We are tasked with solving the following initial value problem:

$y'(x^2 + 2) + 4x(y^2 + 2y + 1) = 0, \quad y(1) = -1$

Step 1: Simplify the given equation

First, let's observe that $y^2 + 2y + 1$ can be factored as $(y+1)^2$. So the equation becomes:

$y'(x^2 + 2) + 4x(y + 1)^2 = 0$

Step 2: Solve for $y'$

Rearrange the equation to solve for $y'$:

$y'(x^2 + 2) = -4x(y + 1)^2$

$y' = \frac{-4x(y + 1)^2}{x^2 + 2}$

Step 3: Separate the variables

We can now separate the variables $y$ and $x$. Write the equation as:

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

Rearrange to separate the $x$-terms and $y$-terms:

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

Step 4: Integrate both sides

Now we integrate both sides of the equation.

Left-hand side:

The integral of $\frac{1}{(y + 1)^2}$ with respect to $y$ is:

$\int \frac{dy}{(y + 1)^2} = -\frac{1}{y + 1}$

Right-hand side:

For the right-hand side, we need to integrate $\frac{-4x}{x^2 + 2}$. This can be done by recognizing it as a standard integral:

$\int \frac{-4x}{x^2 + 2} \, dx = -2 \ln(x^2 + 2)$

Thus, we now have:

$-\frac{1}{y + 1} = -2 \ln(x^2 + 2) + C$

Step 5: Solve for $y$

To isolate $y$, first simplify the equation:

$\frac{1}{y + 1} = 2 \ln(x^2 + 2) - C$

Now, take the reciprocal of both sides:

$y + 1 = \frac{1}{2 \ln(x^2 + 2) - C}$

Finally, subtract 1 from both sides:

$y = \frac{1}{2 \ln(x^2 + 2) - C} - 1$

Step 6: Use the initial condition to find $C$

We are given the initial condition $y(1) = -1$. Substitute $x = 1$ and $y = -1$ into the equation:

$-1 = \frac{1}{2 \ln(1^2 + 2) - C} - 1$

Simplify the logarithm:

$-1 = \frac{1}{2 \ln(3) - C} - 1$

Add 1 to both sides:

$0 = \frac{1}{2 \ln(3) - C}$

This means that the denominator must be infinite, which happens when $C = 2 \ln(3)$. Thus, the value of $C$ is:

$C = 2 \ln(3)$

Step 7: Write the final solution

Substituting $C = 2 \ln(3)$ into the equation for $y$, we get the solution:

$y = \frac{1}{2 \ln(x^2 + 2) - 2 \ln(3)} - 1$

Step 8: Interval of validity

The solution is valid as long as the denominator is non-zero, which means:

$2 \ln(x^2 + 2) - 2 \ln(3) \neq 0$

Simplifying this:

$\ln(x^2 + 2) \neq \ln(3)$

$x^2 + 2 \neq 3$

$x^2 \neq 1$

Thus, $x \neq \pm 1$. Since we are given $y(1) = -1$, the interval of validity must exclude $x = -1$, and the solution is valid for $x > 0$.

Final Answer:

The solution is:

$y = \frac{1}{2 \ln(x^2 + 2) - 2 \ln(3)} - 1$

with an interval of validity $(0, \infty)$.

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Would you like further explanation on any step, or any clarifications? Here's a list of related questions for practice:

1. How do you handle integrating rational expressions in differential equations?
2. What is the method to find the interval of validity for a differential equation solution?
3. How do logarithmic integrals appear in solving differential equations?
4. What happens when a term in a solution causes division by zero?
5. How can we check the correctness of a solution using the initial condition?

Tip: When solving separable differential equations, always check if the variables are properly separated before integrating.