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The given differential equation is:

\[
x^2 y'' - 4x y' - 6y = 0
\]

with the initial conditions \( y(1) = -7 \) and \( y'(1) = 6 \).

This is a second-order linear differential equation with variable coefficients. Let's solve it step by step.

### Step 1: Try a solution of the form \( y = x^r \)

Assume a solution of the form \( y(x) = x^r \). Then,

- \( y' = r x^{r-1} \)
- \( y'' = r(r-1) x^{r-2} \)

Substitute these into the original equation:

\[
x^2 \cdot r(r-1) x^{r-2} - 4x \cdot r x^{r-1} - 6x^r = 0
\]

Simplifying each term:

\[
r(r-1) x^r - 4r x^r - 6x^r = 0
\]

Factoring out \( x^r \):

\[
x^r \left( r(r-1) - 4r - 6 \right) = 0
\]

For non-trivial solutions (\( x^r \neq 0 \)), we need the expression in parentheses to be zero:

\[
r(r-1) - 4r - 6 = 0
\]

This simplifies to:

\[
r^2 - r - 4r - 6 = 0
\]
\[
r^2 - 5r - 6 = 0
\]

### Step 2: Solve the characteristic equation

The characteristic equation is:

\[
r^2 - 5r - 6 = 0
\]

Using the quadratic formula:

\[
r = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-6)}}{2(1)}
\]
\[
r = \frac{5 \pm \sqrt{25 + 24}}{2}
\]
\[
r = \frac{5 \pm \sqrt{49}}{2}
\]
\[
r = \frac{5 \pm 7}{2}
\]

Thus, the two roots are:

\[
r_1 = \frac{5 + 7}{2} = 6, \quad r_2 = \frac{5 - 7}{2} = -1
\]

### Step 3: General solution

The general solution is a linear combination of the independent solutions \( x^{r_1} \) and \( x^{r_2} \):

\[
y(x) = C_1 x^6 + C_2 x^{-1}
\]

### Step 4: Apply the initial conditions

We have two initial conditions: \( y(1) = -7 \) and \( y'(1) = 6 \).

- **Condition 1: \( y(1) = -7 \)**

Substitute \( x = 1 \) into the general solution:

\[
C_1(1)^6 + C_2(1)^{-1} = -7
\]
\[
C_1 + C_2 = -7 \quad \text{(Equation 1)}
\]

- **Condition 2: \( y'(1) = 6 \)**

First, find the derivative of \( y(x) \):

\[
y'(x) = 6C_1 x^5 - C_2 x^{-2}
\]

Substitute \( x = 1 \):

\[
6C_1(1)^5 - C_2(1)^{-2} = 6
\]
\[
6C_1 - C_2 = 6 \quad \text{(Equation 2)}
\]

### Step 5: Solve the system of equations

From Equation 1: \( C_1 + C_2 = -7 \)

From Equation 2: \( 6C_1 - C_2 = 6 \)

Add these two equations:

\[
(6C_1 - C_2) + (C_1 + C_2) = 6 + (-7)
\]
\[
7C_1 = -1
\]
\[
C_1 = -\frac{1}{7}
\]

Substitute \( C_1 = -\frac{1}{7} \) into Equation 1:

\[
-\frac{1}{7} + C_2 = -7
\]
\[
C_2 = -7 + \frac{1}{7} = -\frac{49}{7} + \frac{1}{7} = -\frac{48}{7}
\]

### Step 6: Final solution

Thus, the solution to the differential equation is:

\[
y(x) = -\frac{1}{7} x^6 - \frac{48}{7} x^{-1}
\]

Would you like a detailed breakdown of any part of the solution or have further questions? Here are five related questions you might find useful:

1. How can you verify the solution of a second-order differential equation?
2. What are other methods to solve second-order linear differential equations with variable coefficients?
3. How do initial conditions affect the constants in the general solution of a differential equation?
4. What role does the characteristic equation play in solving differential equations?
5. How can you handle non-homogeneous differential equations?

**Tip:** For complex second-order equations, using a trial solution like \( y = x^r \) can greatly simplify solving the problem when the equation has variable coefficients.

Find y as a function of x if the equation is x^2 y'' - 4x y' - 6y = 0 with initial conditions y(1) = -7, y'(1) = 6.

Learn how to solve the second-order differential equation x^2 y'' - 4x y' - 6y = 0 using characteristic equations and initial conditions y(1) = -7, y'(1) = 6. Includes detailed steps on solving using trial solutions and quadratic formulas.

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