To solve the given differential equation:

$x^2 y'' + 7x y' + 9y = 25 \ln(x),$

we recognize it as a second-order linear non-homogeneous differential equation. Let's solve it step-by-step.

Step 1: Solve the Homogeneous Equation

The homogeneous part of the equation is: $x^2 y'' + 7x y' + 9y = 0.$ This is a Cauchy-Euler equation. To solve it, we use the substitution $y = x^r$, leading to: $r(r-1) + 7r + 9 = 0.$ Simplify: $r^2 + 6r + 9 = 0.$ This factors as: $(r + 3)^2 = 0.$ Thus, $r = -3$ is a repeated root. The general solution of the homogeneous equation is: $y_h = C_1 x^{-3} + C_2 x^{-3} \ln(x),$ where $C_1$ and $C_2$ are constants.

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Step 2: Solve the Non-Homogeneous Equation

The non-homogeneous term is $25 \ln(x)$. We use the method of undetermined coefficients. Assume a particular solution of the form: $y_p = A \ln(x) + B.$ Substitute $y_p$ into the original equation. Compute derivatives: $y_p' = \frac{A}{x}, \quad y_p'' = -\frac{A}{x^2}.$ Substitute $y_p$, $y_p'$, and $y_p''$ into the differential equation: $x^2 \left(-\frac{A}{x^2}\right) + 7x \left(\frac{A}{x}\right) + 9(A \ln(x) + B) = 25 \ln(x).$ Simplify each term: $-A + 7A + 9A \ln(x) + 9B = 25 \ln(x).$ Combine terms involving $\ln(x)$ and constants: $9A \ln(x) + (6A + 9B) = 25 \ln(x).$ Equating coefficients:

• For $\ln(x)$: $9A = 25 \implies A = \frac{25}{9}$,
• For the constant: $6A + 9B = 0 \implies 6\left(\frac{25}{9}\right) + 9B = 0$.

Solve for $B$: $\frac{150}{9} + 9B = 0 \implies 9B = -\frac{150}{9} \implies B = -\frac{50}{9}.$

Thus, the particular solution is: $y_p = \frac{25}{9} \ln(x) - \frac{50}{9}.$

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Step 3: General Solution

The general solution is the sum of the homogeneous and particular solutions: $y = y_h + y_p = C_1 x^{-3} + C_2 x^{-3} \ln(x) + \frac{25}{9} \ln(x) - \frac{50}{9}.$

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Final Answer:

$y = C_1 x^{-3} + C_2 x^{-3} \ln(x) + \frac{25}{9} \ln(x) - \frac{50}{9}.$

Let me know if you need further clarification or want to explore specific cases! 😊

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Expanding Questions:

1. How do you determine a particular solution for non-homogeneous equations?
2. What is the role of repeated roots in the solution of Cauchy-Euler equations?
3. Why does the substitution $y = x^r$ work for Cauchy-Euler equations?
4. Can we use variation of parameters for this problem? Why or why not?
5. How does the presence of $\ln(x)$ in the non-homogeneous term affect the choice of $y_p$?

Tip: Always check the solution by substituting it back into the original differential equation.