First we have to solve the homogeneous equation 

(3.1)  x²y" − 4xy' + 6y = 0. 

This is an Euler-Cauchy equation, so we look for solutions of the form y = x^m. The characteristic equation for m is 

m(m − 1) − 4m + 6 = 0. 

Note m(m − 1) − 4m + 6 = m² − 5m + 6 = (m − 2)(m − 3), so the roots are m = 2 and m = 3. Thus, we have basic solutions y₁ = x² and y₂ = x³ for the homogeneous equation (3.1). 

To solve the equation in the problem, we put it in standard form by dividing both sides by x². This gives us

y" -/xy' +  6/x2y = 1/x.

We look for a particular solution of this equation of the form y = u(x)y₁(x)+ v(x)y₂(x). The equations for u and v are 

u'x² + v'x³ = 0 

2u'x + 3v'x² = 1/x. 

If we multiply the top equation by −2 and the bottom equation by x, the system becomes 

−2u'x² − 2v'x³ = 0 

2u'x² + 3v'x³ = 1.

Adding these equations gives v'x³ = 1, so v' = 1/x³. Substituting this is the first equation gives u' = −1/x². Easy integrations then give

u = 1/x

v = − 1/2x² 

Plugging into the formula y = u(x)y¹(x) + v(x)y²(x) for our particular solution gives 

y = (1/x)x² − (1/2x²)x³ = 1/2x. 

Adding on the general solution of the homogeneous equation, we find 

y = 1/2x + C₁x² + C₂x³ 

for the general solution of the inhomogeneous equation.