We are given the following differential equation:

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

Step-by-step Solution:

1. Rearrange the equation

We can first divide both sides of the equation by $\sqrt{1 + y^2} \cdot (4 + x^2)$, which simplifies it slightly:

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

This makes it clearer to see that the equation is separable.

2. Separate variables

Now, we group all terms involving $x$ on one side and terms involving $y$ on the other side:

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

3. Integrate both sides

Integrate both sides with respect to their respective variables.

• The integral on the left side: $\int \frac{x}{4 + x^2} \, dx$ This can be done by using substitution. Let $u = 4 + x^2$, then $du = 2x \, dx$, so we get: $\frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| = \frac{1}{2} \ln|4 + x^2|$

• The integral on the right side: $\int \frac{y}{\sqrt{1 + y^2}} \, dy$ This can be solved using substitution as well. Let $v = 1 + y^2$, then $dv = 2y \, dy$, giving: $\frac{1}{2} \int \frac{dv}{\sqrt{v}} = \sqrt{v} = \sqrt{1 + y^2}$

4. Combine the results

After integrating both sides, we have:

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

Multiply both sides by 2 to simplify:

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

5. Exponentiate both sides

To remove the logarithm, exponentiate both sides:

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

Since $e^{C'}$ is just a constant, we can rewrite the equation as:

$4 + x^2 = C \cdot e^{-2\sqrt{1 + y^2}}$

This is the implicit solution to the differential equation.

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Would you like more details on any of the steps? Here are some related questions to expand your understanding:

1. What is a separable differential equation and how do you solve it?
2. How does substitution work in integrals, and why was it used here?
3. What are the steps to solve logarithmic differential equations?
4. Can you explain the integration process for $\int \frac{x}{4 + x^2} \, dx$ in more detail?
5. How can you interpret the constant $C$ in the general solution?

Tip: When integrating functions involving square roots and rational functions, substitution is a powerful tool. Always look for a substitution that simplifies the terms inside the integrand.