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The problem presented is a second-order linear differential equation:

\[
x^2 y'' - 3x y' + 4y = \ln(x) \quad \text{for} \quad x > 0
\]

This is a non-homogeneous linear differential equation. The general approach to solving such an equation involves two steps:

1. **Solve the homogeneous equation** \( x^2 y'' - 3x y' + 4y = 0 \) to find the complementary solution \( y_c(x) \).
2. **Find a particular solution** \( y_p(x) \) to the non-homogeneous equation \( x^2 y'' - 3x y' + 4y = \ln(x) \).

Let's proceed step by step:

### 1. Solve the homogeneous equation:

The homogeneous equation is:
\[
x^2 y'' - 3x y' + 4y = 0
\]

This is a Cauchy-Euler equation. To solve it, assume a solution of the form:
\[
y(x) = x^r
\]

Substitute \( y(x) = x^r \) into the homogeneous equation:
\[
y' = r x^{r-1}, \quad y'' = r(r-1) x^{r-2}
\]

Substituting into the homogeneous equation:
\[
x^2 \cdot r(r-1) x^{r-2} - 3x \cdot r x^{r-1} + 4x^r = 0
\]
\[
r(r-1) x^r - 3r x^r + 4x^r = 0
\]
\[
r(r-1) - 3r + 4 = 0
\]
Simplifying:
\[
r^2 - r - 3r + 4 = 0
\]
\[
r^2 - 4r + 4 = 0
\]
\[
(r-2)^2 = 0
\]
Thus, \( r = 2 \) is a double root.

The general solution for the homogeneous equation is then:
\[
y_c(x) = (C_1 + C_2 \ln(x)) x^2
\]
where \( C_1 \) and \( C_2 \) are constants.

### 2. Find a particular solution:

To find a particular solution to the non-homogeneous equation:
\[
x^2 y'' - 3x y' + 4y = \ln(x)
\]

We can use the method of **variation of parameters** or **undetermined coefficients**. Since the right-hand side is \( \ln(x) \), we propose a particular solution of the form:
\[
y_p(x) = A \ln(x) + B
\]

Substitute \( y_p(x) \) into the non-homogeneous equation to determine \( A \) and \( B \). I will now perform this step.

Let me calculate this.The general solution to the differential equation:

\[
x^2 y'' - 3x y' + 4y = \ln(x)
\]

is:

\[
y(x) = C_1 x^2 + C_2 x^2 \ln(x) + \frac{1}{4} \ln(x) + \frac{1}{4}
\]

where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions (if given).

This solution consists of the complementary (homogeneous) part:
\[
y_c(x) = C_1 x^2 + C_2 x^2 \ln(x)
\]
and the particular solution:
\[
y_p(x) = \frac{1}{4} \ln(x) + \frac{1}{4}
\]

Would you like further details on any specific part, or do you have any other questions?

Here are 5 related questions you might want to explore:

1. How can initial conditions be used to determine \(C_1\) and \(C_2\)?
2. What method would you use to solve this differential equation if the right-hand side was \(x^3\) instead of \( \ln(x) \)?
3. Can you explain how variation of parameters works in solving non-homogeneous differential equations?
4. How does the method of undetermined coefficients differ from variation of parameters?
5. What are the specific applications of Cauchy-Euler equations in physics and engineering?

**Tip:** Always check the form of the non-homogeneous term to choose the appropriate method for finding a particular solution (e.g., undetermined coefficients or variation of parameters).

Berechnen Sie die allgemeine Lösung der folgenden DGL: x^2 y'' - 3x y' + 4y = ln(x), x > 0

Learn how to find the general solution to the second-order non-homogeneous Cauchy-Euler differential equation x^2 y'' - 3x y' + 4y = ln(x). This step-by-step guide covers solving both the homogeneous equation and finding the particular solution using undetermined coefficients.

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