The given differential equation is:

$\frac{x^2}{y^2 - 5} \frac{dy}{dx} = \frac{1}{2y}$

with the initial condition $y(1) = \sqrt{6}$.

Step 1: Separate the variables

We start by rearranging the equation to separate $y$ and $x$.

Multiply both sides by $y$ and rearrange terms:

$\frac{x^2}{y^2 - 5} \, dy = \frac{dx}{2y}$

Now multiply both sides by $2y$ and get:

$2y \frac{x^2}{y^2 - 5} \, dy = dx$

Separate the variables:

$\frac{2y}{y^2 - 5} \, dy = \frac{dx}{x^2}$

Step 2: Integrate both sides

Now integrate both sides separately.

1. On the left side, we use substitution. Let $u = y^2 - 5$, so $du = 2y \, dy$.

$\int \frac{2y}{y^2 - 5} \, dy = \int \frac{du}{u} = \ln|u| = \ln|y^2 - 5|$

2. On the right side, integrate $\frac{1}{x^2}$:

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

Thus, the equation becomes:

$\ln|y^2 - 5| = -\frac{1}{x} + C$

Step 3: Solve for $y$

Exponentiate both sides to eliminate the logarithm:

$|y^2 - 5| = e^{-\frac{1}{x} + C} = A e^{-\frac{1}{x}}$

where $A = e^C$ is a constant. So:

$y^2 - 5 = A e^{-\frac{1}{x}}$

Now solve for $y^2$:

$y^2 = A e^{-\frac{1}{x}} + 5$

Thus, the general solution is:

$y = \pm \sqrt{A e^{-\frac{1}{x}} + 5}$

Step 4: Apply the initial condition

We are given $y(1) = \sqrt{6}$. Substituting $x = 1$ and $y = \sqrt{6}$ into the solution:

$\sqrt{6}^2 = A e^{-\frac{1}{1}} + 5$

$6 = A e^{-1} + 5$

$1 = A e^{-1}$

Thus, $A = e$. Therefore, the particular solution becomes:

$y = \pm \sqrt{e \cdot e^{-\frac{1}{x}} + 5}$

$y = \pm \sqrt{e^{1 - \frac{1}{x}} + 5}$

Since $y(1) = \sqrt{6}$ is positive, we choose the positive root:

$y = \sqrt{e^{1 - \frac{1}{x}} + 5}$

This is the particular solution to the differential equation.

Would you like a further breakdown of any steps? Here are some related questions you might consider:

1. How do we determine the constant $A$ for different initial conditions?
2. Can you explain the method of separation of variables in more detail?
3. How does the solution change if a different initial condition is provided?
4. What are the key properties of exponential functions used in solving differential equations?
5. How do you verify that this solution satisfies the original differential equation?

Tip: Always double-check your work by substituting the particular solution into the original equation to verify that it satisfies both the equation and the initial condition.