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.