We will use the Reduction of Order method to find a second linearly independent solution $y_2(x)$ to the given differential equation:

$x^2 y'' + 2x y' - 2y = 0$

Step 1: Assume the form of $y_2(x)$

Since we already know one solution, $y_1(x) = x$, we assume the second solution has the form:

$y_2(x) = v(x) y_1(x) = v(x) x$

where $v(x)$ is an unknown function to be determined.

Step 2: Compute First and Second Derivatives

Using the product rule,

$y_2' = v' x + v$

$y_2'' = v'' x + 2 v'$

Step 3: Substitute into the Differential Equation

Substituting $y_2, y_2', y_2''$ into:

$x^2 y_2'' + 2x y_2' - 2y_2 = 0$

$x^2 (v'' x + 2 v') + 2x (v' x + v) - 2(v x) = 0$

Expanding:

$x^3 v'' + 2x^2 v' + 2x^2 v' + 2x v - 2x v = 0$

$x^3 v'' + 4x^2 v' = 0$

Step 4: Solve for $v(x)$

Dividing by $x^2$ (for $x \neq 0$):

$x v'' + 4 v' = 0$

This is a first-order linear equation in $v'$. Let $u = v'$, so $u' = v''$, and rewrite:

$x u' + 4u = 0$

This is a separable equation:

$\frac{du}{dx} = -\frac{4}{x} u$

Separating variables:

$\frac{du}{u} = -\frac{4}{x} dx$

Integrating:

$\ln |u| = -4 \ln |x| + C$

$u = C x^{-4}$

Since $u = v'$, integrating again:

$v = \int C x^{-4} dx$

$v = \frac{C}{-3} x^{-3} = -\frac{C}{3} x^{-3}$

Step 5: Compute $y_2(x)$

$y_2 = v x = -\frac{C}{3} x^{-3} \cdot x$

$y_2 = -\frac{C}{3} x^{-2}$

For simplicity, let $C = 3$, so:

$y_2 = -x^{-2} = -\frac{1}{x^2}$

Step 6: Fundamental Set of Solutions

The general solution is:

$y(x) = C_1 x + C_2 \frac{1}{x^2}$

where $C_1$ and $C_2$ are arbitrary constants.

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Would you like a deeper explanation for any step?

Here are some related questions:

1. What is the general method for reduction of order?
2. How can we confirm that $y_1(x)$ and $y_2(x)$ are linearly independent?
3. Can we solve this equation using another method, such as the Frobenius method?
4. What are the conditions for applying the reduction of order technique?
5. How do we choose the function $v(x)$ when applying the reduction of order?

Tip:

Always check if your second solution is linearly independent of the first by using the Wronskian determinant:

$W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}$