To solve the second-order differential equation

$x^3 y'' - x y' + y = 0, \quad x > 0$

using the method of reduction of order, given that $y_1(x) = x$ is a known solution, we proceed as follows:

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Step 1: Assume the form of the second solution

The method of reduction of order assumes that the second solution $y_2(x)$ has the form: $y_2(x) = v(x)y_1(x) = v(x)x,$ where $v(x)$ is a function to be determined.

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Step 2: Substitute $y_2(x)$ into the original equation

The derivatives of $y_2(x)$ are: $y_2'(x) = v'(x)x + v(x),$ $y_2''(x) = v''(x)x + 2v'(x).$

Substitute $y_2(x)$, $y_2'(x)$, and $y_2''(x)$ into the original equation: $x^3 y_2'' - x y_2' + y_2 = 0.$ Substitute $y_2''(x) = v''(x)x + 2v'(x)$, $y_2'(x) = v'(x)x + v(x)$, and $y_2(x) = v(x)x$: $x^3 \big(v''(x)x + 2v'(x)\big) - x\big(v'(x)x + v(x)\big) + v(x)x = 0.$

Expand each term:

1. $x^3 y_2'' = x^3 (v''(x)x + 2v'(x)) = x^4 v''(x) + 2x^3 v'(x)$,
2. $-x y_2' = -x (v'(x)x + v(x)) = -x^2 v'(x) - x v(x)$,
3. $+ y_2 = v(x)x$.

Combining terms: $x^4 v''(x) + 2x^3 v'(x) - x^2 v'(x) - x v(x) + v(x)x = 0.$

Simplify: $x^4 v''(x) + (2x^3 - x^2)v'(x) = 0.$

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Step 3: Simplify the equation

Factor out $x^2$ from the terms: $x^2 \big(x^2 v''(x) + (2x - 1)v'(x)\big) = 0.$

Since $x > 0$, divide through by $x^2$: $x^2 v''(x) + (2x - 1)v'(x) = 0.$

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Step 4: Solve the reduced equation

This is a first-order linear differential equation in $v'(x)$. Let $u(x) = v'(x)$, so $u'(x) = v''(x)$. Substitute $u(x)$ into the equation: $x^2 u'(x) + (2x - 1)u(x) = 0.$

This is a first-order linear differential equation. Rewriting: $u'(x) + \frac{(2x - 1)}{x^2}u(x) = 0.$

Use the integrating factor method: The integrating factor is: $\mu(x) = e^{\int \frac{2x - 1}{x^2} \, dx}.$ Simplify the integral: $\int \frac{2x - 1}{x^2} \, dx = \int \frac{2}{x} \, dx - \int \frac{1}{x^2} \, dx = 2\ln|x| + \frac{1}{x}.$

Thus, the integrating factor is: $\mu(x) = e^{2\ln|x| + \frac{1}{x}} = x^2 e^{1/x}.$

Multiplying through by $\mu(x)$: $x^2 e^{1/x} u'(x) + \frac{(2x - 1)}{x^2} \cdot x^2 e^{1/x} u(x) = 0.$

Simplify: $\frac{d}{dx}\big(u(x) x^2 e^{1/x}\big) = 0.$

Integrate: $u(x) x^2 e^{1/x} = C_1,$ $u(x) = \frac{C_1}{x^2 e^{1/x}}.$

Since $u(x) = v'(x)$, we have: $v'(x) = \frac{C_1}{x^2 e^{1/x}}.$

Integrate $v'(x)$ to find $v(x)$: $v(x) = \int \frac{C_1}{x^2 e^{1/x}} \, dx.$

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Step 5: General solution

Once $v(x)$ is determined, the general solution is: $y(x) = C_1 x + C_2 v(x)x.$

If you'd like, we can further simplify or solve the integral for $v(x)$. Would you like details on this part?

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Related Questions:

1. What are the main applications of the method of reduction of order in solving differential equations?
2. How does one verify that a solution satisfies a second-order differential equation?
3. What is the significance of the integrating factor in solving first-order linear ODEs?
4. Can we use reduction of order for non-linear differential equations? Why or why not?
5. What is the geometric interpretation of the solutions to second-order differential equations?

Tip: Always verify solutions by substituting them back into the original differential equation. This ensures correctness.