Differential equation help

x³(dy/dx) + (2-3x²)y = x³

First order linear differential equation of the form: dy/dx + p(x) y = q(x)
has solution y = (∫ u(x) q(x) dx + C) / u(x)
where integrating factor u(x) = e^(∫ p(x) dx)

x³(dy/dx) + (2-3x²)y = x³
dy/dx + (2-3x²)/x³ y = 1

This is now in the appropriate form, with p(x) = (2-3x²)/x³ and q(x) = 1

∫ p(x) dx = ∫ (2-3x²)/x³ dx
.............. = ∫ (2/x³ - 3/x) dx
.............. = -1/x² - 3 ln(x) . . . . . . we don't bother with constant of integration yet
.............. = -1/x² - ln(x³)

u(x) = e^(∫ p(x) dx))
....... = e^(-1/x² - ln(x³))
....... = e^(-1/x²) / e^ln(x³)
....... = e^(-x⁻²) / x³

∫ u(x) q(x) dx = ∫ e^(-x⁻²) / x³ dx

Substitute:
v = -x⁻²
dv = 2x⁻³ dv

∫ u(x) q(x) dx = 1/2 ∫ 2 e^(-x⁻²) / x³ dx
...................... = 1/2 ∫ e^(-x⁻²) * 2x⁻³ dx
...................... = 1/2 ∫ e^v dv
...................... = 1/2 e^v
...................... = 1/2 e^(-x⁻²)

y = (∫ u(x) q(x) dx + C) / u(x)
y = (1/2 e^(-x⁻²) + C) / (e^(-x⁻²) / x³)
y = x³/2 + Cx³ e^(x⁻²)

y = x³/2 + Cx³ e^(1/x²)

This is linear. Put it in standard form (assume x = 0 is not in the domain of interest)

dy/dx + (2/x^3 - 3/x) y = 1.

The integrating factor is µ(x) = exp[ ∫ (2/x^3 - 3/x) dx ] = x^(-3) e^(-1/x²). Multiply through by the integrating factor. The left hand side will be the derivative of the product µ(x)y(x).

d/dx [x^(-3) e^(-1/x²) y] = x^(-3) e^(-1/x²).

Integrate both sides. Use a simple substitution u = - 1/x², so du = 2x^(-3) dx.

x^(-3) e^(-1/x²) y = ½ e^(-1/x²) + C.

Solve for y.

y(x) = ½ x^3 + Cx^3 e^(1/x²).

x³(dy/dx) + (2-3x²)y = x³

divide both sides by x^3

dy/dx + y( 2/x^3-3/x) = 1 ---------(1)

This is dy/dx + p(x) y = q(x) from

p(x)= (2/x^3-3/x)
q(x)=1

The integrating factor is e^∫p(x) dx = e^∫ (2/x^3-3/x) dx = e^(-1/x^2-3ln x)

Multiply both sides of (1) by e^(-1/x^2-3ln x)

e^(-1/x^2-3ln x) dy/dx + y(2/x^3-3/x) e^(-1/x^2-3ln x) = e^(-1/x^2-3ln x) ----(2)

The left-hand side is d/dx [ y e^(-1/x^2-3ln x)]

Integrating both sides, equation (2) becomes
∫ d/dx [ y e^(-1/x^2-3ln x)] = ∫ e^(-1/x^2-3ln x) dx

y e^(-1/x^2-3ln x) = ∫ e^(-1/x^2-3ln x) dx ------(3)

y e^(-1/x^2-3ln x) = (1/2) e^(-1/x^2) + C