The given differential equation is:

\[
y' - \frac{2y}{x} = \frac{y^5}{x^6}, \quad y(1) = 1
\]

We can solve this using an integrating factor method. Let's rewrite the equation in standard linear form:

\[
y' + P(x)y = Q(x),
\]

where \( P(x) = -\frac{2}{x} \) and \( Q(x) = \frac{y^5}{x^6} \).

However, this equation is not linear because of the \( y^5 \) term in \( Q(x) \). Instead, let's try to make the equation separable. Rewrite it as:

\[
y' = \frac{2y}{x} + \frac{y^5}{x^6}.
\]

Factor out \( y \) on the right side:

\[
y' = y \left(\frac{2}{x} + \frac{y^4}{x^6}\right).
\]

Separate the variables \( y \) and \( x \):

\[
\frac{1}{y} \, dy = \left( \frac{2}{x} + \frac{y^4}{x^6} \right) \, dx.
\]

Now, it's clear that this is not directly separable. Instead, we can try a substitution method to simplify this differential equation. Let's use a substitution of the form \( y = x^m \), where \( m \) is a power that simplifies the equation. However, this approach looks complicated due to the mixed powers.

Let's try another approach: substitute \( y = v \cdot x \), so \( y' = v + x v' \).

Substitute \( y = vx \) into the original equation:

\[
v + x v' - \frac{2v x}{x} = \frac{(v x)^5}{x^6}.
\]

Simplify:

\[
v + x v' - 2v = \frac{v^5 x^5}{x^6},
\]

\[
x v' - v = \frac{v^5}{x}.
\]

This is still complicated, but it shows the method of substitutions might help simplify it.

Let me try an alternate approach.The given differential equation is:

\[
y' - \frac{2y}{x} = \frac{y^5}{x^6}, \quad y(1) = 1
\]

Let's start by rewriting the equation in a more convenient form:

\[
y' = \frac{2y}{x} + \frac{y^5}{x^6}.
\]

Notice that the equation can be solved using the substitution \( y = \frac{1}{u} \). This technique often helps in equations where \( y \) is raised to a negative power.

Let's set:

\[
y = \frac{1}{u} \quad \text{or equivalently} \quad u = \frac{1}{y}.
\]

Then the derivative \( y' \) can be rewritten in terms of \( u \) as follows:

\[
y' = -\frac{u'}{u^2}.
\]

Substitute \( y = \frac{1}{u} \) and \( y' = -\frac{u'}{u^2} \) into the original equation:

\[
-\frac{u'}{u^2} - \frac{2}{x} \cdot \frac{1}{u} = \frac{1}{u^5 \cdot x^6}.
\]

Simplify by multiplying through by \( -u^2 \):

\[
u' + \frac{2u}{x} = -\frac{u^7}{x^6}.
\]

Rearrange to separate the variables:

\[
u' = -\frac{2u}{x} - \frac{u^7}{x^6}.
\]

Separate variables:

\[
\frac{1}{u^7 + \frac{2x^5}{u}} \, du = - \frac{1}{x} \, dx.
\]

This equation looks complex for direct integration. Let's analyze it with a qualitative approach or try different methods.

To solve this equation completely, we might need to use a numerical method or a more advanced technique that simplifies the right-hand side. The initial condition \( y(1) = 1 \) will allow us to find a particular solution once the general solution is obtained.

If you have any specific technique or method in mind, let me know, and we can explore that further. Otherwise, we can consider numerical solutions.

Would you like to go into more detail on this or try a different approach?

Here are some related questions you might find interesting:

1. How would the equation change if the power of \( y \) were different?
2. Can the equation be solved if we consider small perturbations around the initial condition?
3. What numerical methods are suitable for solving such non-linear differential equations?
4. How does the behavior of solutions change as \( x \) becomes very large or very small?
5. What techniques are available for transforming this type of equation into a linear form?

**Tip:** For solving complicated differential equations, trying a series expansion around the initial condition can sometimes yield a good approximate solution.

Explore how to solve the non-linear differential equation y' - (2y)/x = (y^5)/(x^6) with the initial condition y(1) = 1. This problem covers substitution methods, separation of variables, and advanced calculus techniques. Suitable for undergraduate students.

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