Verified answer

xy′ = y + x^3 + 3x^2 −2x

y' - y/x = x^2 + 3x - 2 (for x ≠ 0)

This is a linear, first-order ODE. The integration factor, p(x) is given by:

p(x) = exp(INTEGRAL of {-1/x dx})

p(x) = exp( -ln(x)) = exp(ln(1/x)) = 1/x

The solution is then given by:

y(x) = x*INTEGRAL of {(1/x)*(x^2 + 3x - 2) dx)})

y(x) = x*INTEGRAL of {(x + 3 - 2/x) dx)})

y(x) = x*[(x^2)/2 + 3x - 2ln(x) + c]

y(x) = (x^3)/2 + 3x^2 - 2x*ln(x) - c*x

x y ' - y = x² ( y / x ) ' = x³ + 3 x² - 2 / x ===> y / x = x² / 2 + 3 x - 2 ln | x | + C ===>

y = x³ / 2 + 3 x² - 2 x ln | x | + C x

{y[x] = x (4 x + C[1] - 2 Log[x])