Looks like the differential equation is
x² y'' + x y' - y = ln(x)
and the known solution is
y = x - ln(x)
Take the first derivatives of the solution:
y' = 1 - 1/x
y'' = 1/x²
Substitute y and its derivatives into the DE:
x² (1/x²) + x (1 - 1/x) - (x - ln(x)) = ln(x)
If y is a valid solution, then this equation should reduce to an identity.
1 + x - 1 - x + ln(x) = ln(x)
0 = 0 ✓
