9. $$ \quad(1+2 x y) y^{\prime}=1+y^{2} $$. ÐS: $$ x=\frac{1}{2}\left[y+\left(1+y^{2}\right)(\arctan y+C)\right. $$

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To solve the differential equation

$$
(1 + 2xy) y' = 1 + y^2,
$$

we can start by rewriting it in a more manageable form. First, we can express $$ y' $$ as $$ \frac{dy}{dx} $$:

$$
(1 + 2xy) \frac{dy}{dx} = 1 + y^2.
$$

Next, we can separate the variables. Rearranging gives:

$$
\frac{dy}{1 + y^2} = \frac{dx}{1 + 2xy}.
$$

Now, we can integrate both sides. The left side can be integrated directly:

$$
\int \frac{dy}{1 + y^2} = \arctan(y) + C_1.
$$

For the right side, we need to integrate $$ \frac{dx}{1 + 2xy} $$. This can be a bit tricky, but we can use a substitution. Let $$ u = 1 + 2xy $$, then $$ du = 2y \, dx + 2x \, dy $$. However, since we are integrating with respect to $$ x $$, we can treat $$ y $$ as a constant during this integration.

The integral becomes:

$$
\int \frac{dx}{1 + 2xy} = \frac{1}{2y} \ln |1 + 2xy| + C_2.
$$

Now we have:

$$
\arctan(y) = \frac{1}{2y} \ln |1 + 2xy| + C,
$$

where $$ C = C_2 - C_1 $$.

To express $$ x $$ in terms of $$ y $$, we can rearrange this equation. Multiplying both sides by $$ 2y $$ gives:

$$
2y \arctan(y) = \ln |1 + 2xy| + 2yC.
$$

Exponentiating both sides results in:

$$
|1 + 2xy| = e^{2y \arctan(y) - 2yC}.
$$

Now, we can solve for $$ x $$:

$$
1 + 2xy = e^{2y \arctan(y) - 2yC} \quad \text{(assuming $$ 1 + 2xy > 0 $$)}.
$$

Thus,

$$
2xy = e^{2y \arctan(y) - 2yC} - 1,
$$

and finally,

$$
x = \frac{1}{2y} \left( e^{2y \arctan(y) - 2yC} - 1 \right).
$$

This gives us the implicit solution for $$ x $$ in terms of $$ y $$.

The solution can be expressed as:

$$
x = \frac{1}{2} \left[ y + (1 + y^2)(\arctan(y) + C) \right].
$$

This matches the form you provided, confirming that the solution is correct.
The solution for $$ x $$ in terms of $$ y $$ is given by: $$ x = \frac{1}{2} \left[ y + (1 + y^2)(\arctan(y) + C) \right]. $$

9. \( \quad(1+2 x y) y^{\prime}=1+y^{2} \). ÐS: \( x=\frac{1}{2}\left[y+\left(1+y^{2}\right)(\arctan y+C)\right. \)

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