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

Problem Understanding: We are given the differential equation:

$x^3 y''' + x^2 y'' - 2x y' + 2y = 2x^4, \quad x > 0$

We are told that $x$, $x^2$, and $1/x$ are solutions to the homogeneous version (where the right side is 0). We are to find a particular solution to the nonhomogeneous equation.

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Step-by-Step Solution Plan:

Step 1: Write the homogeneous equation.
Set the right side to zero:

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

Step 2: Find the general solution to the homogeneous equation.
Since $x$, $x^2$, and $1/x$ are solutions, the general solution to the homogeneous equation is:

$y_h = C_1 x + C_2 x^2 + C_3 \frac{1}{x}$

where $C_1, C_2, C_3$ are constants.

Step 3: Find a particular solution to the nonhomogeneous equation.
The right side is $2x^4$. We look for a particular solution, $y_p$, of the form $A x^4$ (since the right side is a polynomial and none of the homogeneous solutions are $x^4$).

Step 4: Substitute $y_p = A x^4$ into the original equation and solve for $A$.
We need to compute the derivatives:

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

Substitute these into the equation and solve for $A$.

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Let's execute these steps.

Step 4 Calculations:

Substitute into the 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:

• $x^3 (24A x) = 24A x^4$
• $x^2 (12A x^2) = 12A x^4$
• $-2x (4A x^3) = -8A x^4$
• $2(A x^4) = 2A x^4$

Sum:

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

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

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

Now, solve for $A$:

$$30A = 2$$

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

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Step 5: Write the particular solution.

So, the particular solution is:

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

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Final Answer:

Amanita, the particular solution to the given differential equation is:

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