We are given the second-order linear differential equation:

\[
x^2 y'' + 2x y' - 42 y = 0
\]

with initial conditions:
\[
y(1) = -8, \quad y'(1) = 0.
\]

### Step 1: Solve the homogeneous differential equation

This is a Cauchy-Euler (or Euler-Cauchy) equation of the form:
\[
x^2 y'' + a x y' + b y = 0.
\]
In our case, \( a = 2 \) and \( b = -42 \).

The general solution to a Cauchy-Euler equation can be found by assuming a solution of the form:
\[
y(x) = x^r,
\]
where \( r \) is a constant. We differentiate \( y(x) \) with respect to \( x \):
\[
y'(x) = r x^{r-1}, \quad y''(x) = r(r-1) x^{r-2}.
\]

### Step 2: Substitute into the differential equation

Substitute \( y(x) = x^r \), \( y'(x) = r x^{r-1} \), and \( y''(x) = r(r-1) x^{r-2} \) into the original equation:
\[
x^2 [r(r-1) x^{r-2}] + 2x [r x^{r-1}] - 42 x^r = 0.
\]
Simplify each term:
\[
r(r-1) x^r + 2r x^r - 42 x^r = 0.
\]
Factor out \( x^r \):
\[
x^r [r(r-1) + 2r - 42] = 0.
\]
We are left with the characteristic equation:
\[
r(r-1) + 2r - 42 = 0.
\]
Simplify:
\[
r^2 + r - 42 = 0.
\]

### Step 3: Solve the characteristic equation

Solve the quadratic equation \( r^2 + r - 42 = 0 \) using the quadratic formula:
\[
r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-42)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 168}}{2} = \frac{-1 \pm \sqrt{169}}{2} = \frac{-1 \pm 13}{2}.
\]
The two roots are:
\[
r_1 = \frac{-1 + 13}{2} = 6, \quad r_2 = \frac{-1 - 13}{2} = -7.
\]

### Step 4: General solution

The general solution to the differential equation is:
\[
y(x) = C_1 x^6 + C_2 x^{-7}.
\]

### Step 5: Apply initial conditions

We now use the initial conditions to determine \( C_1 \) and \( C_2 \).

- **Initial condition 1**: \( y(1) = -8 \)

Substitute \( x = 1 \) into the general solution:
\[
y(1) = C_1 (1)^6 + C_2 (1)^{-7} = C_1 + C_2 = -8.
\]
Thus, we have:
\[
C_1 + C_2 = -8. \quad (1)
\]

- **Initial condition 2**: \( y'(1) = 0 \)

First, differentiate \( y(x) \):
\[
y'(x) = 6C_1 x^5 - 7C_2 x^{-8}.
\]
Substitute \( x = 1 \) into the derivative:
\[
y'(1) = 6C_1 (1)^5 - 7C_2 (1)^{-8} = 6C_1 - 7C_2 = 0.
\]
Thus, we have:
\[
6C_1 - 7C_2 = 0. \quad (2)
\]

### Step 6: Solve the system of equations

We now solve the system of equations:
1. \( C_1 + C_2 = -8 \)
2. \( 6C_1 - 7C_2 = 0 \)

From equation (2), solve for \( C_1 \):
\[
C_1 = \frac{7}{6} C_2.
\]
Substitute this into equation (1):
\[
\frac{7}{6} C_2 + C_2 = -8,
\]
\[
\frac{13}{6} C_2 = -8,
\]
\[
C_2 = -8 \times \frac{6}{13} = -\frac{48}{13}.
\]

Now, substitute \( C_2 = -\frac{48}{13} \) into \( C_1 = \frac{7}{6} C_2 \):
\[
C_1 = \frac{7}{6} \times \left(-\frac{48}{13}\right) = -\frac{7 \times 48}{6 \times 13} = -\frac{336}{78} = -\frac{168}{39}.
\]

### Final solution:

Thus, the solution to the differential equation is:
\[
y(x) = -\frac{168}{39} x^6 - \frac{48}{13} x^{-7}.
\]

Would you like further details or explanations?

### Related Questions:
1. What is a Cauchy-Euler equation, and when is it used?
2. How is the general solution for a Cauchy-Euler equation derived?
3. Can the method used here apply to non-homogeneous equations?
4. What would change if different initial conditions were given?
5. How does the quadratic formula help in solving second-order differential equations?

**Tip**: To solve higher-order differential equations, always try to reduce them to familiar forms like Cauchy-Euler, which simplifies the solution process.

x^2y'' + 2xy' - 42y = 0, y(1) = -8, y'(1) = 0

Step-by-step solution to the Cauchy-Euler differential equation x^2y'' + 2xy' - 42y = 0 with initial conditions y(1) = -8 and y'(1) = 0. Learn how to solve second-order linear differential equations using the characteristic equation and quadratic formula, suitable for undergraduate students.

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