The solution of the differential equation $$ydx - (x + 2y^2)dy = 0$$ is $$x = f(y)$$. If $$f(-1) = 1$$, then $$f(1)$$ is equal to

The given differential equation is $$ y dx - (x + 2y^2) dy = 0 $$. We need to solve for $$ x $$ as a function of $$ y $$, denoted as $$ x = f(y) $$, and then find $$ f(1) $$ given that $$ f(-1) = 1 $$.

First, rearrange the equation to isolate $$ \frac{dx}{dy} $$. Starting from $$ y dx = (x + 2y^2) dy $$, divide both sides by $$ dy $$ to get:

$$ y \frac{dx}{dy} = x + 2y^2 $$

Now, bring all terms involving $$ x $$ to one side:

$$ y \frac{dx}{dy} - x = 2y^2 $$

Divide both sides by $$ y $$ (assuming $$ y \neq 0 $$) to put the equation in standard linear form:

$$ \frac{dx}{dy} - \frac{1}{y} x = 2y $$

This is a linear differential equation of the form $$ \frac{dx}{dy} + P(y) x = Q(y) $$, where $$ P(y) = -\frac{1}{y} $$ and $$ Q(y) = 2y $$. The integrating factor (IF) is calculated as:

$$ \text{IF} = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln |y|} $$

Since $$ e^{-\ln |y|} = \frac{1}{|y|} $$, but in practice, we use $$ \frac{1}{y} $$ for simplicity, as it will work for both positive and negative $$ y $$ domains. Thus, the integrating factor is $$ \frac{1}{y} $$. Multiply both sides of the equation by this integrating factor:

$$ \frac{1}{y} \cdot \frac{dx}{dy} - \frac{1}{y} \cdot \frac{1}{y} x = \frac{1}{y} \cdot 2y $$

Simplify the right side:

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

The left side is the derivative of $$ \frac{x}{y} $$ with respect to $$ y $$:

$$ \frac{d}{dy} \left( \frac{x}{y} \right) = 2 $$

Integrate both sides with respect to $$ y $$:

$$ \int \frac{d}{dy} \left( \frac{x}{y} \right) dy = \int 2 dy $$

$$ \frac{x}{y} = 2y + C $$

where $$ C $$ is the constant of integration. Solve for $$ x $$:

$$ x = 2y^2 + C y $$

Thus, the solution is $$ x = f(y) = 2y^2 + C y $$.

Given the condition $$ f(-1) = 1 $$, substitute $$ y = -1 $$ and $$ x = 1 $$:

$$ 1 = 2(-1)^2 + C(-1) $$

$$ 1 = 2(1) - C $$

$$ 1 = 2 - C $$

Solve for $$ C $$:

$$ C = 2 - 1 = 1 $$

Therefore, the function is $$ f(y) = 2y^2 + y $$.

Now, find $$ f(1) $$:

$$ f(1) = 2(1)^2 + (1) = 2 + 1 = 3 $$

Hence, the correct answer is Option B.