The solution of the differential equation \({x^2}\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} + y = \log x\) is

We are given DE shown below;

\(\frac{{{x^2}{d^2}y}}{{d{x^2}}} = \frac{{xdy}}{{dx}} + y = \log x\)

By putting x = e^z we can replace \(\frac{{x \cdot dy}}{{dx}}\;\& \;\frac{{{x^2}{d^2}y}}{{d{x^2}}}\) as, (D) (D – 1)y – Dy + y = log e^z = z

⇒ (D² – 2D + 1)y = z (Here F(D) = (D - 1)²

⇒ F(D) = (D - 1)² = (1 - D)² has two equal roots Q = 1, 1

So its C⋅F = (C₁ + C₂z)e^z

⇒ C⋅F = (C₁ + C₂ log x)x (∵ x = e^z)

& its P⋅I = [F(D)]^-1⋅z = (1 - D)^-2⋅ z

Since expansion of (1 - D)^-2 = (1 + 2D + 3D² + 4D³ + …)

⇒ P⋅I = [1 + 2D + 3D² + 4D³ + …](z)

P⋅I = (1 + 2D + 3D² + 4D³ + …)(z)

Where \(D \to \frac{d}{{dz}}\)

\(\Rightarrow P \cdot I = \left[ {z + \frac{{2dz}}{{dz}} + 0 + 0 \ldots } \right]\)

⇒ P⋅I = z + 2

⇒ P⋅I = log x + 2 (since x = e^z)

So net solution of QE will be P⋅I + C⋅F

⇒ y = C⋅F + P⋅I = (C₁ + C₂ log x)x + log x + 2