We are given the following equation:

$\sqrt{1 + y^2} \cdot x \, dx + y \cdot (4 + x^2) \, dy = 0$

This is a first-order differential equation. To solve it, we'll attempt to simplify or find an integrating factor or a substitution that can help separate the variables or make the equation easier to solve.

Step 1: Rearrange the equation

We can try to isolate terms involving $dx$ and $dy$ on opposite sides. First, let’s rewrite the equation clearly:

$\sqrt{1 + y^2} \cdot x \, dx = -y \cdot (4 + x^2) \, dy$

Now, divide both sides by $y \cdot \sqrt{1 + y^2}$ to attempt separation of variables:

$\frac{x}{y} \cdot \frac{dx}{4 + x^2} = - \frac{dy}{\sqrt{1 + y^2}}$

Step 2: Integrate both sides

Now, we have a separable equation. Integrate both sides:

1. The left side becomes:

$\int \frac{x}{4 + x^2} \, dx$

This can be solved using a simple substitution. Let $u = 4 + x^2$, so $du = 2x \, dx$, giving:

$\int \frac{1}{2} \cdot \frac{du}{u} = \frac{1}{2} \ln |4 + x^2|$

2. The right side is:

$\int -\frac{dy}{\sqrt{1 + y^2}}$

This is a standard integral, and the result is:

$-\ln |y + \sqrt{y^2 + 1}|$

Step 3: Combine the results

Putting everything together, we have:

$\frac{1}{2} \ln |4 + x^2| = -\ln |y + \sqrt{y^2 + 1}| + C$

Step 4: Simplify the equation

Multiply through by 2:

$\ln |4 + x^2| = -2 \ln |y + \sqrt{y^2 + 1}| + C'$

This can be rewritten as:

$\ln |4 + x^2| + 2 \ln |y + \sqrt{y^2 + 1}| = C'$

Using properties of logarithms, this simplifies to:

$\ln \left( (4 + x^2) \cdot (y + \sqrt{y^2 + 1})^2 \right) = C'$

Finally, exponentiate both sides:

$(4 + x^2) \cdot (y + \sqrt{y^2 + 1})^2 = C$

where $C = e^{C'}$ is a new constant.

This is the general solution to the differential equation.

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Do you want any further details or clarifications?

Here are 5 questions to explore further:

1. How would you handle solving an implicit equation after separation of variables?
2. Can you apply this technique to solve other separable differential equations?
3. What are other methods of solving first-order differential equations?
4. How do you recognize when a differential equation is separable?
5. How does substitution simplify the integration of certain terms?

Tip: When dealing with complex differentials, always check if the equation can be made separable or if an integrating factor is appropriate.