Divide $$x^2 + 4x - 9$$ by $$x + 2$$:

Divide the leading term $$x^2$$ by $$x$$, which gives $$x$$. Multiply $$x$$ by $$x + 2$$: $$x \cdot (x + 2) = x^2 + 2x$$. Subtract this from $$x^2 + 4x - 9$$:

$$(x^2 + 4x - 9) - (x^2 + 2x) = 2x - 9$$

Now, divide $$2x$$ by $$x$$, which gives $$2$$. Multiply $$2$$ by $$x + 2$$: $$2 \cdot (x + 2) = 2x + 4$$. Subtract this from $$2x - 9$$:

$$(2x - 9) - (2x + 4) = -13$$

So, the division yields:

$$\frac{x^2 + 4x - 9}{x + 2} = x + 2 - \frac{13}{x + 2}$$

Thus, the differential equation becomes:

$$\frac{dy}{dx} = x + 2 - \frac{13}{x + 2}$$

Separate the variables:

$$dy = \left( x + 2 - \frac{13}{x + 2} \right) dx$$

Integrate both sides:

$$\int dy = \int \left( x + 2 - \frac{13}{x + 2} \right) dx$$

The left side integrates to $$y$$. For the right side, integrate term by term:

$$\int (x + 2) dx = \frac{x^2}{2} + 2x$$

$$\int -\frac{13}{x + 2} dx = -13 \int \frac{1}{x + 2} dx = -13 \ln |x + 2|$$

Combining these, we get:

$$y = \frac{x^2}{2} + 2x - 13 \ln |x + 2| + C$$

where $$C$$ is the constant of integration.

Apply the initial condition $$y(0) = 0$$. Substitute $$x = 0$$ and $$y = 0$$:

$$0 = \frac{(0)^2}{2} + 2(0) - 13 \ln |0 + 2| + C$$

Simplify:

$$0 = 0 + 0 - 13 \ln 2 + C$$

Solve for $$C$$:

$$C = 13 \ln 2$$

Substitute $$C$$ back into the solution:

$$y(x) = \frac{x^2}{2} + 2x - 13 \ln |x + 2| + 13 \ln 2$$

This can also be written as:

$$y(x) = \frac{x^2}{2} + 2x + 13 \ln \left( \frac{2}{|x + 2|} \right)$$

Now, evaluate $$y(-4)$$:

$$y(-4) = \frac{(-4)^2}{2} + 2(-4) - 13 \ln |-4 + 2| + 13 \ln 2$$

Compute each term:

$$(-4)^2 = 16, \quad \frac{16}{2} = 8$$

$$2 \cdot (-4) = -8$$

$$|-4 + 2| = |-2| = 2, \quad \ln | -4 + 2 | = \ln 2$$

So,

$$y(-4) = 8 - 8 - 13 \ln 2 + 13 \ln 2 = 0 + 0 = 0$$

Hence, $$y(-4) = 0$$.

The options are:

A. $$-1$$

B. $$1$$

C. $$0$$

D. $$2$$

So, the answer is Option C.