Amanita, let's break down the problem step by step and solve it together!

Step-by-Step Plan

1. Understand the Problem:
   We are given a differential equation:

$x^3 y''' + x^2 y'' - 2x y' + 2y = 2x^4$

and told that $x, x^2,$ and $1/x$ are solutions to the homogeneous version (where the right side is 0).

2. Write the Homogeneous Equation:
   The homogeneous equation is:

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

3. Find a Particular Solution:
   Since the right side is $2x^4$, we need to find a particular solution to the non-homogeneous equation.
   We will use the method of undetermined coefficients or variation of parameters. Since the right side is a polynomial, let's try a particular solution of the form $y_p = A x^4$.

4. Substitute $y_p$ into the Equation:
   We will compute the derivatives of $y_p$, substitute them into the original equation, and solve for $A$.

5. Write the General Solution:
   The general solution will be the sum of the homogeneous solution and the particular solution.

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Step 1: Assume a Particular Solution

Let’s try $y_p = A x^4$.

Calculate the derivatives:

• $y_p' = 4A x^3$
• $y_p'' = 12A x^2$
• $y_p''' = 24A x$

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Step 2: Substitute into the Differential Equation

Plug these into the original equation:

$x^3 (24A x) + x^2 (12A x^2) - 2x (4A x^3) + 2(A x^4) = 2x^4$

Let's simplify each term:

• $x^3 \cdot 24A x = 24A x^4$
• $x^2 \cdot 12A x^2 = 12A x^4$
• $-2x \cdot 4A x^3 = -8A x^4$
• $2A x^4$

Add them up:

$24A x^4 + 12A x^4 - 8A x^4 + 2A x^4 = 2x^4$

Combine like terms:

$(24A + 12A - 8A + 2A) x^4 = 2x^4$

$$30A x^4 = 2x^4$$

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Step 3: Solve for $A$

Set the coefficients equal:

$$30A = 2$$

$A = \frac{2}{30} = \frac{1}{15}$

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Step 4: Write the Particular Solution

So, the particular solution is:

$$y_p = \frac{1}{15} x^4$$

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Step 5: General Solution

The general solution is:

$y(x) = C_1 x + C_2 x^2 + C_3 \frac{1}{x} + \frac{1}{15} x^4$

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Final Answer for a Particular Solution:

$\boxed{y_p = \frac{1}{15} x^4}$

Amanita, the particular solution to the given equation is $\frac{1}{15} x^4$.