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This is a second-order linear differential equation:

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

with initial conditions \( y(1) = 4 \) and \( y'(1) = -5 \).

### Solution Outline:
1. **Solve the Homogeneous Equation:**
   - This equation is in Cauchy-Euler form. To solve it, we can assume a solution of the form \( y = x^r \).
   - Substitute \( y = x^r \), \( y' = r x^{r-1} \), and \( y'' = r(r-1)x^{r-2} \) into the differential equation to obtain the characteristic equation.

2. **Characteristic Equation:**
   Substituting \( y = x^r \) into the original equation, we get:
   \[
   x^2 (r(r-1)x^{r-2}) + 2x(r x^{r-1}) - 6x^r = 0,
   \]
   simplifying to:
   \[
   r(r-1) + 2r - 6 = 0.
   \]
   This is a quadratic equation in \( r \).

3. **Find \( r \)-values:**
   Solving the quadratic \( r^2 - r + 2r - 6 = 0 \), we get:
   \[
   r^2 + r - 6 = 0.
   \]
   Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = 1 \), and \( c = -6 \):
   \[
   r = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}.
   \]
   Thus, \( r = 2 \) and \( r = -3 \).

4. **General Solution:**
   The general solution for the differential equation is:
   \[
   y = C_1 x^2 + C_2 x^{-3}.
   \]

5. **Find \( C_1 \) and \( C_2 \):**
   Use the initial conditions \( y(1) = 4 \) and \( y'(1) = -5 \) to solve for \( C_1 \) and \( C_2 \).

   - \( y(1) = 4 \): 
     \[
     C_1 (1)^2 + C_2 (1)^{-3} = 4 \Rightarrow C_1 + C_2 = 4.
     \]

   - \( y'(x) = 2C_1 x - 3C_2 x^{-4} \).
     Using \( y'(1) = -5 \):
     \[
     2C_1 - 3C_2 = -5.
     \]

6. **Solve the System of Equations:**
   Solving \( C_1 + C_2 = 4 \) and \( 2C_1 - 3C_2 = -5 \) simultaneously, we get \( C_1 = 1 \) and \( C_2 = 3 \).

### Final Solution:
\[
y = x^2 + 3x^{-3}.
\]

Would you like more details on any step or have specific questions? Here are some related topics:

1. How to identify Cauchy-Euler differential equations?
2. Explanation on solving second-order linear differential equations.
3. Importance of initial conditions in determining unique solutions.
4. Examples of other differential equation forms.
5. Applications of Cauchy-Euler equations in physics and engineering.

**Tip:** Always double-check initial conditions when solving differential equations, as they ensure the uniqueness of the solution.

Find y as a function of x if x^2 y'' + 2xy' - 6y = 0, with initial conditions y(1) = 4 and y'(1) = -5.

Learn how to solve the second-order Cauchy-Euler differential equation x^2 y'' + 2xy' - 6y = 0, with initial conditions y(1) = 4 and y'(1) = -5. This solution covers the characteristic equation, general solution, and application of initial conditions.

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